Mixing
Associating two item in a list
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up vote
7
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I am comparing two lists for common strings and my code currently works to output items in common from two lists.
list1
['5', 'orange', '20', 'apple', '50', 'blender']
list2
['25', 'blender', '20', 'pear', '40', 'spatula']
Here is my code so far:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
print(item1)
This code would return blender. What I want to do now is to also print the number before the blender in each list to obtain an output similar to:
blender, 50, 25
I have tried to add two new lines to the for loop but did not have the desired output:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
for num1 in list1[0::2]:
for num2 in list2[0::2]:
if item1 == item2:
print(item1, num1, num2)
I know now that making for loops is not the answer. Also trying to call item1[-1] does not work. I am new to Python and need some help with this!
Thank you
python list
asked 45 mins ago
stevesy
383
add a comment |Â
up vote
7
down vote
favorite
I am comparing two lists for common strings and my code currently works to output items in common from two lists.
list1
['5', 'orange', '20', 'apple', '50', 'blender']
list2
['25', 'blender', '20', 'pear', '40', 'spatula']
Here is my code so far:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
print(item1)
This code would return blender. What I want to do now is to also print the number before the blender in each list to obtain an output similar to:
blender, 50, 25
I have tried to add two new lines to the for loop but did not have the desired output:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
for num1 in list1[0::2]:
for num2 in list2[0::2]:
if item1 == item2:
print(item1, num1, num2)
I know now that making for loops is not the answer. Also trying to call item1[-1] does not work. I am new to Python and need some help with this!
Thank you
python list
asked 45 mins ago
stevesy
383
6
This seems like an XY problem. If that is the kind of output you're looking to get, storing information in lists like that is not going to be the best approach. It honestly looks like you want a dictionary with words as keys, and the number before them as values
– user3483203
44 mins ago
Take a look at theenumerateandzipfunctions
– JETM
42 mins ago
Is this your special requirement to iterate over list on indexes 1,3,5.... Not on 0,2,4,6..... Consider if blender comes at index either 0,2,4,6... of any one List then will it not be considered as common element on lists??
– Shivam Seth
35 mins ago
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I am comparing two lists for common strings and my code currently works to output items in common from two lists.
list1
['5', 'orange', '20', 'apple', '50', 'blender']
list2
['25', 'blender', '20', 'pear', '40', 'spatula']
Here is my code so far:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
print(item1)
This code would return blender. What I want to do now is to also print the number before the blender in each list to obtain an output similar to:
blender, 50, 25
I have tried to add two new lines to the for loop but did not have the desired output:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
for num1 in list1[0::2]:
for num2 in list2[0::2]:
if item1 == item2:
print(item1, num1, num2)
I know now that making for loops is not the answer. Also trying to call item1[-1] does not work. I am new to Python and need some help with this!
Thank you
python list
asked 45 mins ago
stevesy
383
I am comparing two lists for common strings and my code currently works to output items in common from two lists.
list1
['5', 'orange', '20', 'apple', '50', 'blender']
list2
['25', 'blender', '20', 'pear', '40', 'spatula']
Here is my code so far:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
print(item1)
This code would return blender. What I want to do now is to also print the number before the blender in each list to obtain an output similar to:
blender, 50, 25
I have tried to add two new lines to the for loop but did not have the desired output:
for item1 in list1[1::2]:
for item2 in list2[1::2]:
for num1 in list1[0::2]:
for num2 in list2[0::2]:
if item1 == item2:
print(item1, num1, num2)
I know now that making for loops is not the answer. Also trying to call item1[-1] does not work. I am new to Python and need some help with this!
Thank you
python list
python list
asked 45 mins ago
stevesy
383
asked 45 mins ago
stevesy
383
asked 45 mins ago
stevesy
383
asked 45 mins ago
stevesy
383
asked 45 mins ago
stevesy
383
383
6
This seems like an XY problem. If that is the kind of output you're looking to get, storing information in lists like that is not going to be the best approach. It honestly looks like you want a dictionary with words as keys, and the number before them as values
– user3483203
44 mins ago
Take a look at theenumerateandzipfunctions
– JETM
42 mins ago
Is this your special requirement to iterate over list on indexes 1,3,5.... Not on 0,2,4,6..... Consider if blender comes at index either 0,2,4,6... of any one List then will it not be considered as common element on lists??
– Shivam Seth
35 mins ago
add a comment |Â
6
This seems like an XY problem. If that is the kind of output you're looking to get, storing information in lists like that is not going to be the best approach. It honestly looks like you want a dictionary with words as keys, and the number before them as values
– user3483203
44 mins ago
Take a look at theenumerateandzipfunctions
– JETM
42 mins ago
Is this your special requirement to iterate over list on indexes 1,3,5.... Not on 0,2,4,6..... Consider if blender comes at index either 0,2,4,6... of any one List then will it not be considered as common element on lists??
– Shivam Seth
35 mins ago
6
6
This seems like an XY problem. If that is the kind of output you're looking to get, storing information in lists like that is not going to be the best approach. It honestly looks like you want a dictionary with words as keys, and the number before them as values
– user3483203
44 mins ago
This seems like an XY problem. If that is the kind of output you're looking to get, storing information in lists like that is not going to be the best approach. It honestly looks like you want a dictionary with words as keys, and the number before them as values
– user3483203
44 mins ago
Take a look at the
enumerate and zip functions– JETM
42 mins ago
Take a look at the
enumerate and zip functions– JETM
42 mins ago
Is this your special requirement to iterate over list on indexes 1,3,5.... Not on 0,2,4,6..... Consider if blender comes at index either 0,2,4,6... of any one List then will it not be considered as common element on lists??
– Shivam Seth
35 mins ago
Is this your special requirement to iterate over list on indexes 1,3,5.... Not on 0,2,4,6..... Consider if blender comes at index either 0,2,4,6... of any one List then will it not be considered as common element on lists??
– Shivam Seth
35 mins ago
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
5
down vote
accepted
You can do it in two ways, either stay with lists (Which is more messy):
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
item_to_print = item1
print(item1, ",", end="")
for index in range(len(list1)):
if list1[index] == item1:
print(list1[index - 1], ",", end="")
for index in range(len(list2)):
if list2[index] == item1:
print(list2[index - 1])
or the better way (in my opinion) with dictionary:
dict1 = 'apple': '20', 'blender': '50', 'orange': '5'
dict2 = 'blender': '25', 'pear': '20', 'spatula': '40'
for item in dict1:
if item in dict2:
print(item, ",", dict1[item], ",", dict2[item])
Both will output:
>> blender , 50 , 25
answered 33 mins ago
MercyDude
376418
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
1
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
add a comment |Â
up vote
4
down vote
I think you'd be better suited using dictionaries for this problem..
dict1 = 'orange': 5, 'apple': 20, 'blender': 50
dict2 = 'blender': 25, 'pear': 20, 'spatula': 40
so to get the outputs you want
>>dict1['blender']
50
>>dict2['blender']
25
So if you want to get the number of blenders from each dictionary in the format you want, you can really just use
print("blender, "+str(dict1['blender'])+", "+str(dict2['blender']))
To take this one step further and output based on what's in dict1 and dict2
for i in dict1.keys(): #dict1.keys() is essentially a list with ['orange','apple','blender']
if i in dict2.keys(): #same deal, but different items in the list
print(str(i)+", "+str(dict1[i])+", "+str(dict2[i])) #this outputs items that are in BOTH lists
else:
print(str(i)+", "+str(dict1[i])) #this outputs items ONLY in dict1
for i in dict2.keys():
if i not in dict1.keys(): #we use not since we already output the matching in the previous loop
print(str(i)+", "+str(dict2[i])) #this outputs items ONLY in dict2
Outputs:
orange, 5
apple, 20
blender, 50, 25
pear, 20
spatula, 40
answered 37 mins ago
J0hn
458314
add a comment |Â
up vote
3
down vote
You're approaching this problem with the wrong Data Structure. You shouldn't keep this data in lists if you're doing lookups between the two. Instead, it would be much easier to use a dictionary here.
Setup
You can use zip to create these two dictionaries:
a = ['5', 'orange', '20', 'apple', '50', 'blender']
b = ['25', 'blender', '20', 'pear', '40', 'spatula']
dct_a = dict(zip(a[1::2], a[::2]))
dct_b = dict(zip(b[1::2], b[::2]))
This will leave you with these two dictionaries:
'orange': '5', 'apple': '20', 'blender': '50'
'blender': '25', 'pear': '20', 'spatula': '40'
This makes every part of your problem easier to solve. For example, to find common keys:
common = dct_a.keys() & dct_b.keys()
# 'blender'
And to find all the values matching each common key:
[(k, dct_a[k], dct_b[k]) for k in common]
Output:
[('blender', '50', '25')]
answered 40 mins ago
user3483203
24.1k62147
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
add a comment |Â
up vote
3
down vote
with enumerate you can solve your problem:
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for n1, item1 in enumerate(list1[1::2]):
for n2, item2 in enumerate(list2[1::2]):
if item1 == item2:
print(list1[n1*2], list2[n2*2], item1)
enumerate returns a tuple where the first element is the count of the iteration, the second the element.
Since enumerate is counting the iteration and you're stepping 2, we need to multiply by 2:orange will be the first iteration, so n1 will be 0, so the previous value is at index 0*2apple will be in the second iteration, so n1 will be 1, so the previous value is at index 1*2blender will be in the third iteration, so n1 will be 2, so the previousl value is at index 2*2.
this means that in for n1, item1 in enumerate(list1[1::2]): n1 and item1 will have this values:
Iteration 1: n1 = 0, item1 = orange, previous_index = 0*2
Iteration 2: n1 = 1, item1 = apple, previous_index = 1*2
Iteration 3: n1 = 2, item1 = blender, previous_index = 2*2
same goes for for n2, item2 in enumerate(list2[1::2])::
Iteration 1: n2 = 0, item2 = blender, previous_index = 0*2
Iteration 2: n2 = 1, item2 = pear, previous_index = 1*2
Iteration 3: n2 = 2, item2 = spatula, previous_index = 2*2
answered 36 mins ago
Gsk
1,76511216
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
add a comment |Â
up vote
3
down vote
Assuming the nouns in your lists are unique (per list), create dictionaries out of your lists first.
In [1]: list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
In [2]: list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
In [3]:
In [3]: dict1 = dict(reversed(pair) for pair in zip(*[iter(list1)]*2))
In [4]: dict2 = dict(reversed(pair) for pair in zip(*[iter(list2)]*2))
In [5]:
In [5]: dict1
Out[5]: 'apple': '20', 'blender': '50', 'orange': '5'
In [6]: dict2
Out[6]: 'blender': '25', 'pear': '20', 'spatula': '40'
This code leverages the grouper recipe from the itertools docs.
Matching two items from your dictionaries is easy as pie.
In [7]: key = 'blender'
In [8]: print(key, dict1[key], dict2[key])
blender 50 25
You could even build a dictionary that holds the common keys from dict1 and dict2 and a list of the values.
In [12]: common = dict1.keys() & dict2
In [13]: c:[dict1[c], dict2[c]] for c in common
Out[13]: 'blender': ['50', '25']
For an arbitrary number of dicts, you can abstract this further.
In [28]: from functools import reduce
In [29]: from operator import and_
In [30]:
In [30]: dicts = (dict1, dict2)
In [31]: common = reduce(and_, map(set, dicts))
In [32]: c:[d[c] for d in dicts] for c in common
Out[32]: 'blender': ['50', '25']
answered 40 mins ago
timgeb
36.9k104875
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
add a comment |Â
up vote
2
down vote
I hope this code will work for you.
l1 =['5', 'orange', '20', 'apple', '50', 'blender']
l2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for list1_item in zip(*[iter(l1)]*2):
for list2_item in zip(*[iter(l2)]*2):
if list1_item[1]==list2_item[1]:
print list1_item[1],list1_item[0],list2_item[0]
it will print output
blender 50 25
answered 29 mins ago
Abdul Majeed
1,5211315
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
You can do it in two ways, either stay with lists (Which is more messy):
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
item_to_print = item1
print(item1, ",", end="")
for index in range(len(list1)):
if list1[index] == item1:
print(list1[index - 1], ",", end="")
for index in range(len(list2)):
if list2[index] == item1:
print(list2[index - 1])
or the better way (in my opinion) with dictionary:
dict1 = 'apple': '20', 'blender': '50', 'orange': '5'
dict2 = 'blender': '25', 'pear': '20', 'spatula': '40'
for item in dict1:
if item in dict2:
print(item, ",", dict1[item], ",", dict2[item])
Both will output:
>> blender , 50 , 25
answered 33 mins ago
MercyDude
376418
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
1
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
add a comment |Â
up vote
5
down vote
accepted
You can do it in two ways, either stay with lists (Which is more messy):
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
item_to_print = item1
print(item1, ",", end="")
for index in range(len(list1)):
if list1[index] == item1:
print(list1[index - 1], ",", end="")
for index in range(len(list2)):
if list2[index] == item1:
print(list2[index - 1])
or the better way (in my opinion) with dictionary:
dict1 = 'apple': '20', 'blender': '50', 'orange': '5'
dict2 = 'blender': '25', 'pear': '20', 'spatula': '40'
for item in dict1:
if item in dict2:
print(item, ",", dict1[item], ",", dict2[item])
Both will output:
>> blender , 50 , 25
answered 33 mins ago
MercyDude
376418
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
1
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
You can do it in two ways, either stay with lists (Which is more messy):
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
item_to_print = item1
print(item1, ",", end="")
for index in range(len(list1)):
if list1[index] == item1:
print(list1[index - 1], ",", end="")
for index in range(len(list2)):
if list2[index] == item1:
print(list2[index - 1])
or the better way (in my opinion) with dictionary:
dict1 = 'apple': '20', 'blender': '50', 'orange': '5'
dict2 = 'blender': '25', 'pear': '20', 'spatula': '40'
for item in dict1:
if item in dict2:
print(item, ",", dict1[item], ",", dict2[item])
Both will output:
>> blender , 50 , 25
answered 33 mins ago
MercyDude
376418
You can do it in two ways, either stay with lists (Which is more messy):
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for item1 in list1[1::2]:
for item2 in list2[1::2]:
if item1 == item2:
item_to_print = item1
print(item1, ",", end="")
for index in range(len(list1)):
if list1[index] == item1:
print(list1[index - 1], ",", end="")
for index in range(len(list2)):
if list2[index] == item1:
print(list2[index - 1])
or the better way (in my opinion) with dictionary:
dict1 = 'apple': '20', 'blender': '50', 'orange': '5'
dict2 = 'blender': '25', 'pear': '20', 'spatula': '40'
for item in dict1:
if item in dict2:
print(item, ",", dict1[item], ",", dict2[item])
Both will output:
>> blender , 50 , 25
answered 33 mins ago
MercyDude
376418
answered 33 mins ago
MercyDude
376418
answered 33 mins ago
MercyDude
376418
answered 33 mins ago
MercyDude
376418
376418
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
1
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
add a comment |Â
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
1
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
Thank you, I ended up using this and it works great!
– stevesy
28 mins ago
1
1
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
@stevesy Your'e welcome mate.
– MercyDude
21 mins ago
add a comment |Â
up vote
4
down vote
I think you'd be better suited using dictionaries for this problem..
dict1 = 'orange': 5, 'apple': 20, 'blender': 50
dict2 = 'blender': 25, 'pear': 20, 'spatula': 40
so to get the outputs you want
>>dict1['blender']
50
>>dict2['blender']
25
So if you want to get the number of blenders from each dictionary in the format you want, you can really just use
print("blender, "+str(dict1['blender'])+", "+str(dict2['blender']))
To take this one step further and output based on what's in dict1 and dict2
for i in dict1.keys(): #dict1.keys() is essentially a list with ['orange','apple','blender']
if i in dict2.keys(): #same deal, but different items in the list
print(str(i)+", "+str(dict1[i])+", "+str(dict2[i])) #this outputs items that are in BOTH lists
else:
print(str(i)+", "+str(dict1[i])) #this outputs items ONLY in dict1
for i in dict2.keys():
if i not in dict1.keys(): #we use not since we already output the matching in the previous loop
print(str(i)+", "+str(dict2[i])) #this outputs items ONLY in dict2
Outputs:
orange, 5
apple, 20
blender, 50, 25
pear, 20
spatula, 40
answered 37 mins ago
J0hn
458314
add a comment |Â
up vote
4
down vote
I think you'd be better suited using dictionaries for this problem..
dict1 = 'orange': 5, 'apple': 20, 'blender': 50
dict2 = 'blender': 25, 'pear': 20, 'spatula': 40
so to get the outputs you want
>>dict1['blender']
50
>>dict2['blender']
25
So if you want to get the number of blenders from each dictionary in the format you want, you can really just use
print("blender, "+str(dict1['blender'])+", "+str(dict2['blender']))
To take this one step further and output based on what's in dict1 and dict2
for i in dict1.keys(): #dict1.keys() is essentially a list with ['orange','apple','blender']
if i in dict2.keys(): #same deal, but different items in the list
print(str(i)+", "+str(dict1[i])+", "+str(dict2[i])) #this outputs items that are in BOTH lists
else:
print(str(i)+", "+str(dict1[i])) #this outputs items ONLY in dict1
for i in dict2.keys():
if i not in dict1.keys(): #we use not since we already output the matching in the previous loop
print(str(i)+", "+str(dict2[i])) #this outputs items ONLY in dict2
Outputs:
orange, 5
apple, 20
blender, 50, 25
pear, 20
spatula, 40
answered 37 mins ago
J0hn
458314
add a comment |Â
up vote
4
down vote
up vote
4
down vote
I think you'd be better suited using dictionaries for this problem..
dict1 = 'orange': 5, 'apple': 20, 'blender': 50
dict2 = 'blender': 25, 'pear': 20, 'spatula': 40
so to get the outputs you want
>>dict1['blender']
50
>>dict2['blender']
25
So if you want to get the number of blenders from each dictionary in the format you want, you can really just use
print("blender, "+str(dict1['blender'])+", "+str(dict2['blender']))
To take this one step further and output based on what's in dict1 and dict2
for i in dict1.keys(): #dict1.keys() is essentially a list with ['orange','apple','blender']
if i in dict2.keys(): #same deal, but different items in the list
print(str(i)+", "+str(dict1[i])+", "+str(dict2[i])) #this outputs items that are in BOTH lists
else:
print(str(i)+", "+str(dict1[i])) #this outputs items ONLY in dict1
for i in dict2.keys():
if i not in dict1.keys(): #we use not since we already output the matching in the previous loop
print(str(i)+", "+str(dict2[i])) #this outputs items ONLY in dict2
Outputs:
orange, 5
apple, 20
blender, 50, 25
pear, 20
spatula, 40
answered 37 mins ago
J0hn
458314
I think you'd be better suited using dictionaries for this problem..
dict1 = 'orange': 5, 'apple': 20, 'blender': 50
dict2 = 'blender': 25, 'pear': 20, 'spatula': 40
so to get the outputs you want
>>dict1['blender']
50
>>dict2['blender']
25
So if you want to get the number of blenders from each dictionary in the format you want, you can really just use
print("blender, "+str(dict1['blender'])+", "+str(dict2['blender']))
To take this one step further and output based on what's in dict1 and dict2
for i in dict1.keys(): #dict1.keys() is essentially a list with ['orange','apple','blender']
if i in dict2.keys(): #same deal, but different items in the list
print(str(i)+", "+str(dict1[i])+", "+str(dict2[i])) #this outputs items that are in BOTH lists
else:
print(str(i)+", "+str(dict1[i])) #this outputs items ONLY in dict1
for i in dict2.keys():
if i not in dict1.keys(): #we use not since we already output the matching in the previous loop
print(str(i)+", "+str(dict2[i])) #this outputs items ONLY in dict2
Outputs:
orange, 5
apple, 20
blender, 50, 25
pear, 20
spatula, 40
answered 37 mins ago
J0hn
458314
edited 32 mins ago
answered 37 mins ago
J0hn
458314
answered 37 mins ago
J0hn
458314
answered 37 mins ago
J0hn
458314
458314
add a comment |Â
add a comment |Â
up vote
3
down vote
You're approaching this problem with the wrong Data Structure. You shouldn't keep this data in lists if you're doing lookups between the two. Instead, it would be much easier to use a dictionary here.
Setup
You can use zip to create these two dictionaries:
a = ['5', 'orange', '20', 'apple', '50', 'blender']
b = ['25', 'blender', '20', 'pear', '40', 'spatula']
dct_a = dict(zip(a[1::2], a[::2]))
dct_b = dict(zip(b[1::2], b[::2]))
This will leave you with these two dictionaries:
'orange': '5', 'apple': '20', 'blender': '50'
'blender': '25', 'pear': '20', 'spatula': '40'
This makes every part of your problem easier to solve. For example, to find common keys:
common = dct_a.keys() & dct_b.keys()
# 'blender'
And to find all the values matching each common key:
[(k, dct_a[k], dct_b[k]) for k in common]
Output:
[('blender', '50', '25')]
answered 40 mins ago
user3483203
24.1k62147
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
add a comment |Â
up vote
3
down vote
You're approaching this problem with the wrong Data Structure. You shouldn't keep this data in lists if you're doing lookups between the two. Instead, it would be much easier to use a dictionary here.
Setup
You can use zip to create these two dictionaries:
a = ['5', 'orange', '20', 'apple', '50', 'blender']
b = ['25', 'blender', '20', 'pear', '40', 'spatula']
dct_a = dict(zip(a[1::2], a[::2]))
dct_b = dict(zip(b[1::2], b[::2]))
This will leave you with these two dictionaries:
'orange': '5', 'apple': '20', 'blender': '50'
'blender': '25', 'pear': '20', 'spatula': '40'
This makes every part of your problem easier to solve. For example, to find common keys:
common = dct_a.keys() & dct_b.keys()
# 'blender'
And to find all the values matching each common key:
[(k, dct_a[k], dct_b[k]) for k in common]
Output:
[('blender', '50', '25')]
answered 40 mins ago
user3483203
24.1k62147
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You're approaching this problem with the wrong Data Structure. You shouldn't keep this data in lists if you're doing lookups between the two. Instead, it would be much easier to use a dictionary here.
Setup
You can use zip to create these two dictionaries:
a = ['5', 'orange', '20', 'apple', '50', 'blender']
b = ['25', 'blender', '20', 'pear', '40', 'spatula']
dct_a = dict(zip(a[1::2], a[::2]))
dct_b = dict(zip(b[1::2], b[::2]))
This will leave you with these two dictionaries:
'orange': '5', 'apple': '20', 'blender': '50'
'blender': '25', 'pear': '20', 'spatula': '40'
This makes every part of your problem easier to solve. For example, to find common keys:
common = dct_a.keys() & dct_b.keys()
# 'blender'
And to find all the values matching each common key:
[(k, dct_a[k], dct_b[k]) for k in common]
Output:
[('blender', '50', '25')]
answered 40 mins ago
user3483203
24.1k62147
You're approaching this problem with the wrong Data Structure. You shouldn't keep this data in lists if you're doing lookups between the two. Instead, it would be much easier to use a dictionary here.
Setup
You can use zip to create these two dictionaries:
a = ['5', 'orange', '20', 'apple', '50', 'blender']
b = ['25', 'blender', '20', 'pear', '40', 'spatula']
dct_a = dict(zip(a[1::2], a[::2]))
dct_b = dict(zip(b[1::2], b[::2]))
This will leave you with these two dictionaries:
'orange': '5', 'apple': '20', 'blender': '50'
'blender': '25', 'pear': '20', 'spatula': '40'
This makes every part of your problem easier to solve. For example, to find common keys:
common = dct_a.keys() & dct_b.keys()
# 'blender'
And to find all the values matching each common key:
[(k, dct_a[k], dct_b[k]) for k in common]
Output:
[('blender', '50', '25')]
answered 40 mins ago
user3483203
24.1k62147
answered 40 mins ago
user3483203
24.1k62147
answered 40 mins ago
user3483203
24.1k62147
answered 40 mins ago
user3483203
24.1k62147
24.1k62147
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
add a comment |Â
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
Thank you, I haven't gone over dictionaries yet in my class but this does work!
– stevesy
27 mins ago
add a comment |Â
up vote
3
down vote
with enumerate you can solve your problem:
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for n1, item1 in enumerate(list1[1::2]):
for n2, item2 in enumerate(list2[1::2]):
if item1 == item2:
print(list1[n1*2], list2[n2*2], item1)
enumerate returns a tuple where the first element is the count of the iteration, the second the element.
Since enumerate is counting the iteration and you're stepping 2, we need to multiply by 2:orange will be the first iteration, so n1 will be 0, so the previous value is at index 0*2apple will be in the second iteration, so n1 will be 1, so the previous value is at index 1*2blender will be in the third iteration, so n1 will be 2, so the previousl value is at index 2*2.
this means that in for n1, item1 in enumerate(list1[1::2]): n1 and item1 will have this values:
Iteration 1: n1 = 0, item1 = orange, previous_index = 0*2
Iteration 2: n1 = 1, item1 = apple, previous_index = 1*2
Iteration 3: n1 = 2, item1 = blender, previous_index = 2*2
same goes for for n2, item2 in enumerate(list2[1::2])::
Iteration 1: n2 = 0, item2 = blender, previous_index = 0*2
Iteration 2: n2 = 1, item2 = pear, previous_index = 1*2
Iteration 3: n2 = 2, item2 = spatula, previous_index = 2*2
answered 36 mins ago
Gsk
1,76511216
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
add a comment |Â
up vote
3
down vote
with enumerate you can solve your problem:
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for n1, item1 in enumerate(list1[1::2]):
for n2, item2 in enumerate(list2[1::2]):
if item1 == item2:
print(list1[n1*2], list2[n2*2], item1)
enumerate returns a tuple where the first element is the count of the iteration, the second the element.
Since enumerate is counting the iteration and you're stepping 2, we need to multiply by 2:orange will be the first iteration, so n1 will be 0, so the previous value is at index 0*2apple will be in the second iteration, so n1 will be 1, so the previous value is at index 1*2blender will be in the third iteration, so n1 will be 2, so the previousl value is at index 2*2.
this means that in for n1, item1 in enumerate(list1[1::2]): n1 and item1 will have this values:
Iteration 1: n1 = 0, item1 = orange, previous_index = 0*2
Iteration 2: n1 = 1, item1 = apple, previous_index = 1*2
Iteration 3: n1 = 2, item1 = blender, previous_index = 2*2
same goes for for n2, item2 in enumerate(list2[1::2])::
Iteration 1: n2 = 0, item2 = blender, previous_index = 0*2
Iteration 2: n2 = 1, item2 = pear, previous_index = 1*2
Iteration 3: n2 = 2, item2 = spatula, previous_index = 2*2
answered 36 mins ago
Gsk
1,76511216
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
with enumerate you can solve your problem:
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for n1, item1 in enumerate(list1[1::2]):
for n2, item2 in enumerate(list2[1::2]):
if item1 == item2:
print(list1[n1*2], list2[n2*2], item1)
enumerate returns a tuple where the first element is the count of the iteration, the second the element.
Since enumerate is counting the iteration and you're stepping 2, we need to multiply by 2:orange will be the first iteration, so n1 will be 0, so the previous value is at index 0*2apple will be in the second iteration, so n1 will be 1, so the previous value is at index 1*2blender will be in the third iteration, so n1 will be 2, so the previousl value is at index 2*2.
this means that in for n1, item1 in enumerate(list1[1::2]): n1 and item1 will have this values:
Iteration 1: n1 = 0, item1 = orange, previous_index = 0*2
Iteration 2: n1 = 1, item1 = apple, previous_index = 1*2
Iteration 3: n1 = 2, item1 = blender, previous_index = 2*2
same goes for for n2, item2 in enumerate(list2[1::2])::
Iteration 1: n2 = 0, item2 = blender, previous_index = 0*2
Iteration 2: n2 = 1, item2 = pear, previous_index = 1*2
Iteration 3: n2 = 2, item2 = spatula, previous_index = 2*2
answered 36 mins ago
Gsk
1,76511216
with enumerate you can solve your problem:
list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for n1, item1 in enumerate(list1[1::2]):
for n2, item2 in enumerate(list2[1::2]):
if item1 == item2:
print(list1[n1*2], list2[n2*2], item1)
enumerate returns a tuple where the first element is the count of the iteration, the second the element.
Since enumerate is counting the iteration and you're stepping 2, we need to multiply by 2:orange will be the first iteration, so n1 will be 0, so the previous value is at index 0*2apple will be in the second iteration, so n1 will be 1, so the previous value is at index 1*2blender will be in the third iteration, so n1 will be 2, so the previousl value is at index 2*2.
this means that in for n1, item1 in enumerate(list1[1::2]): n1 and item1 will have this values:
Iteration 1: n1 = 0, item1 = orange, previous_index = 0*2
Iteration 2: n1 = 1, item1 = apple, previous_index = 1*2
Iteration 3: n1 = 2, item1 = blender, previous_index = 2*2
same goes for for n2, item2 in enumerate(list2[1::2])::
Iteration 1: n2 = 0, item2 = blender, previous_index = 0*2
Iteration 2: n2 = 1, item2 = pear, previous_index = 1*2
Iteration 3: n2 = 2, item2 = spatula, previous_index = 2*2
answered 36 mins ago
Gsk
1,76511216
edited 25 mins ago
answered 36 mins ago
Gsk
1,76511216
answered 36 mins ago
Gsk
1,76511216
answered 36 mins ago
Gsk
1,76511216
1,76511216
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
add a comment |Â
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
This is great, I have to read up on the enumerate function some more but this worked!
– stevesy
28 mins ago
add a comment |Â
up vote
3
down vote
Assuming the nouns in your lists are unique (per list), create dictionaries out of your lists first.
In [1]: list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
In [2]: list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
In [3]:
In [3]: dict1 = dict(reversed(pair) for pair in zip(*[iter(list1)]*2))
In [4]: dict2 = dict(reversed(pair) for pair in zip(*[iter(list2)]*2))
In [5]:
In [5]: dict1
Out[5]: 'apple': '20', 'blender': '50', 'orange': '5'
In [6]: dict2
Out[6]: 'blender': '25', 'pear': '20', 'spatula': '40'
This code leverages the grouper recipe from the itertools docs.
Matching two items from your dictionaries is easy as pie.
In [7]: key = 'blender'
In [8]: print(key, dict1[key], dict2[key])
blender 50 25
You could even build a dictionary that holds the common keys from dict1 and dict2 and a list of the values.
In [12]: common = dict1.keys() & dict2
In [13]: c:[dict1[c], dict2[c]] for c in common
Out[13]: 'blender': ['50', '25']
For an arbitrary number of dicts, you can abstract this further.
In [28]: from functools import reduce
In [29]: from operator import and_
In [30]:
In [30]: dicts = (dict1, dict2)
In [31]: common = reduce(and_, map(set, dicts))
In [32]: c:[d[c] for d in dicts] for c in common
Out[32]: 'blender': ['50', '25']
answered 40 mins ago
timgeb
36.9k104875
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
add a comment |Â
up vote
3
down vote
Assuming the nouns in your lists are unique (per list), create dictionaries out of your lists first.
In [1]: list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
In [2]: list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
In [3]:
In [3]: dict1 = dict(reversed(pair) for pair in zip(*[iter(list1)]*2))
In [4]: dict2 = dict(reversed(pair) for pair in zip(*[iter(list2)]*2))
In [5]:
In [5]: dict1
Out[5]: 'apple': '20', 'blender': '50', 'orange': '5'
In [6]: dict2
Out[6]: 'blender': '25', 'pear': '20', 'spatula': '40'
This code leverages the grouper recipe from the itertools docs.
Matching two items from your dictionaries is easy as pie.
In [7]: key = 'blender'
In [8]: print(key, dict1[key], dict2[key])
blender 50 25
You could even build a dictionary that holds the common keys from dict1 and dict2 and a list of the values.
In [12]: common = dict1.keys() & dict2
In [13]: c:[dict1[c], dict2[c]] for c in common
Out[13]: 'blender': ['50', '25']
For an arbitrary number of dicts, you can abstract this further.
In [28]: from functools import reduce
In [29]: from operator import and_
In [30]:
In [30]: dicts = (dict1, dict2)
In [31]: common = reduce(and_, map(set, dicts))
In [32]: c:[d[c] for d in dicts] for c in common
Out[32]: 'blender': ['50', '25']
answered 40 mins ago
timgeb
36.9k104875
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Assuming the nouns in your lists are unique (per list), create dictionaries out of your lists first.
In [1]: list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
In [2]: list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
In [3]:
In [3]: dict1 = dict(reversed(pair) for pair in zip(*[iter(list1)]*2))
In [4]: dict2 = dict(reversed(pair) for pair in zip(*[iter(list2)]*2))
In [5]:
In [5]: dict1
Out[5]: 'apple': '20', 'blender': '50', 'orange': '5'
In [6]: dict2
Out[6]: 'blender': '25', 'pear': '20', 'spatula': '40'
This code leverages the grouper recipe from the itertools docs.
Matching two items from your dictionaries is easy as pie.
In [7]: key = 'blender'
In [8]: print(key, dict1[key], dict2[key])
blender 50 25
You could even build a dictionary that holds the common keys from dict1 and dict2 and a list of the values.
In [12]: common = dict1.keys() & dict2
In [13]: c:[dict1[c], dict2[c]] for c in common
Out[13]: 'blender': ['50', '25']
For an arbitrary number of dicts, you can abstract this further.
In [28]: from functools import reduce
In [29]: from operator import and_
In [30]:
In [30]: dicts = (dict1, dict2)
In [31]: common = reduce(and_, map(set, dicts))
In [32]: c:[d[c] for d in dicts] for c in common
Out[32]: 'blender': ['50', '25']
answered 40 mins ago
timgeb
36.9k104875
Assuming the nouns in your lists are unique (per list), create dictionaries out of your lists first.
In [1]: list1 = ['5', 'orange', '20', 'apple', '50', 'blender']
In [2]: list2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
In [3]:
In [3]: dict1 = dict(reversed(pair) for pair in zip(*[iter(list1)]*2))
In [4]: dict2 = dict(reversed(pair) for pair in zip(*[iter(list2)]*2))
In [5]:
In [5]: dict1
Out[5]: 'apple': '20', 'blender': '50', 'orange': '5'
In [6]: dict2
Out[6]: 'blender': '25', 'pear': '20', 'spatula': '40'
This code leverages the grouper recipe from the itertools docs.
Matching two items from your dictionaries is easy as pie.
In [7]: key = 'blender'
In [8]: print(key, dict1[key], dict2[key])
blender 50 25
You could even build a dictionary that holds the common keys from dict1 and dict2 and a list of the values.
In [12]: common = dict1.keys() & dict2
In [13]: c:[dict1[c], dict2[c]] for c in common
Out[13]: 'blender': ['50', '25']
For an arbitrary number of dicts, you can abstract this further.
In [28]: from functools import reduce
In [29]: from operator import and_
In [30]:
In [30]: dicts = (dict1, dict2)
In [31]: common = reduce(and_, map(set, dicts))
In [32]: c:[d[c] for d in dicts] for c in common
Out[32]: 'blender': ['50', '25']
answered 40 mins ago
timgeb
36.9k104875
edited 24 mins ago
answered 40 mins ago
timgeb
36.9k104875
answered 40 mins ago
timgeb
36.9k104875
answered 40 mins ago
timgeb
36.9k104875
36.9k104875
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
add a comment |Â
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
I wrote a similar variant pastebin.com/MgiLJzZf
– Chris_Rands
37 mins ago
add a comment |Â
up vote
2
down vote
I hope this code will work for you.
l1 =['5', 'orange', '20', 'apple', '50', 'blender']
l2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for list1_item in zip(*[iter(l1)]*2):
for list2_item in zip(*[iter(l2)]*2):
if list1_item[1]==list2_item[1]:
print list1_item[1],list1_item[0],list2_item[0]
it will print output
blender 50 25
answered 29 mins ago
Abdul Majeed
1,5211315
add a comment |Â
up vote
2
down vote
I hope this code will work for you.
l1 =['5', 'orange', '20', 'apple', '50', 'blender']
l2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for list1_item in zip(*[iter(l1)]*2):
for list2_item in zip(*[iter(l2)]*2):
if list1_item[1]==list2_item[1]:
print list1_item[1],list1_item[0],list2_item[0]
it will print output
blender 50 25
answered 29 mins ago
Abdul Majeed
1,5211315
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I hope this code will work for you.
l1 =['5', 'orange', '20', 'apple', '50', 'blender']
l2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for list1_item in zip(*[iter(l1)]*2):
for list2_item in zip(*[iter(l2)]*2):
if list1_item[1]==list2_item[1]:
print list1_item[1],list1_item[0],list2_item[0]
it will print output
blender 50 25
answered 29 mins ago
Abdul Majeed
1,5211315
I hope this code will work for you.
l1 =['5', 'orange', '20', 'apple', '50', 'blender']
l2 = ['25', 'blender', '20', 'pear', '40', 'spatula']
for list1_item in zip(*[iter(l1)]*2):
for list2_item in zip(*[iter(l2)]*2):
if list1_item[1]==list2_item[1]:
print list1_item[1],list1_item[0],list2_item[0]
it will print output
blender 50 25
answered 29 mins ago
Abdul Majeed
1,5211315
answered 29 mins ago
Abdul Majeed
1,5211315
answered 29 mins ago
Abdul Majeed
1,5211315
answered 29 mins ago
Abdul Majeed
1,5211315
1,5211315
add a comment |Â
add a comment |Â
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6
This seems like an XY problem. If that is the kind of output you're looking to get, storing information in lists like that is not going to be the best approach. It honestly looks like you want a dictionary with words as keys, and the number before them as values
– user3483203
44 mins ago
Take a look at the
enumerateandzipfunctions– JETM
42 mins ago
Is this your special requirement to iterate over list on indexes 1,3,5.... Not on 0,2,4,6..... Consider if blender comes at index either 0,2,4,6... of any one List then will it not be considered as common element on lists??
– Shivam Seth
35 mins ago