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How to add count for each unique val in list
Clash Royale CLAN TAG#URR8PPP
up vote
7
down vote
favorite
Assume I have a list.
temp = ['A', 'B', 'A', 'B', 'A', 'B']
I am looking for a way to join the count of the string inside.
Intended Output:
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
I was able to solve it by using a list comprehension but I am looking for a way where I don't have to specify the list [1, 1, 2, 2, 3, 3]. Is it possible?
[j + "_" + str(i) for i, j in zip([1, 1, 2, 2, 3, 3], temp)]
python string list counter
edited 1 hour ago
jpp
77k184491
asked 2 hours ago
Rakesh
764
add a comment |Â
up vote
7
down vote
favorite
Assume I have a list.
temp = ['A', 'B', 'A', 'B', 'A', 'B']
I am looking for a way to join the count of the string inside.
Intended Output:
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
I was able to solve it by using a list comprehension but I am looking for a way where I don't have to specify the list [1, 1, 2, 2, 3, 3]. Is it possible?
[j + "_" + str(i) for i, j in zip([1, 1, 2, 2, 3, 3], temp)]
python string list counter
edited 1 hour ago
jpp
77k184491
asked 2 hours ago
Rakesh
764
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Assume I have a list.
temp = ['A', 'B', 'A', 'B', 'A', 'B']
I am looking for a way to join the count of the string inside.
Intended Output:
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
I was able to solve it by using a list comprehension but I am looking for a way where I don't have to specify the list [1, 1, 2, 2, 3, 3]. Is it possible?
[j + "_" + str(i) for i, j in zip([1, 1, 2, 2, 3, 3], temp)]
python string list counter
edited 1 hour ago
jpp
77k184491
asked 2 hours ago
Rakesh
764
Assume I have a list.
temp = ['A', 'B', 'A', 'B', 'A', 'B']
I am looking for a way to join the count of the string inside.
Intended Output:
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
I was able to solve it by using a list comprehension but I am looking for a way where I don't have to specify the list [1, 1, 2, 2, 3, 3]. Is it possible?
[j + "_" + str(i) for i, j in zip([1, 1, 2, 2, 3, 3], temp)]
python string list counter
python string list counter
edited 1 hour ago
jpp
77k184491
asked 2 hours ago
Rakesh
764
edited 1 hour ago
jpp
77k184491
asked 2 hours ago
Rakesh
764
edited 1 hour ago
jpp
77k184491
edited 1 hour ago
jpp
77k184491
edited 1 hour ago
jpp
77k184491
77k184491
asked 2 hours ago
Rakesh
764
asked 2 hours ago
Rakesh
764
asked 2 hours ago
Rakesh
764
764
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
12
down vote
accepted
You can use collections.defaultdict with a for loop:
from collections import defaultdict
L = ['A', 'B', 'A', 'B', 'A', 'B']
dd = defaultdict(int)
res =
for item in L:
dd[item] += 1
res.append(f'item_dd[item]')
print(res)
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
jpp
77k184491
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
3
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like__setitem__. Trust me, you don't want a listcomp here.
– timgeb
1 hour ago
1
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
2
@Rakesh Well, since you asked, for science...dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normalforloop...
– timgeb
1 hour ago
 |Â
show 3 more comments
up vote
8
down vote
You can use a Counter or a defaultdict(int) to keep track of how many times a character has been seen as you encounter them.
>>> from collections import Counter
>>>
>>> temp = ['A', 'B', 'A', 'B', 'A', 'B']
>>> seen = Counter()
>>>
>>> result =
>>> for c in temp:
...: seen.update(c)
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
Note that seen.update(c) might have unexpected results if you expect strings with more than one character in temp. Demo:
>>> seen = Counter()
>>> seen.update('ABC')
>>> seen
>>> Counter('A': 1, 'B': 1, 'C': 1)
Depending on how you want to count and what kind of data you expect, you might want to use the line
seen[c] += 1
instead of
seen.update(c)
Alternatively, without any imports:
>>> seen =
>>> result =
>>>
>>> for c in temp:
...: seen[c] = seen.get(c, 0) + 1
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
timgeb
40.9k105580
add a comment |Â
up vote
4
down vote
You can use a dictionary (or better yet, a collections.defaultdict) to maintain the counts for each item:
from collections import defaultdict
lst = ['A', 'B', 'A', 'B', 'A', 'B']
lst2 =
d = defaultdict(int)
for item in lst:
d[item] += 1
lst2.append('_'.format(item, d[item]))
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
To use a list comprehension here you'd need some way to update state (i.e. the counters) for each item as you iterate the original list. For that you could use a function with a default argument, e.g.:
def get_count(item, d=defaultdict(int)):
d[item] += 1
return '_'.format(item, d[item])
lst2 = [get_count(item) for item in lst]
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
Eugene Yarmash
80.9k21170255
add a comment |Â
up vote
0
down vote
This is the fastest way to get the desired output in form of a dict.
def countElement(a):
g =
for i in a:
if i in g: g[i] +=1
else: g[i] =1
return g
z = [1,1,1,1,2,2,2,2,3,3,4,5,5,234,23,3,12,3,123,12,31,23,13,2,4,23,42,42,34,234,23,42,34,23,423,42,34,23,423,4,234,23,42,34,23,4,23,423,4,23,4]
print (countElement(z))
answered 32 mins ago
Akash Swain
1
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
You can use collections.defaultdict with a for loop:
from collections import defaultdict
L = ['A', 'B', 'A', 'B', 'A', 'B']
dd = defaultdict(int)
res =
for item in L:
dd[item] += 1
res.append(f'item_dd[item]')
print(res)
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
jpp
77k184491
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
3
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like__setitem__. Trust me, you don't want a listcomp here.
– timgeb
1 hour ago
1
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
2
@Rakesh Well, since you asked, for science...dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normalforloop...
– timgeb
1 hour ago
 |Â
show 3 more comments
up vote
12
down vote
accepted
You can use collections.defaultdict with a for loop:
from collections import defaultdict
L = ['A', 'B', 'A', 'B', 'A', 'B']
dd = defaultdict(int)
res =
for item in L:
dd[item] += 1
res.append(f'item_dd[item]')
print(res)
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
jpp
77k184491
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
3
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like__setitem__. Trust me, you don't want a listcomp here.
– timgeb
1 hour ago
1
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
2
@Rakesh Well, since you asked, for science...dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normalforloop...
– timgeb
1 hour ago
 |Â
show 3 more comments
up vote
12
down vote
accepted
up vote
12
down vote
accepted
You can use collections.defaultdict with a for loop:
from collections import defaultdict
L = ['A', 'B', 'A', 'B', 'A', 'B']
dd = defaultdict(int)
res =
for item in L:
dd[item] += 1
res.append(f'item_dd[item]')
print(res)
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
jpp
77k184491
You can use collections.defaultdict with a for loop:
from collections import defaultdict
L = ['A', 'B', 'A', 'B', 'A', 'B']
dd = defaultdict(int)
res =
for item in L:
dd[item] += 1
res.append(f'item_dd[item]')
print(res)
['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
jpp
77k184491
answered 2 hours ago
jpp
77k184491
answered 2 hours ago
jpp
77k184491
answered 2 hours ago
jpp
77k184491
77k184491
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
3
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like__setitem__. Trust me, you don't want a listcomp here.
– timgeb
1 hour ago
1
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
2
@Rakesh Well, since you asked, for science...dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normalforloop...
– timgeb
1 hour ago
 |Â
show 3 more comments
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
3
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like__setitem__. Trust me, you don't want a listcomp here.
– timgeb
1 hour ago
1
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
2
@Rakesh Well, since you asked, for science...dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normalforloop...
– timgeb
1 hour ago
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
Thank you :). Is it possible to write this in a list comprehension?
– Rakesh
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
@Rakesh yes, but only using very ugly side effects.
– timgeb
1 hour ago
3
3
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like
__setitem__. Trust me, you don't want a listcomp here.– timgeb
1 hour ago
You would have to use boolean expressions inside the comprehension where the evaluation of the truthyness of its parts also calls methods like
__setitem__. Trust me, you don't want a listcomp here.– timgeb
1 hour ago
1
1
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
Agree with @timgeb here, a list comprehension would work if you pre-calculated the index values (which would be overly expensive). But not when you need to calculate them based on prior values.
– jpp
1 hour ago
2
2
@Rakesh Well, since you asked, for science...
dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normal for loop...– timgeb
1 hour ago
@Rakesh Well, since you asked, for science...
dd = defaultdict(int); [dd.__setitem__(c, dd[c] + 1) or '_'.format(c, dd[c]) for c in L]. Please use the normal for loop...– timgeb
1 hour ago
 |Â
show 3 more comments
up vote
8
down vote
You can use a Counter or a defaultdict(int) to keep track of how many times a character has been seen as you encounter them.
>>> from collections import Counter
>>>
>>> temp = ['A', 'B', 'A', 'B', 'A', 'B']
>>> seen = Counter()
>>>
>>> result =
>>> for c in temp:
...: seen.update(c)
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
Note that seen.update(c) might have unexpected results if you expect strings with more than one character in temp. Demo:
>>> seen = Counter()
>>> seen.update('ABC')
>>> seen
>>> Counter('A': 1, 'B': 1, 'C': 1)
Depending on how you want to count and what kind of data you expect, you might want to use the line
seen[c] += 1
instead of
seen.update(c)
Alternatively, without any imports:
>>> seen =
>>> result =
>>>
>>> for c in temp:
...: seen[c] = seen.get(c, 0) + 1
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
timgeb
40.9k105580
add a comment |Â
up vote
8
down vote
You can use a Counter or a defaultdict(int) to keep track of how many times a character has been seen as you encounter them.
>>> from collections import Counter
>>>
>>> temp = ['A', 'B', 'A', 'B', 'A', 'B']
>>> seen = Counter()
>>>
>>> result =
>>> for c in temp:
...: seen.update(c)
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
Note that seen.update(c) might have unexpected results if you expect strings with more than one character in temp. Demo:
>>> seen = Counter()
>>> seen.update('ABC')
>>> seen
>>> Counter('A': 1, 'B': 1, 'C': 1)
Depending on how you want to count and what kind of data you expect, you might want to use the line
seen[c] += 1
instead of
seen.update(c)
Alternatively, without any imports:
>>> seen =
>>> result =
>>>
>>> for c in temp:
...: seen[c] = seen.get(c, 0) + 1
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
timgeb
40.9k105580
add a comment |Â
up vote
8
down vote
up vote
8
down vote
You can use a Counter or a defaultdict(int) to keep track of how many times a character has been seen as you encounter them.
>>> from collections import Counter
>>>
>>> temp = ['A', 'B', 'A', 'B', 'A', 'B']
>>> seen = Counter()
>>>
>>> result =
>>> for c in temp:
...: seen.update(c)
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
Note that seen.update(c) might have unexpected results if you expect strings with more than one character in temp. Demo:
>>> seen = Counter()
>>> seen.update('ABC')
>>> seen
>>> Counter('A': 1, 'B': 1, 'C': 1)
Depending on how you want to count and what kind of data you expect, you might want to use the line
seen[c] += 1
instead of
seen.update(c)
Alternatively, without any imports:
>>> seen =
>>> result =
>>>
>>> for c in temp:
...: seen[c] = seen.get(c, 0) + 1
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
timgeb
40.9k105580
You can use a Counter or a defaultdict(int) to keep track of how many times a character has been seen as you encounter them.
>>> from collections import Counter
>>>
>>> temp = ['A', 'B', 'A', 'B', 'A', 'B']
>>> seen = Counter()
>>>
>>> result =
>>> for c in temp:
...: seen.update(c)
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
Note that seen.update(c) might have unexpected results if you expect strings with more than one character in temp. Demo:
>>> seen = Counter()
>>> seen.update('ABC')
>>> seen
>>> Counter('A': 1, 'B': 1, 'C': 1)
Depending on how you want to count and what kind of data you expect, you might want to use the line
seen[c] += 1
instead of
seen.update(c)
Alternatively, without any imports:
>>> seen =
>>> result =
>>>
>>> for c in temp:
...: seen[c] = seen.get(c, 0) + 1
...: result.append('_'.format(c, seen[c]))
...:
>>> result
>>> ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
timgeb
40.9k105580
edited 2 hours ago
answered 2 hours ago
timgeb
40.9k105580
answered 2 hours ago
timgeb
40.9k105580
answered 2 hours ago
timgeb
40.9k105580
40.9k105580
add a comment |Â
add a comment |Â
up vote
4
down vote
You can use a dictionary (or better yet, a collections.defaultdict) to maintain the counts for each item:
from collections import defaultdict
lst = ['A', 'B', 'A', 'B', 'A', 'B']
lst2 =
d = defaultdict(int)
for item in lst:
d[item] += 1
lst2.append('_'.format(item, d[item]))
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
To use a list comprehension here you'd need some way to update state (i.e. the counters) for each item as you iterate the original list. For that you could use a function with a default argument, e.g.:
def get_count(item, d=defaultdict(int)):
d[item] += 1
return '_'.format(item, d[item])
lst2 = [get_count(item) for item in lst]
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
Eugene Yarmash
80.9k21170255
add a comment |Â
up vote
4
down vote
You can use a dictionary (or better yet, a collections.defaultdict) to maintain the counts for each item:
from collections import defaultdict
lst = ['A', 'B', 'A', 'B', 'A', 'B']
lst2 =
d = defaultdict(int)
for item in lst:
d[item] += 1
lst2.append('_'.format(item, d[item]))
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
To use a list comprehension here you'd need some way to update state (i.e. the counters) for each item as you iterate the original list. For that you could use a function with a default argument, e.g.:
def get_count(item, d=defaultdict(int)):
d[item] += 1
return '_'.format(item, d[item])
lst2 = [get_count(item) for item in lst]
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
Eugene Yarmash
80.9k21170255
add a comment |Â
up vote
4
down vote
up vote
4
down vote
You can use a dictionary (or better yet, a collections.defaultdict) to maintain the counts for each item:
from collections import defaultdict
lst = ['A', 'B', 'A', 'B', 'A', 'B']
lst2 =
d = defaultdict(int)
for item in lst:
d[item] += 1
lst2.append('_'.format(item, d[item]))
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
To use a list comprehension here you'd need some way to update state (i.e. the counters) for each item as you iterate the original list. For that you could use a function with a default argument, e.g.:
def get_count(item, d=defaultdict(int)):
d[item] += 1
return '_'.format(item, d[item])
lst2 = [get_count(item) for item in lst]
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
Eugene Yarmash
80.9k21170255
You can use a dictionary (or better yet, a collections.defaultdict) to maintain the counts for each item:
from collections import defaultdict
lst = ['A', 'B', 'A', 'B', 'A', 'B']
lst2 =
d = defaultdict(int)
for item in lst:
d[item] += 1
lst2.append('_'.format(item, d[item]))
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
To use a list comprehension here you'd need some way to update state (i.e. the counters) for each item as you iterate the original list. For that you could use a function with a default argument, e.g.:
def get_count(item, d=defaultdict(int)):
d[item] += 1
return '_'.format(item, d[item])
lst2 = [get_count(item) for item in lst]
print(lst2) # ['A_1', 'B_1', 'A_2', 'B_2', 'A_3', 'B_3']
answered 2 hours ago
Eugene Yarmash
80.9k21170255
edited 1 hour ago
answered 2 hours ago
Eugene Yarmash
80.9k21170255
answered 2 hours ago
Eugene Yarmash
80.9k21170255
answered 2 hours ago
Eugene Yarmash
80.9k21170255
80.9k21170255
add a comment |Â
add a comment |Â
up vote
0
down vote
This is the fastest way to get the desired output in form of a dict.
def countElement(a):
g =
for i in a:
if i in g: g[i] +=1
else: g[i] =1
return g
z = [1,1,1,1,2,2,2,2,3,3,4,5,5,234,23,3,12,3,123,12,31,23,13,2,4,23,42,42,34,234,23,42,34,23,423,42,34,23,423,4,234,23,42,34,23,4,23,423,4,23,4]
print (countElement(z))
answered 32 mins ago
Akash Swain
1
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Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
This is the fastest way to get the desired output in form of a dict.
def countElement(a):
g =
for i in a:
if i in g: g[i] +=1
else: g[i] =1
return g
z = [1,1,1,1,2,2,2,2,3,3,4,5,5,234,23,3,12,3,123,12,31,23,13,2,4,23,42,42,34,234,23,42,34,23,423,42,34,23,423,4,234,23,42,34,23,4,23,423,4,23,4]
print (countElement(z))
answered 32 mins ago
Akash Swain
1
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is the fastest way to get the desired output in form of a dict.
def countElement(a):
g =
for i in a:
if i in g: g[i] +=1
else: g[i] =1
return g
z = [1,1,1,1,2,2,2,2,3,3,4,5,5,234,23,3,12,3,123,12,31,23,13,2,4,23,42,42,34,234,23,42,34,23,423,42,34,23,423,4,234,23,42,34,23,4,23,423,4,23,4]
print (countElement(z))
answered 32 mins ago
Akash Swain
1
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
This is the fastest way to get the desired output in form of a dict.
def countElement(a):
g =
for i in a:
if i in g: g[i] +=1
else: g[i] =1
return g
z = [1,1,1,1,2,2,2,2,3,3,4,5,5,234,23,3,12,3,123,12,31,23,13,2,4,23,42,42,34,234,23,42,34,23,423,42,34,23,423,4,234,23,42,34,23,4,23,423,4,23,4]
print (countElement(z))
answered 32 mins ago
Akash Swain
1
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 32 mins ago
Akash Swain
1
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 32 mins ago
Akash Swain
1
answered 32 mins ago
Akash Swain
1
1
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Akash Swain is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
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