Mixing
Slicing a list into sublists based on condition
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up vote
7
down vote
favorite
I want to slice this list of numbers:
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
into multiple sublists. The condition for slicing is that the numbers in each sublist should be in increasing order.
So the final result will look like this:
list_1 = [97, 122]
list_2 = [99]
list_3 = [98, 111, 112, 113]
list_4 = [100, 102]
Can anyone help me to solve this problem please? Thanks a lot
python python-3.x list
edited 42 mins ago
jpp
66.8k173984
asked 1 hour ago
Huy Nguyen Bui
655
add a comment |Â
up vote
7
down vote
favorite
I want to slice this list of numbers:
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
into multiple sublists. The condition for slicing is that the numbers in each sublist should be in increasing order.
So the final result will look like this:
list_1 = [97, 122]
list_2 = [99]
list_3 = [98, 111, 112, 113]
list_4 = [100, 102]
Can anyone help me to solve this problem please? Thanks a lot
python python-3.x list
edited 42 mins ago
jpp
66.8k173984
asked 1 hour ago
Huy Nguyen Bui
655
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I want to slice this list of numbers:
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
into multiple sublists. The condition for slicing is that the numbers in each sublist should be in increasing order.
So the final result will look like this:
list_1 = [97, 122]
list_2 = [99]
list_3 = [98, 111, 112, 113]
list_4 = [100, 102]
Can anyone help me to solve this problem please? Thanks a lot
python python-3.x list
edited 42 mins ago
jpp
66.8k173984
asked 1 hour ago
Huy Nguyen Bui
655
I want to slice this list of numbers:
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
into multiple sublists. The condition for slicing is that the numbers in each sublist should be in increasing order.
So the final result will look like this:
list_1 = [97, 122]
list_2 = [99]
list_3 = [98, 111, 112, 113]
list_4 = [100, 102]
Can anyone help me to solve this problem please? Thanks a lot
python python-3.x list
python python-3.x list
edited 42 mins ago
jpp
66.8k173984
asked 1 hour ago
Huy Nguyen Bui
655
edited 42 mins ago
jpp
66.8k173984
asked 1 hour ago
Huy Nguyen Bui
655
edited 42 mins ago
jpp
66.8k173984
edited 42 mins ago
jpp
66.8k173984
edited 42 mins ago
jpp
66.8k173984
66.8k173984
asked 1 hour ago
Huy Nguyen Bui
655
asked 1 hour ago
Huy Nguyen Bui
655
asked 1 hour ago
Huy Nguyen Bui
655
655
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
3
down vote
Here is a one-linear Numpythonic approach:
np.split(arr, np.where(np.diff(arr) < 0)[0] + 1)
Or a similar approach but less efficient in python:
from operator import sub
from itertools import starmap
indices = [0] + [
i+1 for i, j in enumerate(list(
starmap(sub, zip(num_list[1:], num_list)))
) if j < 0] + [len(num_list)
] + [len(num_list)]
result = [num_list[i:j] for i, j in zip(indices, indices[1:])]
Demo:
# Numpy
In [8]: np.split(num_list, np.where(np.diff(num_list) < 0)[0] + 1)
Out[8]:
[array([ 97, 122]),
array([99]),
array([ 98, 111, 112, 113]),
array([100, 102])]
# Python
In [42]: from operator import sub
In [43]: from itertools import starmap
In [44]: indices = [0] + [i+1 for i, j in enumerate(list(starmap(sub, zip(num_list[1:], num_list)))) if j < 0] + [len(num_list)]
In [45]: [num_list[i:j] for i, j in zip(indices, indices[1:])]
Out[45]: [[97, 122], [99], [98, 111, 112, 113], [100, 102]]
Explanation:
Using np.diff() you can get the differences of each item with their next item (up until the last element). Then you can use the vectorized nature of numpy to get the indices of the places where this difference is negative, which can be done with a simple comparison and np.where(). Finally you can simply pass the indices to np.split() to split the array based on those indices.
answered 45 mins ago
Kasrâmvd
75.7k982115
Out of curiosity, what would be the last element of thenp.diff(num_list)array? Because the last element has no next item.
– Guimoute
40 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
add a comment |Â
up vote
2
down vote
I've quickly written one way to do this, I'm sure there are more efficient ways, but this works at least:
num_list =[97, 122, 99, 98, 111, 112, 113, 100, 102]
arrays = [[num_list[0]]] # array of sub-arrays (starts with first value)
for i in range(1, len(num_list)): # go through each element after the first
if num_list[i - 1] < num_list[i]: # If it's larger than the previous
arrays[len(arrays) - 1].append(num_list[i]) # Add it to the last sub-array
else: # otherwise
arrays.append([num_list[i]]) # Make a new sub-array
print(arrays)
Hopefully this helps you a bit :)
answered 57 mins ago
DeltaMarine101
435315
add a comment |Â
up vote
2
down vote
Creating a variable number of variables is not recommended. Use a list of lists or dictionary instead. Here's an example with dict and a generator function:
from itertools import zip_longest
def yield_lists(L):
x =
for i, j in zip_longest(L, L[1:], fillvalue=L[-1]):
x.append(i)
if i > j:
yield x
x =
yield x
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
res = dict(enumerate(yield_lists(num_list), 1))
Resut:
1: [97, 122],
2: [99],
3: [98, 111, 112, 113],
4: [100, 102]
For example, access the second list via res[2].
answered 50 mins ago
jpp
66.8k173984
Ideal solution, generators for the win
– Take_Care_
41 mins ago
add a comment |Â
up vote
0
down vote
All nice solutions here. Maybe this one will be easier to understand for some ppl?
def increasing(a, b):
return a < b
def seq_split(lst, cond):
sublst = [lst[0]]
for item in lst[1:]:
if cond(sublst[-1], item):
sublst.append(item)
else:
yield sublst
sublst = [item]
if sublst:
yield sublst
list(seq_split(num_list, increasing))
answered 13 mins ago
code22
825
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Here is a one-linear Numpythonic approach:
np.split(arr, np.where(np.diff(arr) < 0)[0] + 1)
Or a similar approach but less efficient in python:
from operator import sub
from itertools import starmap
indices = [0] + [
i+1 for i, j in enumerate(list(
starmap(sub, zip(num_list[1:], num_list)))
) if j < 0] + [len(num_list)
] + [len(num_list)]
result = [num_list[i:j] for i, j in zip(indices, indices[1:])]
Demo:
# Numpy
In [8]: np.split(num_list, np.where(np.diff(num_list) < 0)[0] + 1)
Out[8]:
[array([ 97, 122]),
array([99]),
array([ 98, 111, 112, 113]),
array([100, 102])]
# Python
In [42]: from operator import sub
In [43]: from itertools import starmap
In [44]: indices = [0] + [i+1 for i, j in enumerate(list(starmap(sub, zip(num_list[1:], num_list)))) if j < 0] + [len(num_list)]
In [45]: [num_list[i:j] for i, j in zip(indices, indices[1:])]
Out[45]: [[97, 122], [99], [98, 111, 112, 113], [100, 102]]
Explanation:
Using np.diff() you can get the differences of each item with their next item (up until the last element). Then you can use the vectorized nature of numpy to get the indices of the places where this difference is negative, which can be done with a simple comparison and np.where(). Finally you can simply pass the indices to np.split() to split the array based on those indices.
answered 45 mins ago
Kasrâmvd
75.7k982115
Out of curiosity, what would be the last element of thenp.diff(num_list)array? Because the last element has no next item.
– Guimoute
40 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
add a comment |Â
up vote
3
down vote
Here is a one-linear Numpythonic approach:
np.split(arr, np.where(np.diff(arr) < 0)[0] + 1)
Or a similar approach but less efficient in python:
from operator import sub
from itertools import starmap
indices = [0] + [
i+1 for i, j in enumerate(list(
starmap(sub, zip(num_list[1:], num_list)))
) if j < 0] + [len(num_list)
] + [len(num_list)]
result = [num_list[i:j] for i, j in zip(indices, indices[1:])]
Demo:
# Numpy
In [8]: np.split(num_list, np.where(np.diff(num_list) < 0)[0] + 1)
Out[8]:
[array([ 97, 122]),
array([99]),
array([ 98, 111, 112, 113]),
array([100, 102])]
# Python
In [42]: from operator import sub
In [43]: from itertools import starmap
In [44]: indices = [0] + [i+1 for i, j in enumerate(list(starmap(sub, zip(num_list[1:], num_list)))) if j < 0] + [len(num_list)]
In [45]: [num_list[i:j] for i, j in zip(indices, indices[1:])]
Out[45]: [[97, 122], [99], [98, 111, 112, 113], [100, 102]]
Explanation:
Using np.diff() you can get the differences of each item with their next item (up until the last element). Then you can use the vectorized nature of numpy to get the indices of the places where this difference is negative, which can be done with a simple comparison and np.where(). Finally you can simply pass the indices to np.split() to split the array based on those indices.
answered 45 mins ago
Kasrâmvd
75.7k982115
Out of curiosity, what would be the last element of thenp.diff(num_list)array? Because the last element has no next item.
– Guimoute
40 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here is a one-linear Numpythonic approach:
np.split(arr, np.where(np.diff(arr) < 0)[0] + 1)
Or a similar approach but less efficient in python:
from operator import sub
from itertools import starmap
indices = [0] + [
i+1 for i, j in enumerate(list(
starmap(sub, zip(num_list[1:], num_list)))
) if j < 0] + [len(num_list)
] + [len(num_list)]
result = [num_list[i:j] for i, j in zip(indices, indices[1:])]
Demo:
# Numpy
In [8]: np.split(num_list, np.where(np.diff(num_list) < 0)[0] + 1)
Out[8]:
[array([ 97, 122]),
array([99]),
array([ 98, 111, 112, 113]),
array([100, 102])]
# Python
In [42]: from operator import sub
In [43]: from itertools import starmap
In [44]: indices = [0] + [i+1 for i, j in enumerate(list(starmap(sub, zip(num_list[1:], num_list)))) if j < 0] + [len(num_list)]
In [45]: [num_list[i:j] for i, j in zip(indices, indices[1:])]
Out[45]: [[97, 122], [99], [98, 111, 112, 113], [100, 102]]
Explanation:
Using np.diff() you can get the differences of each item with their next item (up until the last element). Then you can use the vectorized nature of numpy to get the indices of the places where this difference is negative, which can be done with a simple comparison and np.where(). Finally you can simply pass the indices to np.split() to split the array based on those indices.
answered 45 mins ago
Kasrâmvd
75.7k982115
Here is a one-linear Numpythonic approach:
np.split(arr, np.where(np.diff(arr) < 0)[0] + 1)
Or a similar approach but less efficient in python:
from operator import sub
from itertools import starmap
indices = [0] + [
i+1 for i, j in enumerate(list(
starmap(sub, zip(num_list[1:], num_list)))
) if j < 0] + [len(num_list)
] + [len(num_list)]
result = [num_list[i:j] for i, j in zip(indices, indices[1:])]
Demo:
# Numpy
In [8]: np.split(num_list, np.where(np.diff(num_list) < 0)[0] + 1)
Out[8]:
[array([ 97, 122]),
array([99]),
array([ 98, 111, 112, 113]),
array([100, 102])]
# Python
In [42]: from operator import sub
In [43]: from itertools import starmap
In [44]: indices = [0] + [i+1 for i, j in enumerate(list(starmap(sub, zip(num_list[1:], num_list)))) if j < 0] + [len(num_list)]
In [45]: [num_list[i:j] for i, j in zip(indices, indices[1:])]
Out[45]: [[97, 122], [99], [98, 111, 112, 113], [100, 102]]
Explanation:
Using np.diff() you can get the differences of each item with their next item (up until the last element). Then you can use the vectorized nature of numpy to get the indices of the places where this difference is negative, which can be done with a simple comparison and np.where(). Finally you can simply pass the indices to np.split() to split the array based on those indices.
answered 45 mins ago
Kasrâmvd
75.7k982115
edited 19 mins ago
answered 45 mins ago
Kasrâmvd
75.7k982115
answered 45 mins ago
Kasrâmvd
75.7k982115
answered 45 mins ago
Kasrâmvd
75.7k982115
75.7k982115
Out of curiosity, what would be the last element of thenp.diff(num_list)array? Because the last element has no next item.
– Guimoute
40 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
add a comment |Â
Out of curiosity, what would be the last element of thenp.diff(num_list)array? Because the last element has no next item.
– Guimoute
40 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
Out of curiosity, what would be the last element of the
np.diff(num_list) array? Because the last element has no next item.– Guimoute
40 mins ago
Out of curiosity, what would be the last element of the
np.diff(num_list) array? Because the last element has no next item.– Guimoute
40 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
@Guimoute There's no diff for last element. Just added that to the answer.
– Kasrâmvd
39 mins ago
add a comment |Â
up vote
2
down vote
I've quickly written one way to do this, I'm sure there are more efficient ways, but this works at least:
num_list =[97, 122, 99, 98, 111, 112, 113, 100, 102]
arrays = [[num_list[0]]] # array of sub-arrays (starts with first value)
for i in range(1, len(num_list)): # go through each element after the first
if num_list[i - 1] < num_list[i]: # If it's larger than the previous
arrays[len(arrays) - 1].append(num_list[i]) # Add it to the last sub-array
else: # otherwise
arrays.append([num_list[i]]) # Make a new sub-array
print(arrays)
Hopefully this helps you a bit :)
answered 57 mins ago
DeltaMarine101
435315
add a comment |Â
up vote
2
down vote
I've quickly written one way to do this, I'm sure there are more efficient ways, but this works at least:
num_list =[97, 122, 99, 98, 111, 112, 113, 100, 102]
arrays = [[num_list[0]]] # array of sub-arrays (starts with first value)
for i in range(1, len(num_list)): # go through each element after the first
if num_list[i - 1] < num_list[i]: # If it's larger than the previous
arrays[len(arrays) - 1].append(num_list[i]) # Add it to the last sub-array
else: # otherwise
arrays.append([num_list[i]]) # Make a new sub-array
print(arrays)
Hopefully this helps you a bit :)
answered 57 mins ago
DeltaMarine101
435315
add a comment |Â
up vote
2
down vote
up vote
2
down vote
I've quickly written one way to do this, I'm sure there are more efficient ways, but this works at least:
num_list =[97, 122, 99, 98, 111, 112, 113, 100, 102]
arrays = [[num_list[0]]] # array of sub-arrays (starts with first value)
for i in range(1, len(num_list)): # go through each element after the first
if num_list[i - 1] < num_list[i]: # If it's larger than the previous
arrays[len(arrays) - 1].append(num_list[i]) # Add it to the last sub-array
else: # otherwise
arrays.append([num_list[i]]) # Make a new sub-array
print(arrays)
Hopefully this helps you a bit :)
answered 57 mins ago
DeltaMarine101
435315
I've quickly written one way to do this, I'm sure there are more efficient ways, but this works at least:
num_list =[97, 122, 99, 98, 111, 112, 113, 100, 102]
arrays = [[num_list[0]]] # array of sub-arrays (starts with first value)
for i in range(1, len(num_list)): # go through each element after the first
if num_list[i - 1] < num_list[i]: # If it's larger than the previous
arrays[len(arrays) - 1].append(num_list[i]) # Add it to the last sub-array
else: # otherwise
arrays.append([num_list[i]]) # Make a new sub-array
print(arrays)
Hopefully this helps you a bit :)
answered 57 mins ago
DeltaMarine101
435315
answered 57 mins ago
DeltaMarine101
435315
answered 57 mins ago
DeltaMarine101
435315
answered 57 mins ago
DeltaMarine101
435315
435315
add a comment |Â
add a comment |Â
up vote
2
down vote
Creating a variable number of variables is not recommended. Use a list of lists or dictionary instead. Here's an example with dict and a generator function:
from itertools import zip_longest
def yield_lists(L):
x =
for i, j in zip_longest(L, L[1:], fillvalue=L[-1]):
x.append(i)
if i > j:
yield x
x =
yield x
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
res = dict(enumerate(yield_lists(num_list), 1))
Resut:
1: [97, 122],
2: [99],
3: [98, 111, 112, 113],
4: [100, 102]
For example, access the second list via res[2].
answered 50 mins ago
jpp
66.8k173984
Ideal solution, generators for the win
– Take_Care_
41 mins ago
add a comment |Â
up vote
2
down vote
Creating a variable number of variables is not recommended. Use a list of lists or dictionary instead. Here's an example with dict and a generator function:
from itertools import zip_longest
def yield_lists(L):
x =
for i, j in zip_longest(L, L[1:], fillvalue=L[-1]):
x.append(i)
if i > j:
yield x
x =
yield x
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
res = dict(enumerate(yield_lists(num_list), 1))
Resut:
1: [97, 122],
2: [99],
3: [98, 111, 112, 113],
4: [100, 102]
For example, access the second list via res[2].
answered 50 mins ago
jpp
66.8k173984
Ideal solution, generators for the win
– Take_Care_
41 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Creating a variable number of variables is not recommended. Use a list of lists or dictionary instead. Here's an example with dict and a generator function:
from itertools import zip_longest
def yield_lists(L):
x =
for i, j in zip_longest(L, L[1:], fillvalue=L[-1]):
x.append(i)
if i > j:
yield x
x =
yield x
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
res = dict(enumerate(yield_lists(num_list), 1))
Resut:
1: [97, 122],
2: [99],
3: [98, 111, 112, 113],
4: [100, 102]
For example, access the second list via res[2].
answered 50 mins ago
jpp
66.8k173984
Creating a variable number of variables is not recommended. Use a list of lists or dictionary instead. Here's an example with dict and a generator function:
from itertools import zip_longest
def yield_lists(L):
x =
for i, j in zip_longest(L, L[1:], fillvalue=L[-1]):
x.append(i)
if i > j:
yield x
x =
yield x
num_list = [97, 122, 99, 98, 111, 112, 113, 100, 102]
res = dict(enumerate(yield_lists(num_list), 1))
Resut:
1: [97, 122],
2: [99],
3: [98, 111, 112, 113],
4: [100, 102]
For example, access the second list via res[2].
answered 50 mins ago
jpp
66.8k173984
edited 38 mins ago
answered 50 mins ago
jpp
66.8k173984
answered 50 mins ago
jpp
66.8k173984
answered 50 mins ago
jpp
66.8k173984
66.8k173984
Ideal solution, generators for the win
– Take_Care_
41 mins ago
add a comment |Â
Ideal solution, generators for the win
– Take_Care_
41 mins ago
Ideal solution, generators for the win
– Take_Care_
41 mins ago
Ideal solution, generators for the win
– Take_Care_
41 mins ago
add a comment |Â
up vote
0
down vote
All nice solutions here. Maybe this one will be easier to understand for some ppl?
def increasing(a, b):
return a < b
def seq_split(lst, cond):
sublst = [lst[0]]
for item in lst[1:]:
if cond(sublst[-1], item):
sublst.append(item)
else:
yield sublst
sublst = [item]
if sublst:
yield sublst
list(seq_split(num_list, increasing))
answered 13 mins ago
code22
825
add a comment |Â
up vote
0
down vote
All nice solutions here. Maybe this one will be easier to understand for some ppl?
def increasing(a, b):
return a < b
def seq_split(lst, cond):
sublst = [lst[0]]
for item in lst[1:]:
if cond(sublst[-1], item):
sublst.append(item)
else:
yield sublst
sublst = [item]
if sublst:
yield sublst
list(seq_split(num_list, increasing))
answered 13 mins ago
code22
825
add a comment |Â
up vote
0
down vote
up vote
0
down vote
All nice solutions here. Maybe this one will be easier to understand for some ppl?
def increasing(a, b):
return a < b
def seq_split(lst, cond):
sublst = [lst[0]]
for item in lst[1:]:
if cond(sublst[-1], item):
sublst.append(item)
else:
yield sublst
sublst = [item]
if sublst:
yield sublst
list(seq_split(num_list, increasing))
answered 13 mins ago
code22
825
All nice solutions here. Maybe this one will be easier to understand for some ppl?
def increasing(a, b):
return a < b
def seq_split(lst, cond):
sublst = [lst[0]]
for item in lst[1:]:
if cond(sublst[-1], item):
sublst.append(item)
else:
yield sublst
sublst = [item]
if sublst:
yield sublst
list(seq_split(num_list, increasing))
answered 13 mins ago
code22
825
answered 13 mins ago
code22
825
answered 13 mins ago
code22
825
answered 13 mins ago
code22
825
825
add a comment |Â
add a comment |Â
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