Split a string to K substrings using List Comprehension
1
I have a long string with me like this
s = 'abcdabcdabcdabcdabcdefghi'
I want to split it to K substrings, where each substring must be at least of length 1 ie non-empty. I want all such possible combinations.
The output I am expecting must be like following if K is 3
[['abcda', 'bcdabcdabcda', 'bcdefghi'], [.....], [....], ... ]
I wanted to do this with list comprehension but I am stuck. Is it possible to implement.? Are there any other faster alternatives.?
python
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
add a comment |
1
I have a long string with me like this
s = 'abcdabcdabcdabcdabcdefghi'
I want to split it to K substrings, where each substring must be at least of length 1 ie non-empty. I want all such possible combinations.
The output I am expecting must be like following if K is 3
[['abcda', 'bcdabcdabcda', 'bcdefghi'], [.....], [....], ... ]
I wanted to do this with list comprehension but I am stuck. Is it possible to implement.? Are there any other faster alternatives.?
python
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
2
Why does your expected output with K=3 not start with['a','b','cdabcdabcdabcdabcdefghi']?
– usr2564301
Nov 25 '18 at 10:48
I just gave a random sample. The order doesn't matter
– Sreeram TP
Nov 25 '18 at 10:49
Do you want every possible combination ofKsublists?
– SuperShoot
Nov 25 '18 at 10:49
I want to get every possible combination of K sublists. I am looking for some faster method. List comprehension most preferably.
– Sreeram TP
Nov 25 '18 at 10:51
add a comment |
1
1
1
1
I have a long string with me like this
s = 'abcdabcdabcdabcdabcdefghi'
I want to split it to K substrings, where each substring must be at least of length 1 ie non-empty. I want all such possible combinations.
The output I am expecting must be like following if K is 3
[['abcda', 'bcdabcdabcda', 'bcdefghi'], [.....], [....], ... ]
I wanted to do this with list comprehension but I am stuck. Is it possible to implement.? Are there any other faster alternatives.?
python
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
I have a long string with me like this
s = 'abcdabcdabcdabcdabcdefghi'
I want to split it to K substrings, where each substring must be at least of length 1 ie non-empty. I want all such possible combinations.
The output I am expecting must be like following if K is 3
[['abcda', 'bcdabcdabcda', 'bcdefghi'], [.....], [....], ... ]
I wanted to do this with list comprehension but I am stuck. Is it possible to implement.? Are there any other faster alternatives.?
python
python
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
asked Nov 25 '18 at 10:44
Sreeram TPSreeram TP
2,98731438
2,98731438
2
Why does your expected output with K=3 not start with['a','b','cdabcdabcdabcdabcdefghi']?
– usr2564301
Nov 25 '18 at 10:48
I just gave a random sample. The order doesn't matter
– Sreeram TP
Nov 25 '18 at 10:49
Do you want every possible combination ofKsublists?
– SuperShoot
Nov 25 '18 at 10:49
I want to get every possible combination of K sublists. I am looking for some faster method. List comprehension most preferably.
– Sreeram TP
Nov 25 '18 at 10:51
add a comment |
2
Why does your expected output with K=3 not start with['a','b','cdabcdabcdabcdabcdefghi']?
– usr2564301
Nov 25 '18 at 10:48
I just gave a random sample. The order doesn't matter
– Sreeram TP
Nov 25 '18 at 10:49
Do you want every possible combination ofKsublists?
– SuperShoot
Nov 25 '18 at 10:49
I want to get every possible combination of K sublists. I am looking for some faster method. List comprehension most preferably.
– Sreeram TP
Nov 25 '18 at 10:51
2
2
Why does your expected output with K=3 not start with
['a','b','cdabcdabcdabcdabcdefghi']?– usr2564301
Nov 25 '18 at 10:48
Why does your expected output with K=3 not start with
['a','b','cdabcdabcdabcdabcdefghi']?– usr2564301
Nov 25 '18 at 10:48
I just gave a random sample. The order doesn't matter
– Sreeram TP
Nov 25 '18 at 10:49
I just gave a random sample. The order doesn't matter
– Sreeram TP
Nov 25 '18 at 10:49
Do you want every possible combination of
K sublists?– SuperShoot
Nov 25 '18 at 10:49
Do you want every possible combination of
K sublists?– SuperShoot
Nov 25 '18 at 10:49
I want to get every possible combination of K sublists. I am looking for some faster method. List comprehension most preferably.
– Sreeram TP
Nov 25 '18 at 10:51
I want to get every possible combination of K sublists. I am looking for some faster method. List comprehension most preferably.
– Sreeram TP
Nov 25 '18 at 10:51
add a comment |
2 Answers
2
active
oldest
votes
2
Using itertools.combinations, you can get separation index pairs:
>>> s = 'abcdef'
>>> k = 3
>>> list(combinations(range(1, len(s)), k-1))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]
using that index pair to get string slices
(1, 2)-> (s[:1],s[1:2],s[2:])
(1, 3)-> (s[:1],s[1:3],s[3:])- ...
(4, 5)-> (s[:4],s[4:5],s[5:])
>>> from itertools import combinations
>>> s = 'abcdef'
>>> k = 3
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'cdef'], ['a', 'bc', 'def'], ..., ['abcd', 'e', 'f']]
>>> k = 4
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'c', 'def'], ['a', 'b', 'cd', 'ef'], ..., ['abc', 'd', 'e', 'f']]
s[:1] == s[0:1] == s[None:1]s[2:] == s[2:len(s)] == s[2:None]
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add anotherforloop(s).
– falsetru
Nov 25 '18 at 10:59
add a comment |
0
You could find all slices of your list so that none of sliced parts will be empty without any extended libraries like so:
s = 'abcd'
substrings =
# find slice of first part - from a|bcd to ab|cd
for first_slice in range(len(s)-2):
# find slice of second and last part, for bcd - from b|cd to bc|d
# for cd - just c|d
for second_slice in range(first_slice+1, len(s)-1):
substrings.append([s[:first_slice+1], s[first_slice+1: second_slice+1], s[second_slice+1:]])
print(substrings) # -> [['a', 'b', 'cd'], ['a', 'bc', 'd'], ['ab', 'c', 'd']]
s = 'abcdabcdabcdabcdabcdefghi'
print(len(substrings)) # -> 276
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Using itertools.combinations, you can get separation index pairs:
>>> s = 'abcdef'
>>> k = 3
>>> list(combinations(range(1, len(s)), k-1))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]
using that index pair to get string slices
(1, 2)-> (s[:1],s[1:2],s[2:])
(1, 3)-> (s[:1],s[1:3],s[3:])- ...
(4, 5)-> (s[:4],s[4:5],s[5:])
>>> from itertools import combinations
>>> s = 'abcdef'
>>> k = 3
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'cdef'], ['a', 'bc', 'def'], ..., ['abcd', 'e', 'f']]
>>> k = 4
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'c', 'def'], ['a', 'b', 'cd', 'ef'], ..., ['abc', 'd', 'e', 'f']]
s[:1] == s[0:1] == s[None:1]s[2:] == s[2:len(s)] == s[2:None]
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add anotherforloop(s).
– falsetru
Nov 25 '18 at 10:59
add a comment |
2
Using itertools.combinations, you can get separation index pairs:
>>> s = 'abcdef'
>>> k = 3
>>> list(combinations(range(1, len(s)), k-1))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]
using that index pair to get string slices
(1, 2)-> (s[:1],s[1:2],s[2:])
(1, 3)-> (s[:1],s[1:3],s[3:])- ...
(4, 5)-> (s[:4],s[4:5],s[5:])
>>> from itertools import combinations
>>> s = 'abcdef'
>>> k = 3
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'cdef'], ['a', 'bc', 'def'], ..., ['abcd', 'e', 'f']]
>>> k = 4
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'c', 'def'], ['a', 'b', 'cd', 'ef'], ..., ['abc', 'd', 'e', 'f']]
s[:1] == s[0:1] == s[None:1]s[2:] == s[2:len(s)] == s[2:None]
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add anotherforloop(s).
– falsetru
Nov 25 '18 at 10:59
add a comment |
2
2
2
Using itertools.combinations, you can get separation index pairs:
>>> s = 'abcdef'
>>> k = 3
>>> list(combinations(range(1, len(s)), k-1))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]
using that index pair to get string slices
(1, 2)-> (s[:1],s[1:2],s[2:])
(1, 3)-> (s[:1],s[1:3],s[3:])- ...
(4, 5)-> (s[:4],s[4:5],s[5:])
>>> from itertools import combinations
>>> s = 'abcdef'
>>> k = 3
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'cdef'], ['a', 'bc', 'def'], ..., ['abcd', 'e', 'f']]
>>> k = 4
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'c', 'def'], ['a', 'b', 'cd', 'ef'], ..., ['abc', 'd', 'e', 'f']]
s[:1] == s[0:1] == s[None:1]s[2:] == s[2:len(s)] == s[2:None]
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
Using itertools.combinations, you can get separation index pairs:
>>> s = 'abcdef'
>>> k = 3
>>> list(combinations(range(1, len(s)), k-1))
[(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)]
using that index pair to get string slices
(1, 2)-> (s[:1],s[1:2],s[2:])
(1, 3)-> (s[:1],s[1:3],s[3:])- ...
(4, 5)-> (s[:4],s[4:5],s[5:])
>>> from itertools import combinations
>>> s = 'abcdef'
>>> k = 3
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'cdef'], ['a', 'bc', 'def'], ..., ['abcd', 'e', 'f']]
>>> k = 4
>>> [[s[i:j] for i, j in zip((None,) + idxs, idxs + (None,))]
... for idxs in combinations(range(1, len(s)), k-1)]
[['a', 'b', 'c', 'def'], ['a', 'b', 'cd', 'ef'], ..., ['abc', 'd', 'e', 'f']]
s[:1] == s[0:1] == s[None:1]s[2:] == s[2:len(s)] == s[2:None]
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
edited Nov 25 '18 at 11:05
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
answered Nov 25 '18 at 10:53
falsetrufalsetru
250k34443436
250k34443436
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add anotherforloop(s).
– falsetru
Nov 25 '18 at 10:59
add a comment |
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add anotherforloop(s).
– falsetru
Nov 25 '18 at 10:59
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
Seems cool. Will this be faster than the other solution.?
– Sreeram TP
Nov 25 '18 at 10:57
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add another
for loop(s).– falsetru
Nov 25 '18 at 10:59
@SreeramTP, Time complexity is same as far as I know. But other solution (Filip Młynarski) works only for k=3 case. To go future you need to add another
for loop(s).– falsetru
Nov 25 '18 at 10:59
add a comment |
0
You could find all slices of your list so that none of sliced parts will be empty without any extended libraries like so:
s = 'abcd'
substrings =
# find slice of first part - from a|bcd to ab|cd
for first_slice in range(len(s)-2):
# find slice of second and last part, for bcd - from b|cd to bc|d
# for cd - just c|d
for second_slice in range(first_slice+1, len(s)-1):
substrings.append([s[:first_slice+1], s[first_slice+1: second_slice+1], s[second_slice+1:]])
print(substrings) # -> [['a', 'b', 'cd'], ['a', 'bc', 'd'], ['ab', 'c', 'd']]
s = 'abcdabcdabcdabcdabcdefghi'
print(len(substrings)) # -> 276
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
add a comment |
0
You could find all slices of your list so that none of sliced parts will be empty without any extended libraries like so:
s = 'abcd'
substrings =
# find slice of first part - from a|bcd to ab|cd
for first_slice in range(len(s)-2):
# find slice of second and last part, for bcd - from b|cd to bc|d
# for cd - just c|d
for second_slice in range(first_slice+1, len(s)-1):
substrings.append([s[:first_slice+1], s[first_slice+1: second_slice+1], s[second_slice+1:]])
print(substrings) # -> [['a', 'b', 'cd'], ['a', 'bc', 'd'], ['ab', 'c', 'd']]
s = 'abcdabcdabcdabcdabcdefghi'
print(len(substrings)) # -> 276
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
add a comment |
0
0
0
You could find all slices of your list so that none of sliced parts will be empty without any extended libraries like so:
s = 'abcd'
substrings =
# find slice of first part - from a|bcd to ab|cd
for first_slice in range(len(s)-2):
# find slice of second and last part, for bcd - from b|cd to bc|d
# for cd - just c|d
for second_slice in range(first_slice+1, len(s)-1):
substrings.append([s[:first_slice+1], s[first_slice+1: second_slice+1], s[second_slice+1:]])
print(substrings) # -> [['a', 'b', 'cd'], ['a', 'bc', 'd'], ['ab', 'c', 'd']]
s = 'abcdabcdabcdabcdabcdefghi'
print(len(substrings)) # -> 276
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
You could find all slices of your list so that none of sliced parts will be empty without any extended libraries like so:
s = 'abcd'
substrings =
# find slice of first part - from a|bcd to ab|cd
for first_slice in range(len(s)-2):
# find slice of second and last part, for bcd - from b|cd to bc|d
# for cd - just c|d
for second_slice in range(first_slice+1, len(s)-1):
substrings.append([s[:first_slice+1], s[first_slice+1: second_slice+1], s[second_slice+1:]])
print(substrings) # -> [['a', 'b', 'cd'], ['a', 'bc', 'd'], ['ab', 'c', 'd']]
s = 'abcdabcdabcdabcdabcdefghi'
print(len(substrings)) # -> 276
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
edited Nov 25 '18 at 11:00
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
answered Nov 25 '18 at 10:53
Filip MłynarskiFilip Młynarski
1,7961413
1,7961413
add a comment |
add a comment |
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2
Why does your expected output with K=3 not start with
['a','b','cdabcdabcdabcdabcdefghi']?– usr2564301
Nov 25 '18 at 10:48
I just gave a random sample. The order doesn't matter
– Sreeram TP
Nov 25 '18 at 10:49
Do you want every possible combination of
Ksublists?– SuperShoot
Nov 25 '18 at 10:49
I want to get every possible combination of K sublists. I am looking for some faster method. List comprehension most preferably.
– Sreeram TP
Nov 25 '18 at 10:51