How to efficiently split a list that has a certain periodicity, into multiple lists?

user_4143312 0
Viewed 59

For example the original list:

['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']

We want to split the list into lists started with 'a' and ended with 'a', like the following:

['a','b','c','a']

['a','d','e','a']

['a','b','e','f','j','a']

['a','c','a']

The final ouput can also be a list of lists. I have tried a double for loop approach with 'a' as the condition, but this is inefficient and not pythonic.

python-3.x list python
5 Answers

One possible solution is using re (regex)

import re

l = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
r = [list(f"a{_}a") for _ in re.findall("(?<=a)[^a]+(?=a)", "".join(l))]
print(r)
# [['a', 'b', 'c', 'a'], ['a', 'd', 'e', 'a'], ['a', 'b', 'e', 'f', 'j', 'a'], ['a', 'c', 'a']]
user_85065480

You can do this in one loop:

lst = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']

out = [[]]
for i in lst:
    if i == 'a':
        out[-1].append(i)
        out.append([])
    out[-1].append(i)
out = out[1:] if out[-1][-1] == 'a' else out[1:-1]

Also using numpy.split:

out = [ary.tolist() + ['a'] for ary in np.split(lst, np.where(np.array(lst) == 'a')[0])[1:-1]]

Output:

[['a', 'b', 'c', 'a'], ['a', 'd', 'e', 'a'], ['a', 'b', 'e', 'f', 'j', 'a'], ['a', 'c', 'a']]
Admin1

Firstly you can store the indices of 'a' from the list.

oList = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']

idx_a = list()

for idx, char in enumerate(oList):
    if char == 'a':
        idx_a.append(idx)

Then for every consecutive indices you can get the sub-list and store it in a list

ans = [oList[idx_a[x]:idx_a[x + 1] + 1] for x in range(len(idx_a))]

You can also get more such lists if you take in-between indices also.

user_73731440

You can do this with a single iteration and a simple state machine:

original_list = list('kabcadeabefjacab')

multiple_lists = []
for c in original_list:
    if multiple_lists:
        multiple_lists[-1].append(c)
    if c == 'a':
        multiple_lists.append([c])
if multiple_lists[-1][-1] != 'a':
    multiple_lists.pop()

print(multiple_lists)

[['a', 'b', 'c', 'a'], ['a', 'd', 'e', 'a'], ['a', 'b', 'e', 'f', 'j', 'a'], ['a', 'c', 'a']]
user_37997590

We can use str.split() to split the list once we str.join() it to a string, and then use a f-string to add back the stripped "a"s. Note that even if the list starts/ends with an "a", this the split list will have an empty string representing the substring before the split, so our unpacking logic that discards the first + last subsequences will still work as intended.

def split(data):
    _, *subseqs, _ = "".join(data).split("a")
    return [list(f"a{seq}a") for seq in subseqs]

Output:

>>> from pprint import pprint
>>> testdata = ['k','a','b','c','a','d','e','a','b','e','f','j','a','c','a','b']
>>> pprint(split(testdata))
[['a', 'b', 'c', 'a'],
 ['a', 'd', 'e', 'a'],
 ['a', 'b', 'e', 'f', 'j', 'a'],
 ['a', 'c', 'a']]
user_62210240
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