Python: How to split a list based on a specific element

Question

If we have the following list in Python

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends", "."]

How do I split this to get a list which contains elements that finish with the full stop? So i want to get the following elements in my new list:

["I","am","good","."]
["I","like","you","."]
["we","are","not","friends","."]

My attempts so far:

cleaned_sentence = []
a = 0
while a < len(sentence):
    current_word = sentence[a]
    if current_word == "." and len(cleaned_sentence) == 0:
        cleaned_sentence.append(sentence[0:sentence.index(".")+1])
        a += 1
    elif current_word == "." and len(cleaned_sentence) > 0:
        sub_list = sentence[sentence.index(".")+1:-1]
        sub_list.append(sentence[-1])
        cleaned_sentence.append(sub_list[0:sentence.index(".")+1])
        a += 1
    else:
        a += 1

for each in cleaned_sentence:
    print(each)

Running this on sentence produces

['I', 'am', 'good', '.']
['I', 'like', 'you', '.']
['I', 'like', 'you', '.']

Possible duplicate of [Split List By Value and Keep Separators](https://stackoverflow.com/questions/45189895/split-list-by-value-and-keep-separators) — Canh, Oct 01 '18 at 12:22

blhsing · Accepted Answer · 2018-10-01T12:32:35.293

16

You can use itertools.groupby:

from itertools import groupby
i = (list(g) for _, g in groupby(sentence, key='.'.__ne__))
print([a + b for a, b in zip(i, i)])

This outputs:

[['I', 'am', 'good', '.'], ['I', 'like', 'you', '.'], ['we', 'are', 'not', 'friends', '.']]

If your list doesn't always end with '.' then you can use itertools.zip_longest instead:

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends"]
i = (list(g) for _, g in groupby(sentence, key='.'.__ne__))
print([a + b for a, b in zip_longest(i, i, fillvalue=[])])

This outputs:

[['I', 'am', 'good', '.'], ['I', 'like', 'you', '.'], ['we', 'are', 'not', 'friends']]

edited Oct 01 '18 at 12:32

answered Oct 01 '18 at 12:24

blhsing

91,368
6
71
106

1

I like `groupby` but I think using `operator.ne` is better than the dunder methods – Chris_Rands Oct 01 '18 at 12:27
`iter` should not be necessary. The generator expression constitutes an iterator already – user2390182 Oct 01 '18 at 12:27
also this assumes the last sentence ends with a full-stop otherwise the entire last sentence is missed (this might be a fair assumption) – Chris_Rands Oct 01 '18 at 12:27
1

@Chris_Rands You would have to wrap `operator.ne` with `partial` or use a lambda (which kinda sucks :) ), but I generally agree. – user2390182 Oct 01 '18 at 12:29
1

@Chris_Rands Indeed. I've updated my answer with a solution for that scenario then. – blhsing Oct 01 '18 at 12:33
@schwobaseggl Indeed. I've updated my answer so it uses the generator expression as an iterator. Thanks. – blhsing Oct 01 '18 at 12:34

Willem Van Onsem · Answer 2 · 2018-10-01T12:29:59.083

We can do this in two stages: first calculating the indices where the dots are located, and then making slices, like:

idxs = [i for i, v in enumerate(sentence, 1) if v == '.']   # calculating indices

result = [sentence[i:j] for i, j in zip([0]+idxs, idxs)]    # splitting accordingly

This then yields:

>>> [sentence[i:j] for i, j in zip([0]+idxs, idxs)]
[['I', 'am', 'good', '.'], ['I', 'like', 'you', '.'], ['we', 'are', 'not', 'friends', '.']]

You can then for example print the individual elements with:

for sub in [sentence[i:j] for i, j in zip([0]+idxs, idxs)]:
    print(sub)

This then will print:

>>> idxs = [i for i, v in enumerate(sentence, 1) if v == '.']
>>> for sub in [sentence[i:j] for i, j in zip([0]+idxs, idxs)]:
...     print(sub)
...
['I', 'am', 'good', '.']
['I', 'like', 'you', '.']
['we', 'are', 'not', 'friends', '.']

score 2 · Answer 3 · answered Oct 01 '18 at 12:30

2

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends", "."]

output = []
temp = []
for item in sentence:
    temp.append(item)
    if item == '.':
        output.append(temp)
        temp = []
if temp:
    output.append(temp)

print(output)

answered Oct 01 '18 at 12:30

apatniv

1,771
10
13

Brute force but still works. But not the python way. Better to use functionalities given in the python. – Eswar Oct 01 '18 at 12:37
1

@Eswar: The solution is not brute-force, it is the best you could do and it only loops once. I think you refer to the fact that it contains more lines and doesn't use any libraries. IMO, which I find to be more readable in this case. Readability can mean different things to different folks. – apatniv Oct 01 '18 at 12:50

score 1 · Answer 4 · answered Oct 01 '18 at 12:20

Using a simple iteration.

Demo:

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends", "."]
last = len(sentence) - 1
result = [[]]
for i, v in enumerate(sentence):
    if v == ".":
        result[-1].append(".")
        if i != last:
            result.append([])
    else:
        result[-1].append(v)
print(result)

Output:

[['I', 'am', 'good', '.'], ['I', 'like', 'you', '.'], ['we', 'are', 'not', 'friends', '.']]

gboffi · Answer 5 · 2018-10-01T13:17:47.557

This answer aims to be the simplest one...

The data

sentences = ["I", "am", "good", ".",
            "I", "like", "you", ".",
            "We", "are", "not", "friends", "."]

We initialize the output list and represent that we are starting a new sentence

l, start = [], 1

We loop on the data list, using w to address the current word

if we are at the start of a new sentence we clear the flag and add an empty list to the tail of the output list
we append the current word to the last sublist (note that ① we are guaranteed that there is at least a last sublist (do you like alliterations?) and ② every word gets appended)
if we are at the end — we have met a "." — we raise again the flag.

Note the single comment…

for w in sentences:
    if start: start = l.append([]) # l.append() returns None, that is falsey...
    l[-1].append(w)
    if w == ".": start = 1

score 0 · Answer 6 · answered Jun 13 '22 at 00:57

You could do this by joining the elements together into a string and then splitting the string back again using a regex:

import re

sentence = ["I", "am", "good", ".", "I", "like", "you", ".", "we", "are", "not", "friends", "."]
result = [m.split('\0') for m in re.findall(r'(?<=\0).*?\.(?=\0|$)', '\0'.join(['.']+sentence))]

Output:

[
 ['I', 'am', 'good', '.'],
 ['I', 'like', 'you', '.'],
 ['we', 'are', 'not', 'friends', '.']
]

Python: How to split a list based on a specific element

6 Answers6

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