Given this list:

['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.', 'Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']

How could i return lists separated by a full stop? returning this:

['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.'][ 'Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']

Thanks for any help

CodePudding user response：

You can use a simple loop:

l = ['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.', 'Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']

out = [[]]
for i, e in enumerate(l):
    out[-1].append(e)
    if e == '.' and i 1 != len(l):
        out.append([])

output:

[['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.'],
 ['Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.'],
]

CodePudding user response：

I would create a generator:

def sentences_generator(lst):
    sentence = []
    for w in lst:
        sentence.append(w)
        if w == '.':
            yield sentence
            sentence = []

And then I would use it in this way (lst is your list of words):

sentences = list(sentences_generator(lst))

CodePudding user response：

[s.strip().split()   ["."] for s in " ".join(l).split(".") if s]

1. join with spaces
2. split on periods
3. include only those items that are not empty
4. include items that are a joined list of
   • stripped sentences
   • resplit by spaces
   • appended by a last item period

output:

[['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.'], ['Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']]

CodePudding user response：

Use the join-split method, as detailed in

CodePudding user response：

You could do it using itertools.groupby. The group numbers would be formed by alternating True/False based on items being the stop value or not.

from itertools import groupby

R = [list(g) ['.'] for d,g in groupby(L,lambda n:n!='.') if d]

print(R)

[['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 
  'very', 'pythonly', '.'],
 ['Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']]

Note that there will always be a '.' at the end of each group in the output even if the last item doesn't end with a '.'. Also, this will not output empty groups between two '.' items.

For a more generalized solution that includes empty groups and doesn't add any extra '.', you can use accumulate to obtain a new group number after each '.' item encountered:

from itertools import chain, accumulate, groupby

group = chain([0],accumulate(c=='.' for c in L))
R = [list(g) for _,g in groupby(L,lambda _:next(group))]

CodePudding user response：

Oneliner using more_itertools.split_after:

result = list(split_after(lst, '.'.__eq__))

Full demo:

from more_itertools import split_after

lst = ['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.', 'Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']

result = list(split_after(lst, '.'.__eq__))

print(result)

Output:

[['Pythoners', 'are', 'very', 'intelligent', 'and', 'work', 'very', 'pythonly', '.'], ['Now', 'they', 'are', 'pythoning', 'their', 'way', 'to', 'success', '.']]