Python: List containing sublist of strings

Question

I have a list of strings s as follows:

s = ['Hello', 'world', '!', 'How', 'are', 'you', '?', 'Have', 'a', 'good', 'day', '.']

I want this list to be split into sublists. Whenever there's a ?!.\n a new sublist is formed as follows:

final = [['Hello', 'world', '!'],
         ['How', 'are', 'you', '?'],
         ['Have', 'a', 'good', 'day', '.']]

I tried this:

x = 0
for i in range(len(s)):
    if s[i] in ('!','?','.','\n'):
         final = s[x: x+i]
    x = i+1

final stores my output. Not getting the way it should be. Any suggestions?

mkiever · Accepted Answer · 2016-02-28 17:10:07Z

2

You were not that far away:

x=0
final=[]
for i in range(len(s)):
    if s[i] in ('!','?','.','\n'):
        final.append(s[x:i+1])
        x=i+1

Only a bit of indexing problem and making final a list to collect all partial lists.

answered Feb 28, 2016 at 17:10

mkiever

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2 Comments

Copperfield Over a year ago

I recommend the use of enumerate to get the index instead, and in general is better to use a set for membership testing because of its constant time vs the linear search in tuples or list

mkiever Over a year ago

@Copperfield: Indeed enumerate is nicer for indexing, but would change the structure of the code. I wanted to stay as close as possible to the original structure.

gtlambert · Accepted Answer · 2016-02-28 17:01:24Z

1

You could use the following:

s = ['Hello', 'world', '!', 'How', 'are', 'you', '?', 'Have', 'a', 'good', 'day', '.']
letters = ['!', '?', '.']

idxes = [idx for idx, val in enumerate(s) if val in letters]
idxes = [-1] + idxes
answer = [s[idxes[i]+1:idxes[i+1]+1] for i in range(len(idxes[:-1]))]
print(answer)

Output

[['Hello', 'world', '!'], ['How', 'are', 'you', '?'], ['Have', 'a', 'good', 'day', '.']]

This uses a list comprehension with the built in enumerate function to extract the idxes of s where a punctuation mark occurs. It then uses another list comprehension to construct a list of sublists by slicing the s using the values of idxes.

answered Feb 28, 2016 at 17:01

gtlambert

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Comments

jfsturtz · Accepted Answer · 2016-02-28 17:01:31Z

1

s = ['Hello', 'world', '!', 'How', 'are', 'you', '?', 'Have', 'a', 'good', 'day', '.']
final = []
b = []
for x in s:
    b.append(x)
    if x in ('.', '?', '!', '\n'):
        final.append(b)
        b = []

answered Feb 28, 2016 at 17:01

jfsturtz

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Comments

Saiful Azad · Accepted Answer · 2016-02-28 17:21:15Z

0

1 Let final is an empty array.

2 While loop is true when is not empty and index < len(s).

3 append to final array with 0 to postion+1 words

4 shrink your main string s.

5 incr the index value

final = []
i =0
while len(s) and i<len(s):
    if s[i] in ('!','?','.','\n'):
         final.append( s[:i+1])
         s  = s[i+1:]
    i +=1  
print(final)

edited Feb 28, 2016 at 17:21

answered Feb 28, 2016 at 17:13

Saiful Azad

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Comments

roman · Accepted Answer · 2016-02-28 17:42:03Z

0

I'm not really often use python, but in your case I think you can also try to create a generator from you initial list, so you don't have to store list of lists:

>>> from itertools import chain
>>> def func(s):
...     g = iter(s)
...     def inner_func(g):
...         for x in g:
...             yield x
...             if x in ('.', '?', '!', '\n'):
...                 break
...     while True:
...         try:
...             f = g.next()
...         except StopIteration:
...             break
...         else:
...             yield inner_func(chain([f], g))
>>> [[y for y in x] for x in func(s)]
[['Hello', 'world', '!'], ['How', 'are', 'you', '?'], ['Have', 'a', 'good', 'day', '.']]

answered Feb 28, 2016 at 17:42

roman

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