Split a string and save them to a list with Python

Question

I have a string that I inserted a space into it in all different positions and saved them to a list. Now this list of strings with space in them, I want to split those strings and put the output in one list, when am doing this, it happens that am having multiple list inside:

This is the code am working on:

var ='sans'
res = [var[:i]+' '+var[i:] for i in range(len(var))]
// The previous line: AM adding a space to see maybe that would generate other words
cor = [res[i].split() for i in range (len(res))]

And this is the output am getting:

>>> cor
[['sans'], ['s', 'ans'], ['sa', 'ns'], ['san', 's']]

What am expecting:

>>> cor
    ['sans', 's', 'ans', 'sa', 'ns', 'san', 's']

Am new to python, I don't know what am missing.

Thanks

pault · Accepted Answer · 2017-12-30 01:07:20Z

6

An alternative approach:

cor = " ".join(res).split()

Output:

['sans', 's', 'ans', 'sa', 'ns', 'san', 's']

Explanation

" ".join(res) will join the individual strings in res with a space in between them. Then calling .split() will split this string on whitespace back into a list.

EDIT: A second approach that doesn't involve the intermediate variable res, although this one isn't quite as easy on the eyes:

cor = [var[:i/2+1] if i%2==1 else var[i/2:] for i in range(2*len(var)-1)]

Basically you flip between building substrings from the front and the back.

edited Dec 30, 2017 at 1:07

answered Dec 30, 2017 at 0:44

pault

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19 Comments

Stefan Pochmann Over a year ago

@roganjosh Compared to yours maybe :-P

roganjosh Over a year ago

@StefanPochmann haha, oh come on, this is gonna have to come down to a timeit because this answer beats even a list comp IMO in simplicity :P

pault Over a year ago

@StefanPochmann's answer is likely the best (he's got my upvote) but double list comprehensions can be difficult to understand for someone thats "new to python."

Stefan Pochmann Over a year ago

@roganjosh But it's much faster than ours.

Stefan Pochmann Over a year ago

@roganjosh Yeah you made a big error there. I'm not spelled with "ph" :-). Also, there's no timing code, it's just the solutions.

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Stefan Pochmann · Accepted Answer · 2017-12-30 00:48:52Z

3

First of all, your

[res[i].split() for i in range (len(res))]

is a complicated unpythonic way to do the same as this:

[r.split() for r in res]

Now... the problem is that you treat r.split() as your end result. You should instead use it as a source to treat it further:

[s for r in res for s in r.split()]

answered Dec 30, 2017 at 0:48

Stefan Pochmann

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2 Comments

RoadRunner Over a year ago

This approach seems to be quite fast also, as seen here.

Stefan Pochmann Over a year ago

@RoadRunner With var ='sans' * 1000 and n = 40 it looks better: ideone.com/6dC8Rb. But interestingly, Dekel's is even faster then. That didn't seem right, so I tested them locally independently (also with Python 3.5) and mine was faster there. Then I moved mine to the end and lo and behold, it became the fastest: ideone.com/yYXRrn. Then I moved yours to the end and it became almost the fastest: ideone.com/sN5Gnn. I guess I'll have to stop benchmarking on ideone. Seems like later tests benefit from earlier tests increasing the process priority or so.

Dekel · Accepted Answer · 2017-12-30 00:45:23Z

2

If you have a list

cor = [['sans'], ['s', 'ans'], ['sa', 'ns'], ['san', 's']]

And you want to flatten it, you can use the following:

flat = [x for y in cor for x in y]

The output will be:

['sans', 's', 'ans', 'sa', 'ns', 'san', 's']

You can also make that directly with the res variable:

cor = [x for y in [res[i].split() for i in range (len(res))] for x in y]

edited Dec 30, 2017 at 0:45

answered Dec 30, 2017 at 0:43

Dekel

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3 Comments

Stefan Pochmann Over a year ago

Would be better to not build that cor in the first place.

Dekel Over a year ago

@StefanPochmann you are right, added an example for that as well

Stefan Pochmann Over a year ago

Nah, that's not what I meant. You're still building that list (even in the exact same bad way).

RoadRunner · Accepted Answer · 2017-12-30 02:09:27Z

1

You coud always use map() to split each string in res:

list(map(str.split, res))

Which gives:

[['sans'], ['s', 'ans'], ['sa', 'ns'], ['san', 's']]

Then you can use itertools.chain.from_iterable to flatten the list:

list(chain.from_iterable(map(str.split, res)))

Which Outputs:

['sans', 's', 'ans', 'sa', 'ns', 'san', 's']

edited Dec 30, 2017 at 2:09

answered Dec 30, 2017 at 2:01

RoadRunner

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Comments

user9158931 · Accepted Answer · 2018-01-01 03:24:01Z

0

You could do something like this in one line without importing any module :

var ='sans'

final=[]
list(map(lambda x:list(map(lambda y:final.append(y),x)),[(var[i:]+' '+var[:i]).split() for i in range(0,len(var))]))
print(final)

output:

['sans', 'ans', 's', 'ns', 'sa', 's', 'san']

answered Jan 1, 2018 at 3:24

user9158931

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