Flexible splitting of an input string in Python function

Question

In Python, what would be the most elegant way of converting this structure:

['group1, group2, group3']

into this structure:

['group1', 'group2', 'group3']

and potentially back.

What is needed is a function that can take either a list of a string of comma separated values (first case) or a list of strings of the same values (second case) and handle them as one and the same: a list of strings.

In pseudocode:

x = ['group1, group2, group3']
y = ['group1', 'group2', 'group3']

f(x) <==> f(y)    <- equivalent behavior

Also, if using split() as per suggestions:

Is there a way to make the delimiter space insensitive or conditional or a regex: I'd like to get to the ['group1', 'group2', 'group3'] result in either ['group1, group2, group3'] or ['group1,group2,group3'] or even this ['group1, group2,group3'] (or a combination thereof) as an input?

A bit more clarification:

>>> single_string = False
>>> a = ['group1', 'group2', 'group3','group4']
>>> [t.strip() for t in [a][0].split(',')] if single_string else a
['group1', 'group2', 'group3', 'group4']
>>> single_string = True
>>> b = ['group1,group2, group3,  group4']
>>> [t.strip() for t in [b][0].split(',')] if single_string else b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute 'split'
>>>

Basically, I'm looking for the most elegant Python conditional expression that would result in the same output both in case of a and b above:['group1', 'group2', 'group3', 'group4'].

hiro protagonist · Accepted Answer · 2018-10-11 05:01:15Z

3

you could use split and join to get from one representation to the other:

print('group1, group2, group3'.split(', '))
# ['group1', 'group2', 'group3']
print(', '.join(['group1', 'group2', 'group3']))
# group1, group2, group3

in a function you could do this:

def f(items, index, separator=', '):
    if isinstance(items, str):
        items = items.split(separator)
    return items[index]


print(f('group1, group2, group3', 1))        # group2
print(f(['group1', 'group2', 'group3'], 1))  # group2

edited Oct 11, 2018 at 5:01

answered Oct 11, 2018 at 4:55

hiro protagonist

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2 Comments

Simeon Leyzerzon Over a year ago

can a separator be conditional? ', ' or ','. I guess a regex is more appropriate?

hiro protagonist Over a year ago

sure, if you need to split several things re.split should work.

user2390182 · Accepted Answer · 2018-10-11 06:26:51Z

1

You were almost there in your edit:

def f(lst):
    return [t.strip() for t in lst[0].split(',')] if len(lst) == 1 else lst

>>> a = ['group1', 'group2', 'group3','group4']
>>> b = ['group1,group2, group3,  group4']

>>> f(a)
['group1', 'group2', 'group3', 'group4']
>>> f(b)
['group1', 'group2', 'group3', 'group4']

answered Oct 11, 2018 at 6:26

user2390182

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Comments

Stephen Rauch · Accepted Answer · 2018-10-11 04:54:03Z

1

How about using split():

Code:

['group1, group2, group3'][0].split(', ')

Test Code:

x = ['group1, group2, group3']    
print(x[0].split(', '))

Results:

['group1', 'group2', 'group3']

To return:

To return use str.join()

y = x[0].split(', ')
print(y)

xx = [', '.join(y)]
print(xx)

Results:

['group1', 'group2', 'group3']
['group1, group2, group3']

answered Oct 11, 2018 at 4:54

Stephen Rauch♦

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3 Comments

Simeon Leyzerzon Over a year ago

Thanks. Is there a way to make the delimiter space insensitive: I'd like to get to the ['group1', 'group2', 'group3'] result in either ['group1, group2, group3'] or ['group1,group2,group3'] or even this ['group1, group2,group3'] (or a combination thereof) as an input?

user2390182 Over a year ago

@SimeonLeyzerzon You can split on the bare comma and strip the tokens: y = [t.strip() for t in x[0].split(',').

Simeon Leyzerzon Over a year ago

@schwobaseggl Your comment seems to be the closest to what I'm after - you may want to make it into an official answer, I've updated my question with a bit more clarification if you care to take another look. Thanks.

Sohaib Farooqi · Accepted Answer · 2018-10-11 05:19:12Z

1

You can define a function using str.split on , followed by flattening of result using itertools.chain

>>> from itertools import chain
>>> f = lambda x: list(chain(*[a.split(', ') for a in x]))

>>> f(['group1', 'group2', 'group3'])
>>> ['group1', 'group2', 'group3']

>>> f(['group1, group2, group3'])
>>> ['group1', 'group2', 'group3']

In case of variable separator and optional space, regex would be a better choice

>>> import re
>>> f = lambda x, sep: list(chain(*[re.split('{}\s*'.format(sep), a) for a in x]))
>>> f(['group1, group2,group3'],',')
>>> ['group1', 'group2', 'group3']

edited Oct 11, 2018 at 5:19

answered Oct 11, 2018 at 5:01

Sohaib Farooqi

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Comments

Anil Kumar Gupta · Accepted Answer · 2018-10-11 05:10:37Z

0

split() and join() can try .

case1 = ['group1, group2, group3']
output = [ele.split(",") for ele in case1]

Output for case-1

[['group1', ' group2', ' group3']]

Case-2

case2 = [['group1', ' group2', ' group3']]
output = [", ".join(ele) for ele in case2]

Output for case-2

['group1, group2, group3']

answered Oct 11, 2018 at 5:10

Anil Kumar Gupta

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Comments

vash_the_stampede · Accepted Answer · 2018-10-11 06:09:18Z

0

By using str.replace() to eliminate all white spaces you can use a formula that will work on all cases presented.

lst = ['group1, group2, group3']

res = lst[0].replace(' ', '').split(',')
# ['group1', 'group2', 'group3']

answered Oct 11, 2018 at 6:09

vash_the_stampede

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Comments

Sam Bourne · Accepted Answer · 2018-10-11 06:45:42Z

0

I think it's most clear simply using split and strip.

# various scenarios to test
TESTS = [
    ['group1, group2, group3'],
    ['group1', 'group2', 'group3'],
    ['group1,group2,  group3'],
    ['group1', 'group2, group3'],
    ['group1 ,group2', 'group3'],
]


def normalize(args, delimiter=','):
    results = []
    for arg in args:
        results.extend(x.strip() for x in arg.split(delimiter))
    return results


for args in TESTS:
    assert normalize(args) == ['group1', 'group2', 'group3']


# test alternative delimiter
assert normalize(['group1 | group2 | group3'], delimiter='|') == ['group1', 'group2', 'group3']

answered Oct 11, 2018 at 6:45

Sam Bourne

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