What is a good way to do some_string.split('') in python? This syntax gives an error:
a = '1111'
a.split('')
ValueError: empty separator
I would like to obtain:
['1', '1', '1', '1']
asked Jun 29, 2013 at 13:14
Saullo G. P. Castro
59.4k28 gold badges191 silver badges244 bronze badges
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5Do you really need to do that? Most of the time you can treat a string in the same way you treat a list anyway.arshajii– arshajii2013-06-29 13:16:03 +00:00Commented Jun 29, 2013 at 13:16
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3I tried to do that and it gave this error, so I thought it would be nice to include here as a questionSaullo G. P. Castro– Saullo G. P. Castro2013-06-29 13:19:48 +00:00Commented Jun 29, 2013 at 13:19
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1You might also find this interesting - Alternative way to split a list into groups of n.Cristian Ciupitu– Cristian Ciupitu2013-06-29 13:36:56 +00:00Commented Jun 29, 2013 at 13:36
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@sqpc: No, it would not give "this error", so you didn't try that. Did you try to just do nothing? Not split it at all?Lennart Regebro– Lennart Regebro2013-06-29 13:54:19 +00:00Commented Jun 29, 2013 at 13:54
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2@arshajii contrary to your "most of the time ",'str' object does not support item assignment. This fill the need.brainLoop– brainLoop2018-04-30 16:32:26 +00:00Commented Apr 30, 2018 at 16:32
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Show 6 more comments
6 Answers
Use list():
>>> list('1111')
['1', '1', '1', '1']
Alternatively, you can use map() (Python 2.7 only):
>>> map(None, '1111')
['1', '1', '1', '1']
Time differences:
$ python -m timeit "list('1111')"
1000000 loops, best of 3: 0.483 usec per loop
$ python -m timeit "map(None, '1111')"
1000000 loops, best of 3: 0.431 usec per loop
Comments
One can cast strings to list directly
>>> list('1111')
['1', '1', '1', '1']
or using list comprehensions
>>> [i for i in '1111']
['1', '1', '1', '1']
second way can be useful if one wants to split strings for substrings more than 1 symbol length
>>> some_string = '12345'
>>> [some_string[i:i+2] for i in range(0, len(some_string), 2)]
['12', '34', '5']
Comments
Strings are iterables and can be indexed, hence you don't really need to split it at all:
>>> for char in '11111':
... print char
...
1
1
1
1
1
>>> '11111'[4]
'1'
You can "split" it with a call to list, but it doesn't make much difference:
>>> for char in list('11111'):
... print char
...
1
1
1
1
1
>>> list('11111')[4]
'1'
So you only need to do this if your code explicitly expects a list. For example:
>>> list('11111').append('2')
>>> l = list('11111')
>>> l.append(2)
>>> l
['1', '1', '1', '1', '1', 2]
This doesn't work with a straight string:
>>> l.append('2')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'str' object has no attribute 'append'
In that case you would need:
>>> l += '2'
>>> l
'111112'
Comments
Method #1:
s="Amalraj"
l=[i for i in s]
print(l)
Output:
['A', 'm', 'a', 'l', 'r', 'a', 'j']
Method #2:
s="Amalraj"
l=list(s)
print(l)
Output:
['A', 'm', 'a', 'l', 'r', 'a', 'j']
Method #3:
import re; # importing regular expression module
s="Amalraj"
l=re.findall('.',s)
print(l)
Output:
['A', 'm', 'a', 'l', 'r', 'a', 'j']
Comments
If like me you arrived here to split in packets and not char by char, I made this :
>>> n=2
>>> str='423346396677674A446E6838436E304164334A30427874784467567863334942446E41424E413D3D'
>>> [ str[byte:byte+n] for byte in range(0,len(str)) if byte % n == 0 ]
['42', '33', '46', '39', '66', '77', '67', '4A', '44', '6E', '68', '38', '43', '6E', '30', '41', '64', '33', '4A', '30', '42', '78', '74', '78', '44', '67', '56', '78', '63', '33', '49', '42', '44', '6E', '41', '42', '4E', '41', '3D', '3D']
Comments
You can even omit the list() call altogether, with a combination of map + lambda fn and then use unpacking.
>>> s = "efgvdfbdb"
>>> [* map(lambda x: x, s) ]
['e', 'f', 'g', 'v', 'd', 'f', 'b', 'd', 'b']
