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How do I split a string into a list of characters? str.split does not work.

"foobar"    →    ['f', 'o', 'o', 'b', 'a', 'r']
Mateen Ulhaq
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Adrian
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14 Answers14

1068
>>> list("foobar")
['f', 'o', 'o', 'b', 'a', 'r']

Use the list constructor.

Mateen Ulhaq
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user225312
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    In my opinion much better than the ruby method, you can convert between sequence types freely, even better, in C level. – arthurprs Feb 23 '11 at 00:37
  • list constructor is a elegant feature which automatically converts string to character array. Since, String is a homogenous sequence of unicode characters its so cool to be working with Python and creator Guido has made it the better. Loving python for its wonderful capabilities. – Doogle Aug 20 '17 at 06:07
  • I want flag here to not do this ... but anyway if you want callable you could escape this behavior using `cast_method = lambda x: [x]` – madzohan Dec 18 '17 at 20:01
  • @Doogle: Capabilities-wise while String is an object and split() can be called on it, list() is a function so it cannot be called on it. – karatedog Mar 12 '21 at 22:18
87

You take the string and pass it to list()

s = "mystring"
l = list(s)
print l
Senthil Kumaran
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74

You can also do it in this very simple way without list():

>>> [c for c in "foobar"]
['f', 'o', 'o', 'b', 'a', 'r']
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    Welcome to stackoverflow. Would you mind extending the answer a little bit to explain how it solves the problem. – Nagama Inamdar Mar 24 '15 at 06:07
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    This is a mere `for`, there's not much to explain. I think you should read the python tutorial on [data structures](https://docs.python.org/2/tutorial/datastructures.html), especially list comprehension. – WhyNotHugo Apr 03 '15 at 00:52
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    This just means `list(map(lambda c: c, iter("foobar")))`, but more readable and meaningful. – InQβ Dec 12 '17 at 01:26
52

If you want to process your String one character at a time. you have various options.

uhello = u'Hello\u0020World'

Using List comprehension:

print([x for x in uhello])

Output:

['H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd']

Using map:

print(list(map(lambda c2: c2, uhello)))

Output:

['H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd']

Calling Built in list function:

print(list(uhello))

Output:

['H', 'e', 'l', 'l', 'o', ' ', 'W', 'o', 'r', 'l', 'd']

Using for loop:

for c in uhello:
    print(c)

Output:

H
e
l
l
o

W
o
r
l
d
dtasev
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Sid
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32

If you just need an array of chars:

arr = list(str)

If you want to split the str by a particular delimiter:

# str = "temp//temps" will will be ['temp', 'temps']
arr = str.split("//")
22

I explored another two ways to accomplish this task. It may be helpful for someone.

The first one is easy:

In [25]: a = []
In [26]: s = 'foobar'
In [27]: a += s
In [28]: a
Out[28]: ['f', 'o', 'o', 'b', 'a', 'r']

And the second one use map and lambda function. It may be appropriate for more complex tasks:

In [36]: s = 'foobar12'
In [37]: a = map(lambda c: c, s)
In [38]: a
Out[38]: ['f', 'o', 'o', 'b', 'a', 'r', '1', '2']

For example

# isdigit, isspace or another facilities such as regexp may be used
In [40]: a = map(lambda c: c if c.isalpha() else '', s)
In [41]: a
Out[41]: ['f', 'o', 'o', 'b', 'a', 'r', '', '']

See python docs for more methods

Alexey Milogradov
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21

The task boils down to iterating over characters of the string and collecting them into a list. The most naïve solution would look like

result = []
for character in string:
    result.append(character)

Of course, it can be shortened to just

result = [character for character in string]

but there still are shorter solutions that do the same thing.

list constructor can be used to convert any iterable (iterators, lists, tuples, string etc.) to list.

>>> list('abc')
['a', 'b', 'c']

The big plus is that it works the same in both Python 2 and Python 3.

Also, starting from Python 3.5 (thanks to the awesome PEP 448) it's now possible to build a list from any iterable by unpacking it to an empty list literal:

>>> [*'abc']
['a', 'b', 'c']

This is neater, and in some cases more efficient than calling list constructor directly.

I'd advise against using map-based approaches, because map does not return a list in Python 3. See How to use filter, map, and reduce in Python 3.

Community
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vaultah
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  • I think the last proposal is very nice. But I don't see why you revisited some of the other approaches, (most of them) have been posted here already and distract from the _amazing_ python 3.5 solution! – MSeifert Apr 05 '16 at 17:38
18

split() inbuilt function will only separate the value on the basis of certain condition but in the single word, it cannot fulfill the condition. So, it can be solved with the help of list(). It internally calls the Array and it will store the value on the basis of an array.

Suppose,

a = "bottle"
a.split() // will only return the word but not split the every single char.

a = "bottle"
list(a) // will separate ['b','o','t','t','l','e']
tuomastik
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Anshul Singh Suryan
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13

Unpack them:

word = "Paralelepipedo"
print([*word])
enbermudas
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10

To split a string s, the easiest way is to pass it to list(). So,

s = 'abc'
s_l = list(s) #  s_l is now ['a', 'b', 'c']

You can also use a list comprehension, which works but is not as concise as the above:

s_l = [c for c in s]

There are other ways, as well, but these should suffice. Later, if you want to recombine them, a simple call to "".join(s_l) will return your list to all its former glory as a string...

Gary02127
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5

You can use extend method in list operations as well.

>>> list1 = []
>>> list1.extend('somestring')
>>> list1
['s', 'o', 'm', 'e', 's', 't', 'r', 'i', 'n', 'g']
Georgy
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4

If you wish to read only access to the string you can use array notation directly.

Python 2.7.6 (default, Mar 22 2014, 22:59:38) 
[GCC 4.8.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> t = 'my string'
>>> t[1]
'y'

Could be useful for testing without using regexp. Does the string contain an ending newline?

>>> t[-1] == '\n'
False
>>> t = 'my string\n'
>>> t[-1] == '\n'
True
Sylvain
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2
from itertools import chain

string = 'your string'
chain(string)

similar to list(string) but returns a generator that is lazily evaluated at point of use, so memory efficient.

Mohammad Kanan
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minggli
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    Not sure where this would be more useful than the string itself, which is iterable. – Ry- May 21 '20 at 10:13
1

Well, much as I like the list(s) version, here's another more verbose way I found (but it's cool so I thought I'd add it to the fray):

>>> text = "My hovercraft is full of eels"
>>> [text[i] for i in range(len(text))]
['M', 'y', ' ', 'h', 'o', 'v', 'e', 'r', 'c', 'r', 'a', 'f', 't', ' ', 'i', 's', ' ', 'f', 'u', 'l', 'l', ' ', 'o', 'f', ' ', 'e', 'e', 'l', 's']
John Lockwood
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