How do I split a string into a list of characters?

Question

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

"foobar"    →    ['f', 'o', 'o', 'b', 'a', 'r']

Advocation

In Python, strings are already arrays of characters for all purposes except replacement. You can slice them, reference or look up items by index, etc. – Southland

Link to other direction – Wowser

See stackoverflow.com/questions/743806 for splitting the string into words. – Occipital

Answer 1

Use the list constructor:

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

list builds a new list using items obtained by iterating over the input iterable. A string is an iterable -- iterating over it yields a single character at each iteration step.

Pursuant

In my opinion much better than the ruby method, you can convert between sequence types freely, even better, in C level. – Wysocki

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] – Lemmuela

@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. – Sidwell

This does not not work in the latest versions of R anymore, I think. – Reveille

Answer 2

You take the string and pass it to list()

s = "mystring"
l = list(s)
print l

Recept

Answer 3

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

>>> [c for c in "foobar"]
['f', 'o', 'o', 'b', 'a', 'r']

Meader

Welcome to stackoverflow. Would you mind extending the answer a little bit to explain how it solves the problem. – Fenton

This is a mere for, there's not much to explain. I think you should read the python tutorial on data structures, especially list comprehension. – Botts

This just means list(map(lambda c: c, iter("foobar"))), but more readable and meaningful. – Colorado

Answer 4

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

Lactoflavin

Are there differences in the performance characteristics of each of these methods? – Idolum

Answer 5

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("//")

Grimaldi

Answer 6

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

Costrel

The first way is very simple. Are there reasons people would want something more complex? – Sprinkler

Hello! First option is simple indeed. The second one, though, has better potential for handling more complex processing. – Costrel

Answer 7

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.

Desk

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! – Airwaves

Answer 8

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']

Declinometer

Answer 9

Unpack them:

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

Torsk

Answer 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...

Slurry

Answer 11

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']

Punchy

Answer 12

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

Kokaras

Answer 13

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']

Highbrow

camelcase = ''.join([text[i].upper() if i % 2 else text[i].lower() for i in range(len(text))]) – Eous

@Eous - that's actually aLtErNaTiNg case. CamelCase looks likeThis or LikeThis. – Slurry

Answer 14

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.

Supertanker

Not sure where this would be more useful than the string itself, which is iterable. – Individuation

Answer 15

Since strings are iterables, you can also use iterable unpacking to assign to a list. Below, the characters in my_string are unpacked into the my_list list.

my_string = "foobar"
*my_list, = my_string

print(my_list)   # ['f', 'o', 'o', 'b', 'a', 'r']

This is especially useful, if you need to save the first or last character into a separate variable.

first, *rest = "foobar"

print(first)  # f
print(rest)   # ['o', 'o', 'b', 'a', 'r']

Selfappointed

Answer 16

Here is a nice script that will help you find which method is most efficient for your case:

import timeit
from itertools import chain

string = "thisisthestringthatwewanttosplitintoalist"

def getCharList(str):
  return list(str)

def getCharListComp(str):
  return [char for char in str]

def getCharListMap(str):
  return list(map(lambda c: c, str))

def getCharListForLoop(str):
  list = []
  for c in str:
    list.append(c)

def getCharListUnpack(str):
  return [*str]

def getCharListExtend(str):
  list = []
  return list.extend(str)

def getCharListChain(str):
  return chain(str)
 
time_list = timeit.timeit(stmt='getCharList(string)', globals=globals(), number=1)
time_listcomp = timeit.timeit(stmt='getCharListComp(string)', globals=globals(), number=1)
time_listmap = timeit.timeit(stmt='getCharListMap(string)', globals=globals(), number=1)
time_listforloop = timeit.timeit(stmt='getCharListForLoop(string)', globals=globals(), number=1)
time_listunpack = timeit.timeit(stmt='getCharListUnpack(string)', globals=globals(), number=1)
time_listextend = timeit.timeit(stmt='getCharListExtend(string)', globals=globals(), number=1)
time_listchain = timeit.timeit(stmt='getCharListChain(string)', globals=globals(), number=1)

print(f"Execution time using list constructor is {time_list} seconds")
print(f"Execution time using list comprehension is {time_listcomp} seconds")
print(f"Execution time using map is {time_listmap} seconds")
print(f"Execution time using for loop is {time_listforloop} seconds")
print(f"Execution time using unpacking is {time_listunpack} seconds")
print(f"Execution time using extend is {time_listextend} seconds")
print(f"Execution time using chain is {time_listchain} seconds")

Whortleberry

Passing number=1 to timeit.timeit is probably not a good idea as a large number of iterations is required to get a reliable result. The default number=1000000 is a better amount. – Austria

Answer 17

you can use

*var, = othervar

to convert it to a list

The code would look like this:

foo = "foobar"

*foolist, = foo

print(foolist)

['f', 'o', 'o', 'b', 'a', 'r']

Shellyshelman