python – How to convert string representation of list to a list?

The Question :

581 people think this question is useful

I was wondering what the simplest way is to convert a string representation of a list like the following to a list:

x = '[ "A","B","C" , " D"]'

Even in cases where the user puts spaces in between the commas, and spaces inside of the quotes, I need to handle that as well and convert it to:

x = ["A", "B", "C", "D"] 

I know I can strip spaces with strip() and split() and check for non-letter characters. But the code was getting very kludgy. Is there a quick function that I’m not aware of?

The Question Comments :
  • What are you actually trying to accomplish? There is probably a far better way than trying to convert Python list syntax into an actual list…
  • What version of Python are you using?
  • @Nicholas Knight: I am trying to handle user input in a legacy app where all lists were entered as unicode lists with square parenthesis. @Mark Byers , I am using python 2.6 so the ast.literal approach works best

The Answer 1

839 people think this answer is useful
>>> import ast
>>> x = '[ "A","B","C" , " D"]'
>>> x = ast.literal_eval(x)
>>> x
['A', 'B', 'C', ' D']
>>> x = [n.strip() for n in x]
>>> x
['A', 'B', 'C', 'D']


With ast.literal_eval you can safely evaluate an expression node or a string containing a Python literal or container display. The string or node provided may only consist of the following Python literal structures: strings, bytes, numbers, tuples, lists, dicts, booleans, and None.

The Answer 2

99 people think this answer is useful

The json module is a better solution whenever there is a stringified list of dictionaries. The json.loads(your_data) function can be used to convert it to a list.

>>> import json
>>> x = '[ "A","B","C" , " D"]'
>>> json.loads(x)
['A', 'B', 'C', ' D']


>>> x = '[ "A","B","C" , {"D":"E"}]'
>>> json.loads(x)
['A', 'B', 'C', {'D': 'E'}]

The Answer 3

84 people think this answer is useful

The eval is dangerous – you shouldn’t execute user input.

If you have 2.6 or newer, use ast instead of eval:

>>> import ast
>>> ast.literal_eval('["A","B" ,"C" ," D"]')
["A", "B", "C", " D"]

Once you have that, strip the strings.

If you’re on an older version of Python, you can get very close to what you want with a simple regular expression:

>>> x='[  "A",  " B", "C","D "]'
>>> re.findall(r'"\s*([^"]*?)\s*"', x)
['A', 'B', 'C', 'D']

This isn’t as good as the ast solution, for example it doesn’t correctly handle escaped quotes in strings. But it’s simple, doesn’t involve a dangerous eval, and might be good enough for your purpose if you’re on an older Python without ast.

The Answer 4

15 people think this answer is useful
import ast
l = ast.literal_eval('[ "A","B","C" , " D"]')
l = [i.strip() for i in l]

The Answer 5

12 people think this answer is useful

There is a quick solution:

x = eval('[ "A","B","C" , " D"]')

Unwanted whitespaces in the list elements may be removed in this way:

x = [x.strip() for x in eval('[ "A","B","C" , " D"]')]

The Answer 6

11 people think this answer is useful

Inspired from some of the answers above that work with base python packages I compared the performance of a few (using Python 3.7.3):

Method 1: ast

import ast
list(map(str.strip, ast.literal_eval(u'[ "A","B","C" , " D"]')))
# ['A', 'B', 'C', 'D']

import timeit
timeit.timeit(stmt="list(map(str.strip, ast.literal_eval(u'[ \"A\",\"B\",\"C\" , \" D\"]')))", setup='import ast', number=100000)
# 1.292875313000195

Method 2: json

import json
list(map(str.strip, json.loads(u'[ "A","B","C" , " D"]')))
# ['A', 'B', 'C', 'D']

import timeit
timeit.timeit(stmt="list(map(str.strip, json.loads(u'[ \"A\",\"B\",\"C\" , \" D\"]')))", setup='import json', number=100000)
# 0.27833264000014424

Method 3: no import

list(map(str.strip, u'[ "A","B","C" , " D"]'.strip('][').replace('"', '').split(',')))
# ['A', 'B', 'C', 'D']

import timeit
timeit.timeit(stmt="list(map(str.strip, u'[ \"A\",\"B\",\"C\" , \" D\"]'.strip('][').replace('\"', '').split(',')))", number=100000)
# 0.12935059100027502

I was disappointed to see what I considered the method with the worst readability was the method with the best performance… there are tradeoffs to consider when going with the most readable option… for the type of workloads I use python for I usually value readability over a slightly more performant option, but as usual it depends.

The Answer 7

10 people think this answer is useful

If it’s only a one dimensional list, this can be done without importing anything:

>>> x = u'[ "A","B","C" , " D"]'
>>> ls = x.strip('[]').replace('"', '').replace(' ', '').split(',')
>>> ls
['A', 'B', 'C', 'D']

The Answer 8

6 people think this answer is useful

Assuming that all your inputs are lists and that the double quotes in the input actually don’t matter, this can be done with a simple regexp replace. It is a bit perl-y but works like a charm. Note also that the output is now a list of unicode strings, you didn’t specify that you needed that, but it seems to make sense given unicode input.

import re
x = u'[ "A","B","C" , " D"]'
junkers = re.compile('[[" \]]')
result = junkers.sub('', x).split(',')
print result
--->  [u'A', u'B', u'C', u'D']

The junkers variable contains a compiled regexp (for speed) of all characters we don’t want, using ] as a character required some backslash trickery. The re.sub replaces all these characters with nothing, and we split the resulting string at the commas.

Note that this also removes spaces from inside entries u'[“oh no”]’ —> [u’ohno’]. If this is not what you wanted, the regexp needs to be souped up a bit.

The Answer 9

4 people think this answer is useful

If you know that your lists only contain quoted strings, this pyparsing example will give you your list of stripped strings (even preserving the original Unicode-ness).

>>> from pyparsing import *
>>> x =u'[ "A","B","C" , " D"]'
>>> LBR,RBR = map(Suppress,"[]")
>>> qs = quotedString.setParseAction(removeQuotes, lambda t: t[0].strip())
>>> qsList = LBR + delimitedList(qs) + RBR
>>> print qsList.parseString(x).asList()
[u'A', u'B', u'C', u'D']

If your lists can have more datatypes, or even contain lists within lists, then you will need a more complete grammar – like this one on the pyparsing wiki, which will handle tuples, lists, ints, floats, and quoted strings. Will work with Python versions back to 2.4.

The Answer 10

1 people think this answer is useful

To further complete @Ryan ‘s answer using json, one very convenient function to convert unicode is the one posted here:

ex with double or single quotes:

>print byteify(json.loads(u'[ "A","B","C" , " D"]')
>print byteify(json.loads(u"[ 'A','B','C' , ' D']".replace('\'','"')))
['A', 'B', 'C', ' D']
['A', 'B', 'C', ' D']

The Answer 11

1 people think this answer is useful

You may run into such problem while dealing with scraped data stored as Pandas DataFrame.

This solution works like charm if the list of values is present as text.

def textToList(hashtags):
    return hashtags.strip('[]').replace('\'', '').replace(' ', '').split(',')

hashtags = "[ 'A','B','C' , ' D']"
hashtags = textToList(hashtags)

Output: ['A', 'B', 'C', 'D']

No external library required.

The Answer 12

0 people think this answer is useful

I would like to provide a more intuitive patterning solution with regex. The below function takes as input a stringified list containing arbitrary strings.

Stepwise explanation: You remove all whitespacing,bracketing and value_separators (provided they are not part of the values you want to extract, else make the regex more complex). Then you split the cleaned string on single or double quotes and take the non-empty values (or odd indexed values, whatever the preference).

def parse_strlist(sl):
import re
clean = re.sub("[\[\],\s]","",sl)
splitted = re.split("[\'\"]",clean)
values_only = [s for s in splitted if s != '']
return values_only

testsample: “[’21’,”foo” ‘6’, ‘0’, ” A”]”

The Answer 13

0 people think this answer is useful

and with pure python – not importing any libraries

[x for x in  x.split('[')[1].split(']')[0].split('"')[1:-1] if x not in[',',' , ',', ']]

The Answer 14

-1 people think this answer is useful

So, following all the answers I decided to time the most common methods:

from time import time
import re
import json

my_str = str(list(range(19)))

reps = 100000

start = time()
for i in range(0, reps):
    re.findall("\w+", my_str)
print("Regex method:\t", (time() - start) / reps)

start = time()
for i in range(0, reps):
print("json method:\t", (time() - start) / reps)

start = time()
for i in range(0, reps):
print("ast method:\t\t", (time() - start) / reps)

start = time()
for i in range(0, reps):
    [n.strip() for n in my_str]
print("strip method:\t", (time() - start) / reps)

    regex method:    6.391477584838867e-07
    json method:     2.535374164581299e-06
    ast method:      2.4425282478332518e-05
    strip method:    4.983267784118653e-06

So in the end regex wins!

The Answer 15

-1 people think this answer is useful

you can save yourself the .strip() fcn by just slicing off the first and last characters from the string representation of the list (see third line below)

>>> mylist=[1,2,3,4,5,'baloney','alfalfa']
>>> strlist=str(mylist)
['1', ' 2', ' 3', ' 4', ' 5', " 'baloney'", " 'alfalfa'"]
>>> mylistfromstring=(strlist[1:-1].split(', '))
>>> mylistfromstring[3]
>>> for entry in mylistfromstring:
...     print(entry)
...     type(entry)
<class 'str'>
<class 'str'>
<class 'str'>
<class 'str'>
<class 'str'>
<class 'str'>
<class 'str'>

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