How to convert string representation of list to a list

Question:

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?

Asked By: harijay

||

Answers:

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"]')]
Answered By: Alexei Sholik
import ast
l = ast.literal_eval('[ "A","B","C" , " D"]')
l = [i.strip() for i in l]
Answered By: tosh

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.

Answered By: Mark Byers
>>> 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']

ast.literal_eval:

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.

Answered By: Roger Pate

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 in the pyparsing examples directory, which will handle tuples, lists, ints, floats, and quoted strings.

Answered By: PaulMcG

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 it 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.

Answered By: dirkjot

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']

Similarly

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

To further complete Ryan’s answer using JSON, one very convenient function to convert Unicode is in this answer.

Example 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']
Answered By: CptHwK

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”]”

Answered By: Jordy Van Landeghem

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)))
print(my_str)

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):
    json.loads(my_str)
print("JSON method:t", (time() - start) / reps)

start = time()
for i in range(0, reps):
    ast.literal_eval(my_str)
print("AST method:tt", (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!

Answered By: passs

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']
Answered By: ruohola

You can save yourself the .strip() function by just slicing off the first and last characters from the string representation of the list (see the 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]
'4'
>>> for entry in mylistfromstring:
...     print(entry)
...     type(entry)
...
1
<class 'str'>
2
<class 'str'>
3
<class 'str'>
4
<class 'str'>
5
<class 'str'>
'baloney'
<class 'str'>
'alfalfa'
<class 'str'>
Answered By: JCMontalbano

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 trade-offs 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.

Answered By: kinzleb

And with pure Python – not importing any libraries:

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

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.

Answered By: dobydx

This usually happens when you load list stored as string to CSV

If you have your list stored in CSV in form like OP asked:

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

Here is how you can load it back to list:

import csv
with open('YourCSVFile.csv') as csv_file:
    reader = csv.reader(csv_file, delimiter=',')
    rows = list(reader)

listItems = rows[0]

listItems is now list

Answered By: Hrvoje

There isn’t any need to import anything or to evaluate. You can do this in one line for most basic use cases, including the one given in the original question.

One liner

l_x = [i.strip() for i in x[1:-1].replace('"',"").split(',')]

Explanation

x = '[ "A","B","C" , " D"]'
# String indexing to eliminate the brackets.
# Replace, as split will otherwise retain the quotes in the returned list
# Split to convert to a list
l_x = x[1:-1].replace('"',"").split(',')

Outputs:

for i in range(0, len(l_x)):
    print(l_x[i])
# vvvv output vvvvv
'''
 A
B
C
  D
'''
print(type(l_x)) # out: class 'list'
print(len(l_x)) # out: 4

You can parse and clean up this list as needed using list comprehension.

l_x = [i.strip() for i in l_x] # list comprehension to clean up
for i in range(0, len(l_x)):
    print(l_x[i])
# vvvvv output vvvvv
'''
A
B
C
D
'''

Nested lists

If you have nested lists, it does get a bit more annoying. Without using regex (which would simplify the replace), and assuming you want to return a flattened list (and the zen of python says flat is better than nested):

x = '[ "A","B","C" , " D", ["E","F","G"]]'
l_x = x[1:-1].split(',')
l_x = [i
    .replace(']', '')
    .replace('[', '')
    .replace('"', '')
    .strip() for i in l_x
]
# returns ['A', 'B', 'C', 'D', 'E', 'F', 'G']

If you need to retain the nested list it gets a bit uglier, but it can still be done just with regular expressions and list comprehension:

import re

x = '[ "A","B","C" , " D", "["E","F","G"]","Z", "Y", "["H","I","J"]", "K", "L"]'
# Clean it up so the regular expression is simpler
x = x.replace('"', '').replace(' ', '')
# Look ahead for the bracketed text that signifies nested list
l_x = re.split(r',(?=[[A-Za-z0-9',]+])|(?<=]),', x[1:-1])
print(l_x)
# Flatten and split the non nested list items
l_x0 = [item for items in l_x for item in items.split(',') if not '[' in items]
# Convert the nested lists to lists
l_x1 = [
    i[1:-1].split(',') for i in l_x if '[' in i
]
# Add the two lists
l_x = l_x0 + l_x1

This last solution will work on any list stored as a string, nested or not.

Answered By: born_naked

This solution is simpler than some I read in the previous answers, but it requires to match all features of the list.

x = '[ "A","B","C" , " D"]'
[i.strip() for i in x.split('"') if len(i.strip().strip(',').strip(']').strip('['))>0]

Output:

['A', 'B', 'C', 'D']
Answered By: CassAndr

This u can do,

**

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

**
best one is the accepted answer

Though this is not a safe way, the best answer is the accepted one.
wasn’t aware of the eval danger when answer was posted.

Answered By: Tomato Master
Categories: questions Tags: ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.