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?
>>> 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.
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'}]
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.
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"]')]
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.
import ast
l = ast.literal_eval('[ "A","B","C" , " D"]')
l = [i.strip() for i in l]
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']
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.
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.
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.
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.
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.
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
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']
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"]"
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'>
And with pure Python - not importing any libraries:
[x for x in x.split('[')[1].split(']')[0].split('"')[1:-1] if x not in[',',' , ',', ']]
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:\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!
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']
I define the piecewise function
def Li(x):
return piecewise(x, [x < 0, x >= 0], [lambda t: sin(t), lambda t: cos(t)])
And when I evaluate Li(1.0)
The answer is correct
Li(1.0)=array(0.5403023058681398),
But if I write Li(1) the answer is array(0).
I don't understand this behaviour.
This function runs correctly.
def Li(x):
return piecewise(float(x),
[x < 0, x >= 0],
[lambda t: sin(t), lambda t: cos(t)])
It seems that piecewise() converts the return values to the same type as the input so, when an integer is input an integer conversion is performed on the result, which is then returned. Because sine and cosine always return values between −1 and 1 all integer conversions will result in 0, 1 or -1 only - with the vast majority being 0.
>>> x=np.array([0.9])
>>> np.piecewise(x, [True], [float(x)])
array([ 0.9])
>>> x=np.array([1.0])
>>> np.piecewise(x, [True], [float(x)])
array([ 1.])
>>> x=np.array([1])
>>> np.piecewise(x, [True], [float(x)])
array([1])
>>> x=np.array([-1])
>>> np.piecewise(x, [True], [float(x)])
array([-1])
In the above I have explicitly cast the result to float, however, an integer input results in an integer output regardless of the explicit cast. I'd say that this is unexpected and I don't know why piecewise() should do this.
I don't know if you have something more elaborate in mind, however, you don't need piecewise() for this simple case; an if/else will suffice instead:
from math import sin, cos
def Li(t):
return sin(t) if t < 0 else cos(t)
>>> Li(0)
1.0
>>> Li(1)
0.5403023058681398
>>> Li(1.0)
0.5403023058681398
>>> Li(-1.0)
-0.8414709848078965
>>> Li(-1)
-0.8414709848078965
You can wrap the return value in an numpy.array if required.
I am sorry, but this example is taken and modified from
http://docs.scipy.org/doc/numpy/reference/generated/numpy.piecewise.html
But, in fact, using ipython with numpy 1.9
"""
Python 2.7.8 |Anaconda 2.1.0 (64-bit)| (default, Aug 21 2014, 18:22:21)
Type "copyright", "credits" or "license" for more information.
IPython 2.2.0 -- An enhanced Interactive Python.
"""
I have no errors, but "ValueError: too many boolean indices" error appears if I use python 2.7.3 with numpy 1.6
"""
Python 2.7.3 (default, Feb 27 2014, 19:58:35)
"""
I test this function under Linux and Windows and the same error occurs.
Obviously, It is very easy to overcome this situation, but I think that
this behaviour is a mistake in the numpy library.