Convert a string to a list

Question

Let's say I have this string like this:

string = ("['1.345', '1.346', '1.347']")

So it's formatted like a list, but I need to convert it to an actual list object. How can I do that?

• Winston Ewert: Where did the list come from?

Answer 1

You can use literal_eval from the ast module.

>>> string = ("['1.345', '1.346', '1.347']")
>>> import ast
>>> lst = ast.literal_eval(string)
>>> lst
['1.345', '1.346', '1.347']

BTW, you don't need the parentheses around the string. string = "['1.345', '1.346', '1.347']" works just fine.

• Katriel: To clarify, this is (much, much) better than eval because it only evaluates literals, not actual code. So there isn't a security issue with taking arbitrary user input.

Answer 2

string = ("['1.345', '1.346', '1.347']")
lst = eval(string)

• Niklas B.: Argh, imagine this is user input... What else should it be?
• David Robinson: Yes- Praveen's literal_eval is better

Answer 3

It might help if we knew where these strings come from. In general, Praveen's answer is a good one regarding the example you give. I'd like to mention, though, that using doublequotes instead of singlequotes in your string would make the whole thing valid JSON, in which case you could also do:

import json
json.loads('["1.345", "1.346", "1.347"]') # [u'1.345', u'1.346', u'1.347']

Measuring the performance:

import timeit
s = """\
import ast
ast.literal_eval("['1.345', '1.346', '1.347']")
"""
t = timeit.Timer(stmt=s)
print "%.2f usec/pass" % (1000000 * t.timeit(number=100000) / 100000)

Result: 21.25 usec/pass

vs.

import timeit
s = """\
import json
json.loads('["1.345", "1.346", "1.347"]')
"""
t = timeit.Timer(stmt=s)
print "%.2f usec/pass" % (1000000 * t.timeit(number=100000) / 100000)

Result: 6.32 usec/pass

Maybe this is an alternative for you.

• jena: Putting aside the fact that the OP would have to convert the quotes in his strings, why would you think that the json module is less flexible than the ast module in this context?
• Niklas B.: Yeah, actually I have to agree with you, JSON can represent any data that Python literals can. Removed my comment :)

Answer 4

I would strip '[',']'.etc and use split() method on that string..

lst = string.replace('[','').replace(']','').replace("'",'').replace(" ",'').split(',')

I think it's safer than eval if the string is well formatted, but not quite efficient though, because Python has to build a new string with every replace method.