String to list, python

Question

Let's say I have something like this in a file:

[[(A,B), (B,C)],[(x,y), (z,v)]]

I want this as a python list of lists. How do I do that?

In the end, I would like to be able to iterate through the rows and the columns of this array, and get each pair of adjacent values to compare them.

• mgilson: Or ast.literal_eval depending on what A, B, C, ... look like.
• ATOzTOA: Do you need this as output: [[('A','B'), ('B','C')],[('x','y'), ('z','v')]]? Or do you have values for A,B,C...?
• Bakuriu: Much better to avoid eval. I'd prefer using a parser(maybe even ast.parse, or, as mgilson proposed, ast.literal_eval if those letters represent literals).

Answer 1

More esoteric way of doing it:

import yaml
from string import maketrans
s = "[[(A,B), (B,C)],[(x,y), (z,v)]]" 
yaml.load(s.translate(maketrans("()", "[]")))

out:

[[['A', 'B'], ['B', 'C']], [['x', 'y'], ['z', 'v']]]

Answer 2

This works:

>>> import re,ast
>>> st='[[(A,BC), (B,C)],[(x,y), (z,v)]]'
>>> ast.literal_eval(re.sub(r'(\w+)',r"'1円'",st))
[[('A', 'BC'), ('B', 'C')], [('x', 'y'), ('z', 'v')]]

If you really do want a LoLoL rather than a LoLoT (as above), do this:

def rep(match):
 if match.group(1)=='(': return '['
 if match.group(1)==')': return ']'
 return "'{}'".format(match.group(1))
st='[[(A,B), (B,C)],[(x,y), (z,v)]]'
st=re.sub(r'(\w+|[\(\)])', rep,st)
>>> ast.literal_eval(st)
[[['A', 'B'], ['B', 'C']], [['x', 'y'], ['z', 'v']]]

Answer 3

Once you've read the line from the file:

import ast
parsed_list = ast.literal_eval(line)

Answer 4

Pure python...

s = "[[(A,B), (B,C)],[(x,y), (z,v)]]"
print s
s = filter(None, s[1:-1].replace(",[", "").replace("[", "").replace(" ", "").split(']'))
for i,t in enumerate(s):
 t = filter(None, t.replace(",(", "").replace("(", "").split(')'))
 t = [tuple(x.split(",")) for x in t]
 s[i] = t
print s

Output:

>>> 
[[(A,B), (B,C)],[(x,y), (z,v)]]
[[('A', 'B'), ('B', 'C')], [('x', 'y'), ('z', 'v')]]
>>>