Generate special list from file in python

Question

I have a file like this :

one:two:three
four:five:six
seven:height:nine

And so on... What I want is to parse it correctly to obtain this kind of variable:

myVar = [("one", "two", "three"), ("four", "five", "six"), ("seven", "height", "nine")]

Of course, the file isn't stopping at nine, there's a lot more of lines after that.

How can I do that in Python ?

Thanks !

Ashwini Chaudhary · Accepted Answer · 2014-04-07 10:23:56Z

2

Use a list compehension:

with open('filename') as f:
    myVar = [line.rstrip().split(':') for line in f]

If you need a list to tuples then pass line.rstrip().split(':') to tuple():

tuple(line.rstrip().split(':'))

answered Apr 7, 2014 at 10:23

Ashwini Chaudhary

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1 Comment

Worked great thanks ! I was going waaay to complicated for this. Python is such a great language.

thefourtheye · Accepted Answer · 2014-04-07 10:25:04Z

2

The data you are dealing with, looks like delimiter separated. I would recommend using csv.reader, like this

import csv
with open("Input.txt") as in_file:
    reader = csv.reader(in_file, delimiter=':')
    print [row for row in reader]

You can convert this to a tuple, like this

    print [tuple(row) for row in reader]

answered Apr 7, 2014 at 10:25

thefourtheye

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vaultah · Accepted Answer · 2014-04-07 10:36:42Z

2

with open('your file') as f:
    myVar = [ tuple(a.split(':')) for a in f.read().split() ]

print myVar

Outputs

[('one', 'two', 'three'), ('four', 'five', 'six'), ('seven', 'height', 'nine')]

answered Apr 7, 2014 at 10:23

vaultah

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