How to sort records in file

[Lad]

What is the best( easiest)way how to sort a file?
I have a file where each record consists of 3 fields( 3 words) and I
would like to sort records by the first field( word)in each record.
Any idea?
Thanks for help
Lad

 

[Grant Edwards]

What is the best( easiest)way how to sort a file?

$ man sort

I have a file where each record consists of 3 fields( 3 words)
and I would like to sort records by the first field( word)in
each record.

lines = file('myfilename').readlines()
lines.sort()
for l in lines:
print l,

 

[Peter Otten]

$ man sort




lines = file('myfilename').readlines()
lines.sort()
for l in lines:
print l,

Or

lines = file('myfilename').readlines()
lines = [(line.split(None, 1)[0], i, line)
for (i, line) in enumerate(lines)]
lines.sort()
lines = [line for (_, _, line) in lines]
for l in lines:
print l,

if you want to keep the order stable.

Peter

 

[Paul Rubin]

What is the best( easiest)way how to sort a file?
I have a file where each record consists of 3 fields( 3 words) and I
would like to sort records by the first field( word)in each record.

If you're using Linux, the simplest way is with the "sort" command:

sort infile -o outfile

This will sort files of basically unlimited size, as long as you have
enough disk space. The files don't have to fit in memory.

Since you're posting on c.l.py, maybe you're really asking how to read
the file in sorted order in a Python program:

import os
sorted = os.popen("sort infile")

runs the sort command in an external process.

Finally, there's a built-in sorting operation on lists:

def compare(a, b):
def key(line):
return line.split()[0]
return cmp(key(a), key(b))
lines = open("infile").readlines()
lines.sort(compare)

but that requires reading the whole file into memory, etc.