MRAB wrote:
> [snip]
> Simpler would be:
>
> lines = f.readlines()
> lines.sort(key=lambda line: line[ : 3])
>
> or even:
>
> lines = sorted(f.readlines(), key=lambda line: line[ : 3]))
Sure, but a complete newbie (I have this impression about OP) doesn't
have to know about lambda.
I expected my solution to be slower, but it's not (on a file with
100,000 random string lines):
# time ./sort1.py
real 0m0.386s
user 0m0.372s
sys 0m0.014s
# time ./sort2.py
real 0m0.303s
user 0m0.286s
sys 0m0.017s
sort1.py:
#!/usr/bin/python
def sortit(fname):
lines = open(fname).readlines()
lines.sort(key = lambda x: x[:3])
if __name__ == '__main__':
sortit('testfile.txt')
sort2.py:
#!/usr/bin/python
def sortit(fname):
fo = open(fname)
linedict = {}
for line in fo:
key = line[:3]
linedict[key] = line
sortedlines = []
keys = linedict.keys()
keys.sort()
for key in keys:
sortedlines.append(linedict[key])
return sortedlines
if __name__ == '__main__':
sortit('testfile.txt')
Any idea why? After all, I'm "manually" doing quite a lot: allocating
key in a dict, then sorting dict's keys, then iterating over keys and
accessing dict value.
Regards,
mk