Emile van Sebille wrote:
On 3/1/2011 11:49 AM Sean Carolan said...
My advice would be to go read up on the zip() function and the
str.join() function. Then, if you are using python 2.x, go find
itertools.izip. It does the same thing as zip but it's more memory
efficient. With those two you can do it in about two lines or so (and
maybe a few for set up and clarity and such).
This is what I've got so far:
import sys
myfiles = sys.argv[1:]
for i in zip(open(myfiles[0]), open(myfiles[1]), open(myfiles[2])):
print " ".join(i)
How would you:
1. zip an arbitrary number of files in this manner? I hard-coded it
to do only three.
One way:
for i in zip([ open(filename) for filename in myfiles ])
Almost, you need to expand the list:
zip( *[open(filename) for filename in myfiles] )
which is equivalent to pseudocode:
zip(open(myfiles[0]), open(myfiles[1]), ..., open(myfiles[N]))
Another way is:
zip(*map(open, myfiles))
Again, you need to expand the list using *.
2. Strip out trailing spaces and line breaks from the lines in each
file?
Convert the file contents before zip'ing -- so add
def cleanedup(filename):
return [ line.strip() for line in open(filename) ]
Better to do the stripping on demand, rather than all up front.
Especially if the files are huge. Turn the list comprehension [...] into
a generator expression (...):
def cleanedup(filename):
return (line.strip() for line in open(filename))
Then your loop looks like:
for i in zip([ cleanedup(filename) for filename in myfiles ])
--
Steven
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