[Tutor] Re: How can I format rows of irregular sized strings to print in columns?

Bob Gailer bgailer at alum.rpi.edu
Thu May 6 21:43:12 EDT 2004

At 04:31 PM 5/6/2004, Lee Harr wrote:
>>I wrote a little program to look for graphics files and tell me the image
>>size and the file size of each file. The data prints out like this:
>>(292, 240) 35638 defender.bmp
>>(1024, 768) 2359350 evolution3.bmp
>>(78, 76) 17990 GRID1A.bmp
>>How can I make it so that it looks more like this:
>>( 292,  240)    35,638  defender.bmp
>>(1024,  768) 2,359,350  evolution3.bmp
>>(  78,   76)    17,990  GRID1A.bmp
>Please post only in plain text.
>You will want to look at the string formatting operator:
>One handy example:
>>>>a='(292, 240)';b=35638; c='defender.bmp'
>>>>print '%10s %10s %20s' % (a, b, c)
>(292, 240)      35638         defender.bmp

This solution is close, but I think the OP wants the numbers in () to be 
individually aligned in columns. The desired output in proportional font 
doesn't show this, but if you copy/paste into a fixed font editor you will 
see this more clearly.

So the challenge is to parse the numbers out and format them individually. 
I think this is best done using re:

import re
m = re.match(r'\((\d+),\s*(\d+)\)', im.size) # generates a match object 
with 2 groups
# m.groups() is ('292', '240')
'(%4s, %4s)' % m.groups() # is '( 292,  240)'

re can be daunting at first but is well worth the learning curve. So let''s 
break the regular expression \((\d+),\s*(\d+)\) into pieces:
\(      # match a left parentheses
\s*     # match 0 or more "whitespace" characters
(\d+)   # match one or more digits and place them in a group
,       # match a comma
\s*     # match 0 or more "whitespace" characters
(\d+)   # match one or more digits and place them in another group
\)      # match a right parenthesis
The result is placed in a "match object". m.groups() returns all the groups 
in a tuple, which is the desired right argument type for %

Bob Gailer
bgailer at alum.rpi.edu
303 442 2625 home
720 938 2625 cell 
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