On 21/07/2012 20:08, Lipska the Kat wrote:
Greetings Pythoners
A short while back I posted a message that described a task I had set
myself. I wanted to implement the following bash shell script in Python
Here's the script
sort -nr $1 | head -${2:-10}
this script takes a filename and an optional number of lines to display
and sorts the lines in numerical order, printing them to standard out.
if no optional number of lines are input the script prints 10 lines
Here's the file.
50 Parrots
12 Storage Jars
6 Lemon Currys
2 Pythons
14 Spam Fritters
23 Flying Circuses
1 Meaning Of Life
123 Holy Grails
76 Secret Policemans Balls
8 Something Completely Differents
12 Lives of Brian
49 Spatulas
... and here's my very first attempt at a Python program
I'd be interested to know what you think, you can't hurt my feelings
just be brutal (but fair). There is very little error checking as you
can see and I'm sure you can crash the program easily.
'Better' implementations most welcome
#! /usr/bin/env python3.2
import fileinput
from sys import argv
from operator import itemgetter
l=[]
t = tuple
What's the purpose of this line?
filename=argv[1]
lineCount=10
with fileinput.input(files=(filename)) as f:
for line in f:
t=(line.split('\t'))
t[0]=int(t[0])
l.append(t)
l=sorted(l, key=itemgetter(0))
Short is:
l.sort(key=itemgetter(0))
try:
inCount = int(argv[2])
lineCount = inCount
You may as well say:
lineCount = int(argv[2])
except IndexError:
#just catch the error and continue
None
The do-nothing statement is:
pass
for c in range(lineCount):
t=l[c]
If there are fewer than 'lineCount' lines, this will raise IndexError.
You could do this instead:
for t in l[ : lineCount]:
print(t[0], t[1], sep='\t', end='')
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