python strip函数没有给出预期的输出

Question

i have below code in which filenames are FR1.1.csv, FR2.0.csv etc. I am using these names to print in header row but i want to modify these name to FR1.1 , Fr2.0 and so on. Hence i am using strip function to remove .csv. when i have tried it at command prompt its working fine. But when i have added it to main script its not giving output.

for fname in filenames:
    print "fname     : ", fname
    fname.strip('.csv');
    print "after strip fname: ", fname
    headerline.append(fname+' Compile');
    headerline.append(fname+' Run');

output i am getting

fname     :FR1.1.csv
after strip fname: FR1.1.csv

required output-->

fname     :FR1.1.csv
after strip fname: FR1.1

i guess some indentation problem is there in my code after for loop. plesae tell me what is the correct way to achive this.

Answer 1

Strings are immutable, so string methods can't change the original string, they return a new one which you need to assign again:

fname = fname.strip('.csv')   # no semicolons in Python!

But this call doesn't do what you probably expect it to. It will remove all the leading and trailing characters c , s , v and . from your string:

>>> "cross.csv".strip(".csv")
'ro'

So you probably want to do

import re
fname = re.sub(r"\.csv$", "", fname)

Answer 2

Strings are immutable. strip() returns a new string.

>>> "FR1.1.csv".strip('.csv')
'FR1.1'
>>> m = "FR1.1.csv".strip('.csv')
>>> print(m)
FR1.1

You need to do fname = fname.strip('.csv') .

And get rid of the semicolons in the end!

PS - Please see Jon Clement's comment and Tim Pietzcker's answer to know why this code should not be used.

Answer 3

You probably should use os.path for path manipulations:

import os

#...

for fname in filenames:
    print "fname     : ", fname
    fname = os.path.splitext(fname)[0]
    #...

The particular reason why your code fails is provided in other answers.

Answer 4

change

fname.strip('.csv')

with

fname = fname.strip('.csv')