How to pass file name at run time in python

Question

I have a directory which contains 4 .csv files at the moment. I am able retrieve their names using os lib in the code given below:

import os
fileNames = os.listdir(path)

for f in fileNames:
        print(f)

Now I want to pass the file names one by one in my open file command and do the related processing.

how do I pass file name in my command:

file = open(r'C:\Users\hu170f\Documents\TEST1\<filename to be passed>')

You can pass the name as a variable and then attach it to the string there — Manan Adhvaryu
– Manan Adhvaryu, Commented Mar 21, 2022 at 10:25
Have you tried file = open(r'C:\Users\hu170f\Documents\TEST1\'+f)? — Kristian
– Kristian, Commented Mar 21, 2022 at 10:26

balu · Accepted Answer · 2022-03-21 10:47:29Z

1

import os
dir_path = r'D:\text_dirs\\'
fileNames = os.listdir(dir_path)

for fn in fileNames:
    f = open(dir_path+fn, "r")
    print(f.read())

edited Mar 21, 2022 at 10:47

answered Mar 21, 2022 at 10:26

balu

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2 Comments

shikha Over a year ago

If I just have to print the file path along with file the name it is working fine. May be I haven't put up my question correctly. under the for loop I want to open the file one by one using command : file = open(r'<entire filepath>').

balu Over a year ago

@shikha check now . I have opened a dir of txt files and am able to print.

Don CorliJoni · Accepted Answer · 2022-03-21 11:50:05Z

0

import os
path = r"C:\..."
filenames = os.listdir(path)

You can use the map function like a for loop, which only opens one method(The first parameter) and passes one after one element of this list(second parameter) to the method. The lambda function is like a method which executes the open function and returns the _io.TextIOWrapper. files contains a list of _io.TextIOWrapper.

files = list(map(lambda filename: open(f"{path}\{filename}"), filenames))

answered Mar 21, 2022 at 11:50

Don CorliJoni

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