How to open multiple files in loop, in python

Question

I'm new to python. I'm want to open multiple files in Python. I'm can open each of them with open() function. I'm not sure about formatting.

with open("/test/filename.css", "r") as f:
     s = f.readlines()
     print(s)

I'm can open one file, but I'm not sure how to open multiple files. This is the code that I have. In live_filename() function have many files.

inputfiles = live_filename()
    for live in inputfiles:
        with open("/test/............. .css, "r") as f:

I don't know how to put code formatting in space. and I think the live variable is a tuple can't concatenate str. what should i do?

Please edit your question to include a sample input by printing the return value of live_filename(). — Selcuk
– Selcuk, Commented Sep 26, 2019 at 3:00
@lalin Edit your question to show us the result of print(inputfiles) — OverLordGoldDragon
– OverLordGoldDragon, Commented Sep 26, 2019 at 3:06
Possible duplicate of How can I open multiple files using "with open" in Python? — Ramesh
– Ramesh, Commented Sep 26, 2019 at 3:09

OverLordGoldDragon · Accepted Answer · 2019-09-26 03:24:33Z

5

Open each just as you did for one, and then append them to a list:

import os

folderpath = r"D:\my_data" # make sure to put the 'r' in front
filepaths  = [os.path.join(folderpath, name) for name in os.listdir(folderpath)]
all_files = []

for path in filepaths:
    with open(path, 'r') as f:
        file = f.readlines()
        all_files.append(file)

Now, all_files[0] holds the first file loaded, all_files[1] the second, and so on.

UPDATE: for all files in the same folder: first, get the folder path (on Windows, like this). Suppose it's "D:\my_data". Then, you can get all the filepaths of the files as in the script above.

edited Sep 26, 2019 at 3:24

answered Sep 26, 2019 at 3:04

OverLordGoldDragon

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8 Comments

user12117808 Over a year ago

I got it. I make it in function live_filename(). Do I have to do it again?

OverLordGoldDragon Over a year ago

@lalin Just make sure filepaths contains all paths of files you plan to load - you can check via print(filepaths). If so, no need to repeat.

user12117808 Over a year ago

I got it. I will try this. :)

user12117808 Over a year ago

I try this.but error show ==> expected str, bytes or os.PathLike object, not tuple

OverLordGoldDragon Over a year ago

@lalin If you didn't change your live_filename() and it has the same output, then that is the problem. Just don't use it - instead follow my answer instructions step-by-step.

|

MjZac · Accepted Answer · 2019-09-26 04:11:11Z

1

You can do like this:

folder = "..." # Absolute path to folder in which the files reside
files_to_read = [("filename.css","FileName"),( "filename2.css","Filename2")]
for (file, _) in files_to_read:
    filepath = os.path.join(folder, file)
    with open(filepath, "r") as f:
        s = f.readlines()
        print(s)

edited Sep 26, 2019 at 4:11

answered Sep 26, 2019 at 3:03

MjZac

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6 Comments

user12117808 Over a year ago

I got it, but if I have 100 files. what should i do?

OverLordGoldDragon Over a year ago

@lalin Are all these files in the same folder?

user12117808 Over a year ago

@OverLordGoldDragon Yes in the same folder.

MjZac Over a year ago

@lalin Do you want it to be processed in parallel? You are receiving file paths from a function right? Can you show a sample output of live_filename()

user12117808 Over a year ago

@MjZac Yes I receiving file paths from a function. output is : ('home', 'Home') ('fun', 'Fun')

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