How to execute python program and save output in different file names automatically

Question

I have a Python program which processes images and save the resulting output to a file. I am receiving the input and output file names via console using input(). But I want to save the output automatically with the file name as something like "input_file_name_out". Note the string "_out" attached to the input file name

This will help me get user input only for Input file names thereby saving little time for the user since he doesn't need to think everytime what should be the output file name

Sample code snippet

if __name__ == "__main__":

    in_file = input()
    out_file = input()

    processed = process_image(in_file,out_file)

Expected code snippet

if __name__ == "__main__":

    in_file = input()
    # out_file = input() ------Not getting the file name from user thereby supplying only one argument to the below function

        processed = process_image(in_file)

# And within the function
def process_image(img):
    .
    .
    .
    out_file = img+"_out.jpg" ##### What should come here to achieve my requirement ########
    cv2.imwrite(out_file,processed_image)

FObersteiner · Accepted Answer · 2020-02-17 06:59:04Z

2

you could make use of the Path class from the pathlib module:

from pathlib import Path

in_file = Path(input())
insert = '_out'

out_file = in_file.parent / (in_file.stem + insert + in_file.suffix)

Example:

in_file = Path('D:/folder/test.jpg')
insert = '_out'
out_file = in_file.parent / (in_file.stem + insert + in_file.suffix)

print(out_file)
D:\folder\test_out.jpg

answered Feb 17, 2020 at 6:59

FObersteiner

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3 Comments

JKC Over a year ago

while @Redowan's answer is the simplest I have to acknowledge that this is the technically correct answer

FObersteiner Over a year ago

@JKC: as in most cases, more generally applicable solutions come with a bit of overhead ;-) If you're not familiar with pathlib yet, make sure to check it out. For specific cases or quick hacks, an approach like Redowan suggested is also fine I guess.

rednafi Over a year ago

Yeah, this approach generalizes better!!

rednafi · Accepted Answer · 2020-02-17 10:34:10Z

1

All you have to do is,

Take the input_img.jpg that you get from user input
Split the input_img.jpg with delimiter . and that will give you this list: ['input_img', 'jpg']
Pick up the first element and add _out.jpg suffix

def process_image(img):

    # you should split with delimiter ('.'), take the first 
    # element of the list and add "_out.jpg" suffix

    split_name = img.split('.')
    out_file = split_name[0] + "_out" + "." + split_name[1] 
    cv2.imwrite(out_file, processed_image)


if __name__ == "__main__":

    in_file = input()
    processed = process_image(in_file)

edited Feb 17, 2020 at 10:34

answered Feb 17, 2020 at 6:43

rednafi

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4 Comments

JKC Over a year ago

Thanks for your prompt response. This helps

rednafi Over a year ago

The OP asked specifically for .jpg case. Multiple things can be done if the extension is different. One way can be hardcoding .jpg, .png etc. Another way can be asking the user about extension and applying that. Numerous things can be done.

FObersteiner Over a year ago

what I'm trying to point out is that hardcoding '.jpg' makes the code unflexible / very specific. Since you already have the result of .split('.'), why not at least append the original extension?

rednafi Over a year ago

Oh sorry! yeah that's a better approach. My bad! I didn't get that from your earlier comment.

tst · Accepted Answer · 2020-02-17 09:03:50Z

-1

Well that's simple! Just take the "in_file" string you asked for, and concat "_out" to the end and pass it as the out_file.

If you only process ".jpg"

cv2.imwrite(in_file[:-4]+"_out.jpg",processed_image)

if its variable:

cv2.imwrite(in_file[:-4]+"_out"+in_file[-4:],processed_image)

edited Feb 17, 2020 at 9:03

answered Feb 17, 2020 at 6:25

tst

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3 Comments

JKC Over a year ago

But the problem here is the in_file ends with extension as input_file.jpg. So my output file name should be input_file_out.jpg. So I need to first split the input file name and then concat _out and extension

tst Over a year ago

Right. Sorry... updated the answer. You can splice the last 4 characters from the input and concatinate them back, either manually or automatically.

FObersteiner Over a year ago

hardcoding the index -4 is pretty unflexible; what if the file extension has 4 characters, e.g. 'jpeg'?

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