Dynamically insert file name and path for saving csv in pandas

Question

I need to dynamically save my pandas data frame. I have successfully managed to output a CSV file with a static name using the following code:

export_csv = df.to_csv(r'static_name.csv', header=False , index=False)

But I have failed to make this work dynamically. With the code below, I expect to get a file with the name passed into args.save_file_name and .csv suffix. However, I get no result.

import os
import argparse
import pandas as pd

parser = argparse.ArgumentParser()
parser.add_argument('file', help="this is the file you want to open")
parser.add_argument('save_file_name', help="the name of the file you want for the output CSV")

args = parser.parse_args()
print("file name:", args.file) # checking that this worked

...

# export csv
path = os.getcwd()
export_path = path + args.save_file_name + '.csv'
export_csv = df.to_csv(path_or_buf=export_path, header=False, index=False)

probably the file is created in this location os.getcwd() try printing the path and check if file exists — Rakesh
– Rakesh, Commented Aug 23, 2019 at 11:18
@MorrisGevirtz, please, check the answer, and do not forget to accept it in case it is helpful, or add an comment, if it is not. — Jaroslav Bezděk
– Jaroslav Bezděk, Commented Aug 30, 2019 at 8:25

Jaroslav Bezděk · Accepted Answer · 2019-08-23 12:54:32Z

1

I think, that problem is in your export_path variable, that is not made right. The following code should do the job.

export_path = os.path.join(path, args.save_file_name + '.csv')

answered Aug 23, 2019 at 12:54

Jaroslav Bezděk

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