How to Print on the Same Line with Python

Today, I was downloading about 1,600 images and wanted to be able to keep track of how many images I had downloaded so far and what image that I was currently downloading. This is typically very easy with Python, but I decided to print everything on the same line so that I my terminal wouldn't get cluttered with print statements. This is where I encountered my issue.

The format of the string that I was printing started out as

print(f'\rImage\t{count}', end='', flush=True)

where

• \r: is the carriage return which resets the line position to the beginning of the line
• count: the number of the image that I was downloading
• end: the character that print will append to the string that it's printing
• flush: tells print whether or not it should forcibly flush the stream

Basically, what's going to happen is that when the string is printed it will print the string and put the cursor back at the beginning of the line. If this where done in a loop (probably using time.sleep too), then you'd get the effect of the one line counting up. Neat!

import time

for count in range(1000):
    print(f'\rCount\t{count}', end='', flush=True)
    time.sleep(0.01)

We can put this into a function to make it a little easier to read

def print_same_line(msg):
    print('\r', msg, end='', flush=True, sep='')

In the function above, we need to use the sep keyword so that when we use this function a space isn't added between the \r character and our string causing an offset. Now we can use a loop like above and have our code be a bit more readable (we also don't need the \r character at the beginning of our string).

import time

for count in range(1000):
    print_same_line(f'Count\t{count}')
    time.sleep(0.01)

This worked pretty great, but I encountered trouble when I was reading URLs from a file and was printing them to the terminal.

import pathlib
import time
import urllib.request as request

path = pathlib.Path('img')
SLEEP = 0.1 # seconds
urls_file = 'images.txt'

with open(urls_file) as urls:
    for count, url in enumerate(urls):
        msg = '\t'.join(['Image', str(count), url.strip()])
        print_same_line(msg)

        fname = url.strip().split('/')[-1]
        request.urlretrieve(url, path/fname)

        time.sleep(SLEEP)


    print(f'\nDone! Downloaded {count} images')

which resulted in the unexpected output of

Image   1   https://example.com/image-1.jpg
Image   2   https://example.com/image-2.jpg
Image   3   https://example.com/image-3.jpg
Image   4   https://example.com/image-4.jpg
Image   5   https://example.com/image-5.jpg
Image   6   https://example.com/image-6.jpg
...

It took me longer than I'd like to admit to figure out what went wrong. If you look at the line where I grab the file name (fname) from the URL, you'll see that I call the strip method. Why? Each line of the file had a \n character at the end of it. Even though I did it for the file name I did not do it for the printing of the URL. This one little mistake cost me probably 90 to 120 minutes...

Once I fixed it, however, it worked like a charm. Below is the fixed example

import pathlib
import time
import urllib.request as request


path = pathlib.Path('img')
SLEEP = 0.1 # seconds
urls_file = 'images.txt'

with open(urls_file) as urls:
    for count, url in enumerate(urls):
        url = url.strip()
        print_same_line(f'Image\t{count+1}\t{url}')

        fname = url.split('/')[-1]
        request.urlretrieve(url, path/fname)

        time.sleep(SLEEP)


print(f'\nDone! Downloaded {count} images')