Formatting a filename in python

Question

I would like to format a string based on the input. For example if the line count is 1, the output must be image00001.png and if the line count is 400, the output is image00400.png

I use the following to do so

filename = "image%05d.png".format(line_count)

but it doesn't work as expected. What am I missing?

score 0 · Answer 1 · answered Aug 29 '20 at 18:21

You confuse the notation for the substitute operator % and for the .format() method. The right use of the method is filename = "image{:05d}.png".format(line_count). An even better way is to use a format string: filename = f"image{line_count:05d}.png".

score 0 · Answer 2 · answered Aug 29 '20 at 18:22

0

In Python 3 you may use f-strings:

filename = f'image{line_count:05}.png'

answered Aug 29 '20 at 18:22

Franz Diebold

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score 0 · Accepted Answer · answered Aug 29 '20 at 18:22

0

You are mixing different formatting specifications. For .format, use Format string syntax.

line_count = 1
filename = "image{:05d}.png".format(line_count)

or the new f-string method

line_count = 1
filename = f"image{line_count:05d}.png"

answered Aug 29 '20 at 18:22

tdelaney

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score -1 · Answer 4 · edited Aug 29 '20 at 18:37

-1

fname = "image" + str(400).zfill(5) + ".png"
print(fname)

edited Aug 29 '20 at 18:37

sushanth

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answered Aug 29 '20 at 18:24

sambit1991

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The number to zfill tells how many characters wide, the unused MSB characters are filled with zeros – sambit1991 Aug 29 '20 at 18:27

Formatting a filename in python

4 Answers4