How do I format a number with a variable number of digits in Python? [duplicate]

This question already has an answer here:

Say I wanted to display the number 123 with a variable number of padded zeroes on the front.

For example, if I wanted to display it in 5 digits I would have digits = 5 giving me:

If I wanted to display it in 6 digits I would have digits = 6 giving:

How would I do this in Python?

There is a string method called zfill:

>>> '12344'.zfill(10)
0000012344

It will pad the left side of the string with zeros to make the string length N (10 in this case).

If you are using it in a formatted string with the format() method which is preferred over the older style ''% formatting

>>> 'One hundred and twenty three with three leading zeros {0:06}.'.format(123)
'One hundred and twenty three with three leading zeros 000123.'

Here is an example with variable width

>>> '{num:0{width}}'.format(num=123, width=6)
'000123'

You can even specify the fill char as a variable

>>> '{num:{fill}{width}}'.format(num=123, fill='0', width=6)
'000123'

'%0*d' % (5, 123)

With the introduction of formatted string literals ("f-strings" for short) in Python 3.6, it is now possible to access previously defined variables with a briefer syntax:

>>> name = "Fred"
>>> f"He said his name is {name}."
'He said his name is Fred.'

The examples given by John La Rooy can be written as

In [1]: num=123
   ...: fill='0'
   ...: width=6
   ...: f'{num:{fill}{width}}'

Out[1]: '000123'

print "%03d" % (43)

Prints

043

Use string formatting

print '%(#)03d' % {'#': 2}
002
print '%(#)06d' % {'#': 123}
000123

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来源：https://stackoverflow.com/questions/3228865/how-do-i-format-a-number-with-a-variable-number-of-digits-in-python

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