Python: How do I format a number with a variable number of digits?

Question

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: '00123'.

If I wanted to display it in 6 digits I would have digits = 6 giving: '000123'.

How would I do this in Python?

Answer 1

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).

Answer 2

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

Answer 3

print "%03d" % (43)

Prints

043

Answer 4

Use string formatting

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

More info here: link text

Answer 5

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.'

See
http://docs.python.org/library/stdtypes.html#str.format
http://docs.python.org/library/string.html#formatstrings

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'

Answer 6

count=5
num='123'
'0'*(count-len(num))+num