Python String Formatting: Padding Negative Numbers

I would like to format my integers as strings so that, without the sign, they will be zero-padded to have at least two digits. For example I want 1 -1 10 -10 to be 01 -01 10 -10

Solution 1:

you could use str.format, but adding 1 to the size to take the negative number into account here:

l = [1,-1,10,-10,4]

new_l = ["{1:0{0}d}".format(2if x>=0else3,x) for x in l]

print(new_l)

result:

['01', '-01', '10', '-10', '04']

it works because format accepts nested expressions: you can pass the size ({:02d} or {:03d}) as a format item too when saves the hassle of formatting the format string in a first pass.

Solution 2:

According to here, you need a space before the type descriptor, so both

'% d'%(1)

and

'{: d}'.format(1)

result in (notice the space)

' 1'

aligning nicely with the result of the above called with -1:

'-1'

Solution 3:

The most concise, although maybe a bit unreadable, way that I can think of (at least in Python 3) would be a formatted string literal with nested curly braces, similar to the accepted answer. With this, the example from the accepted answer could be written as

l = [1, -1, 10, -10, 4]
new_l = [f"{x:0{2 + (x < 0)}d}" for x in l]
print(new_l)

with the same result:

['01', '-01', '10', '-10', '04']

Solution 4:

Use "{:03d}".format(n). The 0 means leading zeros, the 3 is the field width. May not be exactly what you want:

>>>for n in (-123,-12,-1,0,1,12,123, 1234):...print( '{:03d}'.format(n) )... 
-123
-12
-01
000
001
012
123
1234

Solution 5:

zfill actually pads with zeros at the left and considers negative numbers.

So n.zfill(3 if n < 0 else 2) would do the trick.