Last two digits of decimal number with a mask?

Question

To get the last N digits of a decimal number I usually convert it to a string. For example, for two digits:

n = 129387
int((str(n)[-2:]))
# 87

With binary its relatively straightforward to do this with a mask, such as:

n = 0b00101001
'{:02b}'.format(0b00101001& 0b11)
# '01'
# -- or -- 
'{:02b}'.format(0b00101001& 0x3)
# '01'

Is it possible to apply a mask to a deicimal number to get the last two digits, 87 in the above example?

I think you are referring to the mod operator, e.g. `129387%100 = 87`? The general case is `base**n` where `n` is the number of digits in the base, e.g. binary case `format(0b00101001 % 2**2, '02b')` — AChampion, Aug 15 '20 at 20:55
Unless you convert to string, you **don't have** a "decimal number". — superb rain, Aug 15 '20 at 20:55
@AChampion yes exactly -- that was much easier than I though it would be! — samuelbrody1249, Aug 15 '20 at 20:59

AChampion · Accepted Answer · 2020-08-15T21:07:25.030

2

You are referring to the mod operator, e.g.:

In []:
129387 % 100  # 10**2

Out[]:
87

The general case is base**n where n is the number of digits in the base, e.g. binary case:

In []:
format(0b00101001 % 2**2, '02b')

Out[]:
'01'

In []:
format(0b00101001 % 2**4, '02b')

Out[]:
'1001'

edited Aug 15 '20 at 21:07

answered Aug 15 '20 at 21:01

AChampion

1 Answers1