python human readable large numbers [duplicate]

As I understand it, you only want the ‘most significant’ part. To do so, use floor(log10(abs(n))) to get number of digits and then go from there. Something like this, maybe:

import math

millnames = ['',' Thousand',' Million',' Billion',' Trillion']

def millify(n):
    n = float(n)
    millidx = max(0,min(len(millnames)-1,
                        int(math.floor(0 if n == 0 else math.log10(abs(n))/3))))

    return '{:.0f}{}'.format(n / 10**(3 * millidx), millnames[millidx])

Running the above function for a bunch of different numbers:

for n in (1.23456789 * 10**r for r in range(-2, 19, 1)):
    print('%20.1f: %20s' % (n,millify(n)))
                 0.0:                    0
                 0.1:                    0
                 1.2:                    1
                12.3:                   12
               123.5:                  123
              1234.6:           1 Thousand
             12345.7:          12 Thousand
            123456.8:         123 Thousand
           1234567.9:            1 Million
          12345678.9:           12 Million
         123456789.0:          123 Million
        1234567890.0:            1 Billion
       12345678900.0:           12 Billion
      123456789000.0:          123 Billion
     1234567890000.0:           1 Trillion
    12345678900000.0:          12 Trillion
   123456789000000.0:         123 Trillion
  1234567890000000.0:        1235 Trillion
 12345678899999998.0:       12346 Trillion
123456788999999984.0:      123457 Trillion
1234567890000000000.0:     1234568 Trillion