python human readable large numbers [duplicate]

Answer 1

That number seems pretty human-readable to me. An unfriendly number would be 187289840422780.00. To add commas, you could create your own function or search for one (I found this):

import re

def comma_me(amount):
    orig = amount
    new = re.sub("^(-?\d+)(\d{3})", '\g<1>,\g<2>', amount)
    if orig == new:
        return new
    else:
        return comma_me(new)

f = 12345678
print comma_me(`f`)
Output: 12,345,678

If you want to round a number to make it more readable, there is a python function for that: round().

You could move even further away from the actual data and say "A very high amount" or "Above 100 trillion" using a function that would return a different value based on your programmed benchmarks.

Answer 2

If by 'readable' you mean 'words'; here's a good solution that you can adapt.

http://www.andrew-hoyer.com/experiments/numbers

Answer 3

(Edit: Can't believe I am updating my answer in 2019 and again in 2020! This proves that the next phrase of human evolution is to share knowledge over the internet!)

With Python2.7+ or v3 you just use the "," option in your string formatting:

>>> "{:,}".format(100000000)
'100,000,000'

See PEP 378: Format Specifier for Thousands Separator for more info:

http://www.python.org/dev/peps/pep-0378/

With Python3.6+ you can also use the "_" format:

>>> "{:_}".format(100000000)
'100_000_000'

See PEP 515 for details:

https://www.python.org/dev/peps/pep-0515/

Added in 2014: These days I have the following shell function:

human_readable_numbers () {
    python2.7 -c "print('{:,}').format($1)"
}

So I don't have to squint my eyes on big numbers.

Stop here unless you are unfortunate enough to be working with pre-2.7 code or environment without Python 2.7 when you read this...

You can also do that with locale:

import locale
locale.setlocale(locale.LC_ALL, 'en_US')
locale.format('%d', 2**32, grouping=True)   # returns '4,294,967,296'

Hope this helps!

Answer 4

From here:

def commify(n):
    if n is None: return None
    if type(n) is StringType:
        sepdec = localenv['mon_decimal_point']
    else:
        #if n is python float number we use everytime the dot
        sepdec = '.'
    n = str(n)
    if sepdec in n:
        dollars, cents = n.split(sepdec)
    else:
        dollars, cents = n, None

    r = []
    for i, c in enumerate(reversed(str(dollars))):
        if i and (not (i % 3)):
            r.insert(0, localenv['mon_thousands_sep'])
        r.insert(0, c)
    out = ''.join(r)
    if cents:
        out += localenv['mon_decimal_point'] + cents
    return out

Answer 5

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