Removing Trailing Zeros in Python [duplicate]

Question

I need to find a way to convert the following strings in python:

0.000       => 0 0           => 0 123.45000   => 123.45 0000        => 0 123.4506780 => 123.450678 

and so forth. I tried .rstrip('0').rstrip('.'), but that doesn't work if the input is 0 or 00.

Any ideas? Thanks!

Answer 1

Updated Generalized to maintain precision and handle unseen values:

import decimal import random  def format_number(num):     try:         dec = decimal.Decimal(num)     except:         return 'bad'     tup = dec.as_tuple()     delta = len(tup.digits) + tup.exponent     digits = ''.join(str(d) for d in tup.digits)     if delta <= 0:         zeros = abs(tup.exponent) - len(tup.digits)         val = '0.' + ('0'*zeros) + digits     else:         val = digits[:delta] + ('0'*tup.exponent) + '.' + digits[delta:]     val = val.rstrip('0')     if val[-1] == '.':         val = val[:-1]     if tup.sign:         return '-' + val     return val  # test data NUMS = '''     0.0000      0     0           0     123.45000   123.45     0000        0     123.4506780 123.450678     0.1         0.1     0.001       0.001     0.005000    0.005     .1234       0.1234     1.23e1      12.3     -123.456    -123.456     4.98e10     49800000000     4.9815135   4.9815135     4e30        4000000000000000000000000000000     -0.0000000000004 -0.0000000000004     -.4e-12     -0.0000000000004     -0.11112    -0.11112     1.3.4.5     bad     -1.2.3      bad '''  for num, exp in [s.split() for s in NUMS.split('\n') if s]:     res = format_number(num)     print res     assert exp == res 

Output:

0 0 123.45 0 123.450678 0.1 0.001 0.005 0.1234 12.3 -123.456 49800000000 4.9815135 4000000000000000000000000000000 -0.0000000000004 -0.0000000000004 -0.11112 bad bad