Python Decimals format

Question

WHat is a good way to format a python decimal like this way?

1.00 --> '1'
1.20 --> '1.2'
1.23 --> '1.23'
1.234 --> '1.23'
1.2345 --> '1.23'

Answer 1

If you have Python 2.6 or newer, use format:

'{0:.3g}'.format(num) 

For Python 2.5 or older:

'%.3g'%(num) 

Explanation:

{0}tells format to print the first argument -- in this case, num.

Everything after the colon (:) specifies the format_spec.

.3 sets the precision to 3.

g removes insignificant zeros. See http://en.wikipedia.org/wiki/Printf#fprintf

For example:

tests=[(1.00, '1'),        (1.2, '1.2'),        (1.23, '1.23'),        (1.234, '1.23'),        (1.2345, '1.23')]  for num, answer in tests:     result = '{0:.3g}'.format(num)     if result != answer:         print('Error: {0} --> {1} != {2}'.format(num, result, answer))         exit()     else:         print('{0} --> {1}'.format(num,result)) 

yields

1.0 --> 1 1.2 --> 1.2 1.23 --> 1.23 1.234 --> 1.23 1.2345 --> 1.23 

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Using Python 3.6 or newer, you could use f-strings:

In [40]: num = 1.234; f'{num:.3g}' Out[40]: '1.23' 

Answer 2

Only first part of Justin's answer is correct. Using "%.3g" will not work for all cases as .3 is not the precision, but total number of digits. Try it for numbers like 1000.123 and it breaks.

So, I would use what Justin is suggesting:

>>> ('%.4f' % 12340.123456).rstrip('0').rstrip('.') '12340.1235' >>> ('%.4f' % -400).rstrip('0').rstrip('.') '-400' >>> ('%.4f' % 0).rstrip('0').rstrip('.') '0' >>> ('%.4f' % .1).rstrip('0').rstrip('.') '0.1'