dropping trailing '.0' from floats

Question

I\'m looking for a way to convert numbers to string format, dropping any redundant \'.0\'

The input data is a mix of floats and strings. Desired output:

0

Answer 1

If you only care about 1 decimal place of precision (as in your examples), you can just do:

("%.1f" % i).replace(".0", "")

This will convert the number to a string with 1 decimal place and then remove it if it is a zero:

>>> ("%.1f" % 0).replace(".0", "")
'0'
>>> ("%.1f" % 0.0).replace(".0", "")
'0'
>>> ("%.1f" % 0.1).replace(".0", "")
'0.1'
>>> ("%.1f" % 1.0).replace(".0", "")
'1'
>>> ("%.1f" % 3000.0).replace(".0", "")
'3000'
>>> ("%.1f" % 1.0000001).replace(".0", "")
'1'

Answer 2

FWIW, with Jinja2 where var = 10.3

{{ var|round|int }} will emit integer 10

round(method='floor') and round(method='ceil') are also available.

So that were var2 = 10.9 {{ var|round(method='floor')|int }} will still emit integer 10

Precision can also be controlled using a keyword argument precision=0 of the round function.

ref: http://jinja.pocoo.org/docs/dev/templates/#round

Answer 3

Us the 0 prcision and add a period if you want one. EG "%.0f."

>>> print "%.0f."%1.0
1.
>>> 

Answer 4

See PEP 3101:

'g' - General format. This prints the number as a fixed-point
      number, unless the number is too large, in which case
      it switches to 'e' exponent notation.

Old style (not preferred):

>>> "%g" % float(10)
'10'

New style:

>>> '{0:g}'.format(float(21))
'21'

New style 3.6+:

>>> f'{float(21):g}'
'21'

Answer 5

rstrip doesn't do what you want it to do, it strips any of the characters you give it and not a suffix:

>>> '30000.0'.rstrip('.0')
'3'

Actually, just '%g' % i will do what you want. EDIT: as Robert pointed out in his comment this won't work for large numbers since it uses the default precision of %g which is 6 significant digits.

Since str(i) uses 12 significant digits, I think this will work:

>>> numbers = [ 0.0, 1.0, 0.1, 123456.7 ]
>>> ['%.12g' % n for n in numbers]
['1', '0', '0.1', '123456.7']

Answer 6

(str(i)[-2:] == '.0' and str(i)[:-2] or str(i) for i in ...)