dropping trailing '.0' from floats

Using Python’s string formatting (use str.format() with Python 3.0):

from decimal import Decimal

def format_number(i):
    return '%g' % (Decimal(str(i)))

def floatstrip(x):
    if x == int(x):
        return str(int(x))
    else:
        return str(x)

Be aware, though, that Python represents 0.1 as an imprecise float, on my system 0.10000000000000001 .

from decimal import Decimal
'%g' % (Decimal(str(x)))

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']

>>> '%g' % 0
'0'
>>> '%g' % 0.0
'0'
>>> '%g' % 0.1
'0.1'
>>> '%g' % 1.0
'1'

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

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

>>> x = '1.0'
>>> int(float(x))
1
>>> x = 1
>>> int(float(x))
1

str(x)[-2:] == '.0' and int(x) or x

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'

To print a float that has an integer value as an int:

format = "%d" if f.is_integer() else "%s"
print(format % f)

Example

             0.0 -> 0
             0.1 -> 0.1
            10.0 -> 10
      12345678.9 -> 12345678.9
     123456789.0 -> 123456789
12345678912345.0 -> 12345678912345
12345678912345.6 -> 1.23456789123e+13
  1.000000000001 -> 1.0

So much ugliness out there…

My personal favorite is to convert floats that don’t require to be a float (= when they actually are integers) to int, thus removing the, now useless, trailing 0

(int(i) if i.is_integer() else i for i in lst)

Then you can print them normally.

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

Following code will convert contents of variable no as it is i.e. 45.60. If you use str the output will be 45.6

no = 45.60

strNo = "%.2f" %no

I was dealing with a value from a json dictionary (returned by an API). All the above didnt help me so i constructed by own helper function. It truncates all the trailing zeros.

I Hope it helps someone out there

def remove_zeros(num):
    nums = list(num)
    indexes = (list(reversed(range(len(nums)))))
    for i in indexes:
        if nums[i] == '0':
            del nums[-1]
        else:
            break
    return "".join(nums)

num = "0.000000363000"
print(remove_zeros(num))

prints :

    0.000000363

I did this for removing trailing nonsignificant digits past a certain precision.

c = lambda x: float(int(x * 100)/100.)
c(0.1234)