How to format a minimum number of digits for a float in python

Question

Is there an easy way to format a float so that there is at least one trailing digit, but allows for as many as are present? Desired output would be below:

x = function(1)
x = '1.0'

y = function(1.1)
y = '1.1'

z = function(1.2345)
z = '1.2345'

Using '%.1f' will add the trailing zero, but it will also clip any longer values. I can probably kludge together a small function that checks if the value is an integer and adds a '.0', but that seems un-pythonic.

def function(value):
    if int(value) == value:
        form_string = '%i.0' % value
    else:
        form_string = ('%f' % value).rstrip('0')

    return form_string 

Thanks!

Answer 1

Simply convert to string and check if . is there?

def function(value):
    ret = str(value)
    if '.' not in ret:
        ret += '.0'
    return ret

Anyway, str(float(value)) has the same effect, if you are dealing with small values (ie less than ~10 ¹⁵ ).

Answer 2

Just to add an answer, for a fixed "I want at least one digit after the period" you just have to format it as float:

def parse(a):
    return str(float(a))