Hjdcti

Continuing from the comments above:

Using is_integer():

Example from the docs:

>>> 1.5.is_integer()
False
>>> 1.0.is_integer()
True
>>> 1.4142135623730951.is_integer()
False

Hence:

s = [1.5, 1.0, 2.5, 3.54, 1.0]

print([int(x) if x.is_integer() else x for x in s])

OUTPUT:

[1.5, 1, 2.5, 3.54, 1]

Wrapped in a function:

def func(s):
 return [int(x) if x.is_integer() else x for x in s]

print(func(s))

In case your goal is to convert numbers to a concise string, you could simply use '%g' ("General Format") for formatting:

>>> '%g' % 1.0
'1'
>>> '%g' % 1
'1'
>>> '%g' % 1.5
'1.5'
>>> '%g' % 0.3
'0.3'
>>> '%g' % 0.9999999999
'1'

Python floats are approximations, so something that prints as 1.0 is not necessarily exactly 1.0. If you want to see if something is approximately an integer, use a sufficiently small epsilon value.

EPSILON = 0.0001 # Make this smaller or larger depending on desired accuracy

def func(x):
 if abs(x - round(x)) < EPSILON:
 return round(x)
 else:
 return x

In general, if you're checking whether a float is == to something, that tends to be a code smell, as floating point values are inherently approximate. It's more appropriate in general to check whether a float is near something, within some epsilon range.

Use this function, is_integer is a function from the float class,

You can just use this method to_int to use is_integer to check whether it is a integer (i.e 1.0) or a float (i.e 1.5),

If it is a integer-like float, return int(a), otherwise just return it's original value,

As you see, I am not using elif or else, just directly, because return only returns the first return's value:

def to_int(a):
 if a.is_integer():
 return int(a)
 return a

print(to_int(1.5))
print(to_int(1.0))

Output:

1.5
1

References:

https://python-reference.readthedocs.io/en/latest/docs/float/is_integer.html

https://python-reference.readthedocs.io/en/latest/docs/float

I just thought of another way, using replace, that works because you're only replacing .0, so .5 won't effect,

Only problem is that 1.00 won't work, it will output 10,

But this is still a simple solution:

a = 1.5
print(str(a).replace('.0', ''))
b = 1.0
print(str(b).replace('.0', ''))

Output:

1.5
1

What I used to do in the past in C++ is, lets say you have these variables:

float x = 1.5;
float y = 1.0;

Then you could do something like this:

if(x == (int)x) 
 return 1;
else return 0;

This will return 0 because 1.5 is not equal to 1

if(y == (int)y) 
 return 1;
else return 0;

This will return 1 because 1.0 is equal to 1

Of course your question is about Python and the function is_integer() should work great, I just thought some people might find this useful.

A simple thing you could do is use the modulo operator:

if (myFloat % 1 == 0) // Number is an int
else // numer is not an int

(Note: Not real code (although compilable with some languages)!)

EDIT:
Thanks to "popular demand" here's some Python code (untested):

if myFloat % 1 = 0:
 # Is an integer
 return int(myFloat)
else:
 # Is not an integer
 return myFloat

Again, the code is untested, but I think it should work (not too familiar w/ Python, tbh)