How can I remove ".0" of float numbers?

Question

Say I have a float number. If it is an integer (e.g. 1.0, 9.0, 36.0), I want to remove the ".0 (the decimal point and zero)" and write to stdout. For example, the result will be 1, 9, 36. If the float number is a rational number such as 2.5335 and 95.5893, I want the output to be as it was inputted, meaning 2.5335 and 95.5893. Is there a smart way to do this?

Only whether the number has .0 or other digits at decimal places does matter. I can say my question is this: how can I know whether or not a float number is actually an integer?

I want to use this to decide directory names. For example, if I input 9.0 and 2.45, make two directories, dir9 and dir2.45.

Misconceive

"if the float number is a rational number such as 2.5335 and 95.5893, I want the output to be as it was inputted." What do you mean by this? – Isometrics

Possible duplicate of Python: Remove division decimal – Copperas

Do you trust Python is correct when it tells you a certain 1.0 does not have zeroes all the way, or do you want to check only the first few digits, as in your examples? – Devoted

Are you aware that 0.1 + 0.2 != 0.3 (and other fun consequences of floats being binary rather than decimal)? – Xenolith

I understand 0.1 may not be exactly 0.1, but why cannot 1.0 be precisely 1.0? That 0.1 + 0.2 != 0.3 does not seem to be relevant to this question. Could you explain how it is related to my question? – Misconceive

You have to be more clear about the phrase "I want the output to be as it was inputted", assuming that the input is a decimal rational number. For instance, 123456789.123456769 == 123456789.123456783 is true for my installation (both these floats are printed as 123456789.12345678). On another note, for large "integer" floats, you'll lose digits before the decimal point (for my installation, int(123456789123456789.0) - 123456789123456789 == -5). – Xenolith

Answer 1

Here's a function to format your numbers the way you want them:

def formatNumber(num):
  if num % 1 == 0:
    return int(num)
  else:
    return num

For example:

formatNumber(3.11111)

returns

3.11111

---

formatNumber(3.0)

returns

3

Fidole

Answer 2

you can use string formatting

>>> "%g" % 1.1
'1.1'

>>> "%g" % 1.0
'1'

Tristan

Answer 3

You can combine the 2 previous answers :

formatNumber = lambda n: n if n%1 else int(n)

>>>formatNumber(5)
5

>>>formatNumber(5.23)
5.23

>>>formatNumber(6.0)
6

Prelature

It's possible to define formatNumber = lambda n: int(n) if isinstance(n,float) and n.is_integer() else n – Byers

Answer 4

You can do that with fstrings like

print(f'{1.0:g},{1.2:g}')  # Output: 1,1.2

Clapp

Answer 5

num = 5.0
if float(num) == int(num):
    num = int(num)

Output: 5

Barbet

This works, but I think it would be better to write this as a function. – Lasseter

Answer 6

Further to 'MD. Ferdous Ibne Abu Bakar''s answer, should you have a string this function would work:

val = str(5.0)
def remove_zero_float(val):
    if float(val) == int(float(val)):
        val = int(float(val))
    return val

Output: 5

Apotropaic

Answer 7

just type int(number) example :

int(3.0)

returns

3

Bard

unfortunately, int(3.5) also returns 3 which the OP specifically does not want to have happen. – Histrionic