Round number to nearest integer

Question

I\'ve been trying to round long float numbers like:

32.268907563;
32.268907563;
31.2396694215;
33.6206896552;
...

With no success so far. I

Answer 1

int(round(x))

Will round it and change it to integer

EDIT:

You are not assigning int(round(h)) to any variable. When you call int(round(h)), it returns the integer number but does nothing else; you have to change that line for:

h = int(round(h))

To assign the new value to h

EDIT 2:

As @plowman said in the comments, Python's round() doesn't work as one would normally expect, and that's because the way the number is stored as a variable is usually not the way you see it on screen. There are lots of answers that explain this behavior:

round() doesn't seem to be rounding properly

One way to avoid this problem is to use the Decimal as stated by this answer: https://stackoverflow.com/a/15398691/4345659

In order for this answer to work properly without using extra libraries it would be convenient to use a custom rounding function. After a lot of corrections, I came up with the following solution, that as far as I tested avoided all the storing issues. It is based on using the string representation, obtained with repr() (NOT str()!). It looks hacky but it was the only way I found to solve all the cases. It works with both Python2 and Python3.

def proper_round(num, dec=0):
    num = str(num)[:str(num).index('.')+dec+2]
    if num[-1]>='5':
        return float(num[:-2-(not dec)]+str(int(num[-2-(not dec)])+1))
    return float(num[:-1])

Tests:

>>> print(proper_round(1.0005,3))
1.001
>>> print(proper_round(2.0005,3))
2.001
>>> print(proper_round(3.0005,3))
3.001
>>> print(proper_round(4.0005,3))
4.001
>>> print(proper_round(5.0005,3))
5.001
>>> print(proper_round(1.005,2))
1.01
>>> print(proper_round(2.005,2))
2.01
>>> print(proper_round(3.005,2))
3.01
>>> print(proper_round(4.005,2))
4.01
>>> print(proper_round(5.005,2))
5.01
>>> print(proper_round(1.05,1))
1.1
>>> print(proper_round(2.05,1))
2.1
>>> print(proper_round(3.05,1))
3.1
>>> print(proper_round(4.05,1))
4.1
>>> print(proper_round(5.05,1))
5.1
>>> print(proper_round(1.5))
2.0
>>> print(proper_round(2.5))
3.0
>>> print(proper_round(3.5))
4.0
>>> print(proper_round(4.5))
5.0
>>> print(proper_round(5.5))
6.0
>>> 
>>> print(proper_round(1.000499999999,3))
1.0
>>> print(proper_round(2.000499999999,3))
2.0
>>> print(proper_round(3.000499999999,3))
3.0
>>> print(proper_round(4.000499999999,3))
4.0
>>> print(proper_round(5.000499999999,3))
5.0
>>> print(proper_round(1.00499999999,2))
1.0
>>> print(proper_round(2.00499999999,2))
2.0
>>> print(proper_round(3.00499999999,2))
3.0
>>> print(proper_round(4.00499999999,2))
4.0
>>> print(proper_round(5.00499999999,2))
5.0
>>> print(proper_round(1.0499999999,1))
1.0
>>> print(proper_round(2.0499999999,1))
2.0
>>> print(proper_round(3.0499999999,1))
3.0
>>> print(proper_round(4.0499999999,1))
4.0
>>> print(proper_round(5.0499999999,1))
5.0
>>> print(proper_round(1.499999999))
1.0
>>> print(proper_round(2.499999999))
2.0
>>> print(proper_round(3.499999999))
3.0
>>> print(proper_round(4.499999999))
4.0
>>> print(proper_round(5.499999999))
5.0

Finally, the corrected answer would be:

# Having proper_round defined as previously stated
h = int(proper_round(h))

EDIT 3:

Tests:

>>> proper_round(6.39764125, 2)
6.31 # should be 6.4
>>> proper_round(6.9764125, 1)
6.1  # should be 7

The gotcha here is that the dec-th decimal can be 9 and if the dec+1-th digit >=5 the 9 will become a 0 and a 1 should be carried to the dec-1-th digit.

If we take this into consideration, we get:

def proper_round(num, dec=0):
    num = str(num)[:str(num).index('.')+dec+2]
    if num[-1]>='5':
      a = num[:-2-(not dec)]       # integer part
      b = int(num[-2-(not dec)])+1 # decimal part
      return float(a)+b**(-dec+1) if a and b == 10 else float(a+str(b))
    return float(num[:-1])

In the situation described above b = 10 and the previous version would just concatenate a and b which would result in a concatenation of 10 where the trailing 0 would disappear. This version transforms b to the right decimal place based on dec, as a proper carry.