[FIXED] Round number to nearest integer

Issue

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

32.268907563;
32.268907563;
31.2396694215;
33.6206896552;
...

With no success so far. I tried math.ceil(x), math.floor(x) (although that would round up or down, which is not what I'm looking for) and round(x) which didn't work either (still float numbers).

What could I do?

Code:

for i in widthRange:
    for j in heightRange:
        r, g, b = rgb_im.getpixel((i, j))
        h, s, v = colorsys.rgb_to_hsv(r/255.0, g/255.0, b/255.0)
        h = h * 360
        int(round(h))
        print(h)

Solution

TL;DR:

round(x)

will round it and change it to integer.

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

h = round(h)

to assign the new value to h.

---

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.

One way to avoid this problem is to use the Decimal as stated by this answer.

In order for this answer to work properly without using extra libraries it would be convenient to use a custom rounding function. 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))

---

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.

Answered By - francisco sollima