{"id":2260,"date":"2021-02-20T10:57:24","date_gmt":"2021-02-20T10:57:24","guid":{"rendered":"https:\/\/www.pythontutorial.net\/?page_id=2260"},"modified":"2022-11-24T00:05:07","modified_gmt":"2022-11-24T00:05:07","slug":"python-rounding","status":"publish","type":"page","link":"https:\/\/www.pythontutorial.net\/advanced-python\/python-rounding\/","title":{"rendered":"Python Rounding"},"content":{"rendered":"\n

Summary<\/strong>: in this tutorial, you’ll learn how to use the Python round()<\/code> function to round a number.<\/p>\n\n\n\n

Introduction to the Python round() function #<\/a><\/h2>\n\n\n\n

Rounding means making a number simpler but keeping its value close to its original value. For example, 89 rounded to the nearest ten is 90 because 89 is closer to 90 than to 80.<\/p>\n\n\n\n

To round a number in Python, you use the built-in round()<\/code> function:<\/p>\n\n\n

round(number [, ndigits])<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
The round()<\/code> function rounds the number<\/code> to the closest multiple of 10-ndigits<\/sup>.<\/p>\n\n\n\n
In other words, the round()<\/code> function returns the number<\/code> rounded to ndigits<\/code> precision after the decimal point.<\/p>\n\n\n\n
If  ndigits<\/code> is omitted or None<\/code>, the round()<\/code> will return the nearest integer.<\/p>\n\n\n\n
Python round() function examples #<\/a><\/h2>\n\n\n\n
Let’s take some examples to understand the round()<\/code> function better.<\/p>\n\n\n\n
1) Python round() function examples #<\/a><\/h3>\n\n\n\n
The following example uses the round()<\/code> function without passing the ndigits<\/code>:<\/p>\n\n\n
round(1.25<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
Output:<\/p>\n\n\n
1<\/span><\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
It returns an integer 1<\/code>.<\/p>\n\n\n\n
round(1.25<\/span>, 0<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
Output:<\/p>\n\n\n
1.0<\/span><\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
The following illustrates how the round()<\/code> function works under the hood:<\/p>\n\n\n
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\"\"<\/figure>\n<\/div>\n\n\n
Since ndigits<\/code> is zero, the round()<\/code> function rounds the number 1.25<\/code> to the closet multiple of 10-(0)<\/sup> = 1.<\/p>\n\n\n\n
2) Python round() function example with negative ndigits #<\/a><\/h3>\n\n\n\n
The following example uses the round()<\/code> function with negative ndigits<\/code>:<\/p>\n\n\n
round(15.5<\/span>, -1<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
Because ndigits<\/code> is -1<\/code>, the round()<\/code> function rounds the number 15.5<\/code> to the closest multiple of 20<\/code> (10-(-1)<\/sup>):<\/p>\n\n\n
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\"\"<\/figure>\n<\/div>\n\n\n
Since 15.5<\/code> is situated between 10<\/code> and 20<\/code> (multiple of 10<\/code>), it’s closer to 20<\/code>. Therefore, the round()<\/code> function returns 20<\/code>.<\/p>\n\n\n\n
3) Python round() function example with ties #<\/a><\/h3>\n\n\n\n
When you round a number situated in the middle of two numbers, Python cannot find the closest number.<\/p>\n\n\n\n
For example, if you round the number 1.25<\/code> with n is 1<\/code>. There will be no closest number:<\/p>\n\n\n
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\"\"<\/figure>\n<\/div>\n\n\n
In this case, Python uses the IEEE 754 standard for rounding, called the banker’s rounding.<\/p>\n\n\n\n
In the banker’s rounding, a number is rounded to the nearest value, with ties rounded to the nearest value with an even least significant digit<\/strong>.<\/p>\n\n\n\n
Generally, the least significant digit in a number is the rightmost digit.<\/p>\n\n\n\n
The banker’s rounding comes from the idea that statistically 50% sample of numbers are rounded up and 50% are rounded down.<\/p>\n\n\n\n
For example:<\/p>\n\n\n
round(1.25<\/span>, 1<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
It returns 1.2<\/p>\n\n\n\n
Because the least significant digit of 1.2<\/code> is 2<\/code>, which is even:<\/p>\n\n\n
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\"\"<\/figure>\n<\/div>\n\n\n
Similarly, the rounding of 1.35<\/code> will return 1.4<\/code>:<\/p>\n\n\n
round(1.35<\/span>, 1<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n
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\"\"<\/figure>\n<\/div>\n\n\n
Python uses banker’s rounding but not rounding away from zero because it’s less biased.<\/p>\n\n\n\n
For example, if you average three numbers 1.5<\/code>, 2.5<\/code>, and 3.5<\/code>, the rounding away from zero returns 3<\/code>.0 while the banker’s rounding returns 2.66<\/code>:<\/p>\n\n\n\n
Number<\/th>Banker’s Rounding<\/th>Rounding away from zero<\/th><\/tr><\/thead>
1.5<\/td>2<\/td>2<\/td><\/tr>
2.5<\/td>2<\/td>3<\/td><\/tr>
3.5<\/td>4<\/td>4<\/td><\/tr><\/tbody>
Average<\/strong><\/td>2.66666…<\/td>3.0<\/td><\/tr><\/tfoot><\/table><\/figure>\n\n\n\n
How to round away from zero #<\/a><\/h2>\n\n\n\n
Python doesn’t provide a direct way to round a number away from zero as you might expect. For example:<\/p>\n\n\n\n
Number<\/th>Rounding away from zero<\/th><\/tr><\/thead>
1.2<\/td>1<\/td><\/tr>
1.5<\/td>2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
A common way to round a number away from zero is to use the following expression:<\/p>\n\n\n
int(x + 0.5<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
This expression works correctly for positive numbers. For example:<\/p>\n\n\n
print(int(1.2<\/span> + 0.5<\/span>))\nprint(int(1.5<\/span> + 0.5<\/span>))<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
Output:<\/p>\n\n\n
1<\/span>\n2<\/span><\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
However, it doesn’t work for the negative numbers:<\/p>\n\n\n
print(int(-1.2<\/span> + 0.5<\/span>))\nprint(int(-1.5<\/span> + 0.5<\/span>))<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
Output:<\/p>\n\n\n
0<\/span>\n-1<\/span><\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
For negative numbers, you should subtract 0.5 instead of adding it.<\/p>\n\n\n\n
The following example works properly for the negative numbers:<\/p>\n\n\n
print(int(-1.2<\/span> - 0.5<\/span>))\nprint(int(-1.5<\/span> - 0.5<\/span>))<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
The following defines a helper function that rounds up a number:<\/p>\n\n\n
def<\/span> round_up<\/span>(x)<\/span>:<\/span>\n    if<\/span> x > 0<\/span>:\n        return<\/span> int(x + 0.5<\/span>)\n    return<\/span> int(x - 0.5<\/span>)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
The Python math<\/code> module provides you with a function called copysign()<\/code>:<\/p>\n\n\n
math.copysign(x, y)<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
The copysign()<\/code> function returns the absolute value of x<\/code> but the sign of y<\/code>.<\/p>\n\n\n\n
And you can use this copysign()<\/code> function to develop a round_up()<\/code> function without checking whether x<\/code> is positive or negative:<\/p>\n\n\n
from<\/span> math import<\/span> copysign\n\ndef<\/span> round_up<\/span>(x)<\/span>:<\/span>\n    return<\/span> int(x + copysign(0.5<\/span>, x))<\/code><\/span>Code language:<\/span> Python<\/span> (<\/span>python<\/span>)<\/span><\/small><\/pre>\n\n\n
Summary #<\/a><\/h2>\n\n\n\n
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  • Use the round(number, ndigits)<\/code> function to round a number<\/code> to the ndigits<\/code> precision after the decimal point.<\/li>\n<\/ul>\n
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However, if you pass ndigits<\/code> as zero, the round()<\/code> function returns a float<\/a> 1.0:<\/p>\n\n\n