**Problem Formulation**: Given a float number. How to round the float up in Python?

Here are some examples of what you want to accomplish:

`42.42 --> 43`

`21.00001 --> 22`

`-0.1 --> 0`

Solution: If you have little time, here’s the most straightforward answer:

To round a number up in Python, import the `math`

library with `import math`

, and call `math.ceil(number)`

that returns the ceiling of `number`

, i.e., the smallest integer greater than or equal to `number`

.

In general, there are at least four ways to round a float number `x`

up in Python:

**Round up**: The`math.ceil(x)`

function rounds number`x`

up to the next full integer.**Round up (float representation)**: Alternatively,`numpy.ceil(x)`

rounds up and returns a float representation of the next full integer (e.g.,`2.0`

instead of`2`

).**Round down**: The`math.floor(x)`

function rounds number`x`

down to the next full integer.**Round up and down**: The Python built-in`round(x)`

function rounds`x`

up and down to the closest full integer.**Vanilla Python**: The one-liner expression

rounds`int(x) + ((int(x)!=x) if x>0 else 0)`

`x`

up*without external dependency*. An alternative is the expression`int(x) + bool(x%1)`

to round up positive numbers.

Let’s dive into each of those and more options in the remaining article. I guarantee you’ll get out of it having learned at least a few new Python tricks in the process!

## Method 1: math.ceil()

To round a number up in Python, import the `math`

library with `import math`

, and call `math.ceil(number)`

.

The function returns the ceiling of the specified `number`

that is defined as the smallest integer greater than or equal to `number`

The following code shows how to round the number 42.42 up to 43, 21.00001 to 22, and -0.1 to 0 using the `math.ceil()`

function.

import math print(math.ceil(42.42)) # 43 print(math.ceil(21.00001)) # 22 print(math.ceil(-0.1)) # 0

The following video shows the `math.ceil()`

as well as the `math.floor()`

functions — feel free to watch it to gain a deeper understanding:

## Method 2: np.ceil()

To round a number up in Python, import the NumPy library with `import numpy as np`

, and call `np.ceil(number)`

.

The function returns the ceiling of the specified `number`

that is defined as the smallest integer greater than or equal to `number`

.

The following code shows how to round the number 42.42 up to 43, 21.00001 to 22, and -0.1 to 0 using the `np.ceil()`

function.

import numpy as np print(np.ceil(42.42)) # 43.0 print(np.ceil(21.00001)) # 22.0 print(np.ceil(-0.1)) # 0.0

Both `math.ceil()`

and `np.ceil()`

round up to the next full integer. The difference between `math.ceil()`

and `np.ceil()`

is that the former returns an integer and the latter returns a float value.

## Method 3: int(x) + bool(x%1)

You can also use the following vanilla Python snippet to round a number `x`

up to the next full integer:

- If
`x`

is negative, round up by calling`int(x)`

. - If
`x`

is positive, round up by calling`int(x) + bool(x%1)`

.

**Explanation**: Any non-zero expression passed into the `bool()`

function will yield `True`

which is represented by integer 1. The modulo expression `x%1`

returns the decimal part of `x`

. If it is non-zero, we add `bool(x%1) == 1`

, i.e., we round up. If it is zero, we add `bool(x%1) == 0`

, i.e., we’re already done.

Here’s what this looks like in a simple Python function:

def round_up(x): if x<0: return int(x) return int(x) + bool(x%1) print(round_up(42.42)) # 43 print(round_up(21.00001)) # 22 print(round_up(-0.1)) # 0

You can watch my explainer video on modulo here:

## Method 4: int(x) + int(x)!=x

If you don’t want to import the `math`

module, you can use the one-liner beauty:

`int(x) + ((int(x)!=x) if x>0 else 0)`

This ternary expression rounds up number `x`

to the next full integer. This first cuts off the decimal part using the `int()`

function and then adds one if there is a non-zero decimal part (and it’s a positive number) and zero otherwise.

If the number `x`

is negative, the expression `int(x)`

already rounds up to the next full integer.

def round_up(x): return int(x) + ((int(x)!=x) if x>0 else 0) print(round_up(42.42)) # 43 print(round_up(21.00001)) # 22 print(round_up(-0.1)) # 0

- The
`int()`

built-in function cuts of the decimal part, i.e., rounds down. - The expression
`int(x)!=x`

evaluates to 1 if the decimal part of`x`

is greater than 0. Otherwise, it becomes 0. - This helps us because only if the decimal part is greater than 0, we need to add +1 to the rounded-down number to round it up.
- If the number
`x`

is negative, the expression`int(x)`

already rounds up to the next full integer, so we use the ternary operator`(...) if (...) else (...)`

to account for this condition.

You can watch my introductory video on the **ternary operator** here:

## Method 5: round()

*This method is probably not exactly what you want because it rounds a number up and down, depending on whether the number is closer to the smaller or larger next full integer. However, I’ll still mention it for comprehensibility.*

Pythonβs built-in `round()`

function takes two input arguments:

- a
`number`

and - an optional
`precision`

in decimal digits.

It rounds the number to the given precision and returns the result. The return value has the same type as the input numberβor integer if the `precision`

argument is omitted.

Per default, the precision is set to 0 digits, so `round(3.14)`

results in `3`

.

Here are three examples using the `round()`

function—that show that it doesn’t exactly solve our problem.

import math print(round(42.42)) # 42 print(round(21.00001)) # 21 print(round(-0.1)) # 0

Again, we have a video on the `round()`

function — feel free to watch for maximum learning!

## Method 6: Rounding Up After Integer Division

If the float to be rounded up comes from a division operation `a/b`

, you can also use integer division `a//b`

to round down to the next integer, and increment this by one. Thus, the expression `a//b+1`

rounds the resulting number up if `a`

is not divisible by `b`

, otherwise, the result of `a//b`

would already provide the “rounded-up” semantics.

You can create a simple ternary operator `x if y else z`

to differentiate between those two conditions:

a = int(input('a=')) b = int(input('b=')) rounded_up = a//b + 1 if a%b else a//b print(rounded_up)

The code goes through the following steps:

- Get the input strings from the user using the built-in
`input()`

function. - Convert the inputs to integer values using the built-in
`int()`

function. - Use the modulo operation
`a%b`

to differentiate between`b`

being a divisor of`a`

or not. - If not, the result will have a remainder and you can use integer division
`a//b`

to round down and increment this by one. - If yes, the result won’t have a remainder and you can simply use integer division because it, mathematically, would already be considered to be rounded up.
- You use the ternary operator to pack this logic into a single line of code.

Here’s an example execution that was rounded up:

a=8 b=3 3

And here’s an example execution that wasn’t:

a=8 b=4 2

An alternative one-liner to round up two integers would be the following beauty:

a = int(input('a=')) b = int(input('b=')) rounded_up = a // b + (a % b > 0) print(rounded_up)

The expression `(a % b > 0)`

evaluates to `True`

if `b`

is not a divisor of `a`

, otherwise it evaluates to `False`

. As the Boolean `True`

is represented by the integer value 1 in Python and Boolean `False`

by the integer value 0 in Python, the expression increments only if `b`

is not a divisor of `a`

.

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