# How to Check if a List Has an Even Number of Elements?

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## Problem Formulation

Given a list in Python. How to check if the list has an even number of elements?

Examples:

• `[] --> True`
• ` --> False`
• `[1, 2] --> True`
• `[1, 2, 3] --> False`

Related Article:

## Method 1: len() and Modulo

The most Pythonic way to check if a list has an even number of elements is to use the modulo expression `len(my_list)%2` that returns `1` if the list length is odd and `0` if the list length is even. So to check if a list has an even number of elements use the expression `len(my_list)%2==0`.

Here’s a simple code example:

```def check_even(my_list):
return len(my_list)%2==0

print(check_even([]))
# True

print(check_even())
# False

print(check_even([1, 2]))
# True

print(check_even([1, 2, 3]))
# False
```

As background, feel free to watch the following video on the modulo operator:

The length function is explained in this video and blog article:

A slight variant of this method is the following.

## Method 2: len() and Modulo and bool()

To check if a list has an even number of elements, you can use the modulo expression `len(my_list)%2` that returns `1` if the list length is odd and `0` if the list length is even. So to convert the even value 0 to a boolean, use the built-in `bool()` function around the result and invert the result, i.e., `not bool(len(my_list)%2)`.

Here’s a simple code example:

```def check_even(my_list):
return not bool(len(my_list)%2)

print(check_even([]))
# True

print(check_even())
# False

print(check_even([1, 2]))
# True

print(check_even([1, 2, 3]))
# False
```

As background, you may want to look at this explainer video:

## Method 3: Bitwise AND

You can use the expression `len(my_list)&1` that uses the Bitwise AND operator to return 1 if the list has an even number of elements and 0 otherwise. Now, you simply convert it to a Boolean if needed using the `bool()` function and invert it using the `not` operator: `not bool(len(my_list)&1)`.

Python’s bitwise AND operator `x & y` performs logical AND on each bit position on the binary representations of integers `x` and `y`. Thus, each output bit is 1 if both input bits at the same position are 1, otherwise, it’s 0.

If you run `x & 1`, Python performs logical and with the bit sequence `y=0000...001`. For the result, all positions will be `0` and the last position will be 1 only if `x`‘s last position is already `1` which means it is odd.

After converting it using `bool()`, you still need to invert it using the `not` operator so that it returns `True` if the list has an even number of elements.

Here’s an example:

```def check_even(my_list):
return not bool(len(my_list)&1)

print(check_even([]))
# True

print(check_even())
# False

print(check_even([1, 2]))
# True

print(check_even([1, 2, 3]))
# False
```

Bitwise AND is more efficient than the modulo operator so if performance is an issue for you, you may want to use this third approach.

You may want to watch this video on the Bitwise AND operator:

## Where to Go From Here?

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