search
Search
Map of Data Science
search toc
Thanks for the thanks!
close
account_circle
Profile
exit_to_app
Sign out
search
keyboard_voice
close
Searching Tips
Search for a recipe:
"Creating a table in MySQL"
Search for an API documentation: "@append"
Search for code: "!dataframe"
Apply a tag filter: "#python"
Useful Shortcuts
/ to open search panel
Esc to close search panel
to navigate between search results
d to clear all current filters
Enter to expand content preview Doc Search Code Search Beta SORRY NOTHING FOUND!
mic
Start speaking... Voice search is only supported in Safari and Chrome.
Shrink
Navigate to

# NumPy | diff method

schedule Mar 5, 2023
Last updated
local_offer
PythonNumPy
Tags
map
Check out the interactive map of data science

Numpy's `diff(~)` method computes the difference between each value and its adjacent value in the input array.

# Parameters

1. `a` | `array-like`

The input array.

2. `n` | `int` | `optional`

The number of differences you want to compute recursively. By default, `n=1`. Check out the examples below for clarification.

3. `axis` | `int` | `optional`

The axis along which to compute the differences. For 2D arrays, the allowed values are as follows:

Axis

Meaning

0

Differences will be computed column-wise

1

Differences will be computed row-wise

By default, the axis is equal to the last axis. This means that, for 2D arrays, `axis=1`.

4. `prepend` | `array-like` | `optional`

Values you wish to prepend to the input array a prior to computing the differences.

# Return value

A Numpy array that contains the difference between each value and its adjacent value in the input array.

# Examples

## Basic usage

``` a = np.array([1, 3, 8, 15, 30])np.diff(a) array([ 2, 5, 7, 15]) ```

## Recursively computing differences

Suppose we wanted to compute the differences twice recursively, that is, `n=2`. The `diff(~)` first computes the case when `n=1`, and then performs yet another `diff(~)` on its output.

The case when `n=1`:

``` a = np.array([1, 3, 8, 15, 30])np.diff(a, n=1) array([ 2, 5, 7, 15]) ```

The case when `n=2`:

``` a = np.array([1, 3, 8, 15, 30])np.diff(a, n=2) array([3, 2, 8]) ```

Observe that `n=2` is simply applying the diff method on the output of `n=1`.

## Computing differences for 2D arrays

Consider the following 2D array

``` a = np.array([[1, 3], [8, 15]])a array([[ 1, 3], [ 8, 15]]) ```

### Row-wise

``` np.diff(a) # or axis=1 array([, ]) ```

### Column-wise

``` np.diff(a, axis=0) array([[ 7, 12]]) ```

## Prepending values before computation

``` a = np.array([3, 8, 15, 30])np.diff(a, prepend=1) array([ 2, 5, 7, 15]) ```

Here, we've prepended the value 1 to the `a`, so essentially we're computing the differences of `[1,3,8,15,30]`.

thumb_up
thumb_down
Comment
Citation
Ask a question or leave a feedback...
thumb_up
0
thumb_down
0
chat_bubble_outline
0
settings
Enjoy our search
Hit / to insta-search docs and recipes!