[python] What is :: (double colon) in Python when subscripting sequences?

I know that I can use something like string[3:4] to get a substring in Python, but what does the 3 mean in somesequence[::3]?

When slicing in Python the third parameter is the step. As others mentioned, see Extended Slices for a nice overview.

With this knowledge, [::3] just means that you have not specified any start or end indices for your slice. Since you have specified a step, 3, this will take every third entry of something starting at the first index. For example:

>>> '123123123'[::3]
'111'

Did I miss or nobody mentioned reversing with [::-1] here?

# Operating System List
systems = ['Windows', 'macOS', 'Linux']
print('Original List:', systems)

# Reversing a list  
#Syntax: reversed_list = systems[start:stop:step] 
reversed_list = systems[::-1]

# updated list
print('Updated List:', reversed_list)

source: https://www.programiz.com/python-programming/methods/list/reverse

Python sequence slice addresses can be written as a[start:end:step] and any of start, stop or end can be dropped. a[::3] is every third element of the sequence.

You can also use this notation in your own custom classes to make it do whatever you want

class C(object):
    def __getitem__(self, k):
        return k

# Single argument is passed directly.
assert C()[0] == 0

# Multiple indices generate a tuple.
assert C()[0, 1] == (0, 1)

# Slice notation generates a slice object.
assert C()[1:2:3] == slice(1, 2, 3)

# If you omit any part of the slice notation, it becomes None.
assert C()[:] == slice(None, None, None)
assert C()[::] == slice(None, None, None)
assert C()[1::] == slice(1, None, None)
assert C()[:2:] == slice(None, 2, None)
assert C()[::3] == slice(None, None, 3)

# Tuple with a slice object:
assert C()[:, 1] == (slice(None, None, None), 1)

# Ellipsis class object.
assert C()[...] == Ellipsis

We can then open up slice objects as:

s = slice(1, 2, 3)
assert s.start == 1
assert s.stop == 2
assert s.step == 3

This is notably used in Numpy to slice multi-dimensional arrays in any direction.

Of course, any sane API should use ::3 with the usual "every 3" semantic.

The related Ellipsis is covered further at: What does the Ellipsis object do?

Explanation

s[i:j:k] is, according to the documentation, "slice of s from i to j with step k". When i and j are absent, the whole sequence is assumed and thus s[::k] means "every k-th item".

Examples

First, let's initialize a list:

>>> s = range(20)
>>> s
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]

Let's take every 3^rd item from s:

>>> s[::3]
[0, 3, 6, 9, 12, 15, 18]

Let's take every 3^rd item from s[2:]:

>>> s[2:]
[2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> s[2::3]
[2, 5, 8, 11, 14, 17]

Let's take every 3^rd item from s[5:12]:

>>> s[5:12]
[5, 6, 7, 8, 9, 10, 11]
>>> s[5:12:3]
[5, 8, 11]

Let's take every 3^rd item from s[:10]:

>>> s[:10]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> s[:10:3]
[0, 3, 6, 9]

seq[::n] is a sequence of each n-th item in the entire sequence.

Example:

>>> range(10)[::2]
[0, 2, 4, 6, 8]

The syntax is:

seq[start:end:step]

So you can do (in Python 2):

>>> range(100)[5:18:2]
[5, 7, 9, 11, 13, 15, 17]

The third parameter is the step. So [::3] would return every 3rd element of the list/string.

TL;DR

This visual example will show you how to a neatly select elements in a NumPy Matrix (2 dimensional array) in a pretty entertaining way (I promise). Step 2 below illustrate the usage of that "double colons" :: in question.

(Caution: this is a NumPy array specific example with the aim of illustrating the a use case of "double colons" :: for jumping of elements in multiple axes. This example does not cover native Python data structures like List).

One concrete example to rule them all...

Say we have a NumPy matrix that looks like this:

In [1]: import numpy as np

In [2]: X = np.arange(100).reshape(10,10)

In [3]: X
Out[3]:
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
       [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
       [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
       [70, 71, 72, 73, 74, 75, 76, 77, 78, 79],
       [80, 81, 82, 83, 84, 85, 86, 87, 88, 89],
       [90, 91, 92, 93, 94, 95, 96, 97, 98, 99]])

Say for some reason, your boss wants you to select the following elements:

"But How???"... Read on! (We can do this in a 2-step approach)

Step 1 - Obtain subset

Specify the "start index" and "end index" in both row-wise and column-wise directions.

In code:

In [5]: X2 = X[2:9,3:8]

In [6]: X2
Out[6]:
array([[23, 24, 25, 26, 27],
       [33, 34, 35, 36, 37],
       [43, 44, 45, 46, 47],
       [53, 54, 55, 56, 57],
       [63, 64, 65, 66, 67],
       [73, 74, 75, 76, 77],
       [83, 84, 85, 86, 87]])

Notice now we've just obtained our subset, with the use of simple start and end indexing technique. Next up, how to do that "jumping"... (read on!)

Step 2 - Select elements (with the "jump step" argument)

We can now specify the "jump steps" in both row-wise and column-wise directions (to select elements in a "jumping" way) like this:

In code (note the double colons):

In [7]: X3 = X2[::3, ::2]

In [8]: X3
Out[8]:
array([[23, 25, 27],
       [53, 55, 57],
       [83, 85, 87]])

We have just selected all the elements as required! :)

 Consolidate Step 1 (start and end) and Step 2 ("jumping")

Now we know the concept, we can easily combine step 1 and step 2 into one consolidated step - for compactness:

In [9]: X4 = X[2:9,3:8][::3,::2]

    In [10]: X4
    Out[10]:
    array([[23, 25, 27],
           [53, 55, 57],
           [83, 85, 87]])

Done!

Python uses the :: to separate the End, the Start, and the Step value.