Numpy: find index of the elements within range - python

I have a numpy array of numbers, for example,
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
I would like to find all the indexes of the elements within a specific range. For instance, if the range is (6, 10), the answer should be (3, 4, 5). Is there a built-in function to do this?

You can use np.where to get indices and np.logical_and to set two conditions:
import numpy as np
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
np.where(np.logical_and(a>=6, a<=10))
# returns (array([3, 4, 5]),)

As in #deinonychusaur's reply, but even more compact:
In [7]: np.where((a >= 6) & (a <=10))
Out[7]: (array([3, 4, 5]),)

Summary of the answers
For understanding what is the best answer we can do some timing using the different solution.
Unfortunately, the question was not well-posed so there are answers to different questions, here I try to point the answer to the same question. Given the array:
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
The answer should be the indexes of the elements between a certain range, we assume inclusive, in this case, 6 and 10.
answer = (3, 4, 5)
Corresponding to the values 6,9,10.
To test the best answer we can use this code.
import timeit
setup = """
import numpy as np
import numexpr as ne
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
# or test it with an array of the similar size
# a = np.random.rand(100)*23 # change the number to the an estimate of your array size.
# we define the left and right limit
ll = 6
rl = 10
def sorted_slice(a,l,r):
start = np.searchsorted(a, l, 'left')
end = np.searchsorted(a, r, 'right')
return np.arange(start,end)
"""
functions = ['sorted_slice(a,ll,rl)', # works only for sorted values
'np.where(np.logical_and(a>=ll, a<=rl))[0]',
'np.where((a >= ll) & (a <=rl))[0]',
'np.where((a>=ll)*(a<=rl))[0]',
'np.where(np.vectorize(lambda x: ll <= x <= rl)(a))[0]',
'np.argwhere((a>=ll) & (a<=rl)).T[0]', # we traspose for getting a single row
'np.where(ne.evaluate("(ll <= a) & (a <= rl)"))[0]',]
functions2 = [
'a[np.logical_and(a>=ll, a<=rl)]',
'a[(a>=ll) & (a<=rl)]',
'a[(a>=ll)*(a<=rl)]',
'a[np.vectorize(lambda x: ll <= x <= rl)(a)]',
'a[ne.evaluate("(ll <= a) & (a <= rl)")]',
]
rdict = {}
for i in functions:
rdict[i] = timeit.timeit(i,setup=setup,number=1000)
print("%s -> %s s" %(i,rdict[i]))
print("Sorted:")
for w in sorted(rdict, key=rdict.get):
print(w, rdict[w])
Results
The results are reported in the following plot for a small array (on the top the fastest solution) as noted by #EZLearner they may vary depending on the size of the array. sorted slice could be faster for larger arrays, but it requires your array to be sorted, for arrays with over 10 M of entries ne.evaluate could be an option. Is hence always better to perform this test with an array of the same size as yours:
If instead of the indexes you want to extract the values you can perform the tests using functions2 but the results are almost the same.

I thought I would add this because the a in the example you gave is sorted:
import numpy as np
a = [1, 3, 5, 6, 9, 10, 14, 15, 56]
start = np.searchsorted(a, 6, 'left')
end = np.searchsorted(a, 10, 'right')
rng = np.arange(start, end)
rng
# array([3, 4, 5])

a = np.array([1,2,3,4,5,6,7,8,9])
b = a[(a>2) & (a<8)]

Other way is with:
np.vectorize(lambda x: 6 <= x <= 10)(a)
which returns:
array([False, False, False, True, True, True, False, False, False])
It is sometimes useful for masking time series, vectors, etc.

This code snippet returns all the numbers in a numpy array between two values:
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56] )
a[(a>6)*(a<10)]
It works as following:
(a>6) returns a numpy array with True (1) and False (0), so does (a<10). By multiplying these two together you get an array with either a True, if both statements are True (because 1x1 = 1) or False (because 0x0 = 0 and 1x0 = 0).
The part a[...] returns all values of array a where the array between brackets returns a True statement.
Of course you can make this more complicated by saying for instance
...*(1-a<10)
which is similar to an "and Not" statement.

a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
np.argwhere((a>=6) & (a<=10))

Wanted to add numexpr into the mix:
import numpy as np
import numexpr as ne
a = np.array([1, 3, 5, 6, 9, 10, 14, 15, 56])
np.where(ne.evaluate("(6 <= a) & (a <= 10)"))[0]
# array([3, 4, 5], dtype=int64)
Would only make sense for larger arrays with millions... or if you hitting a memory limits.

This may not be the prettiest, but works for any dimension
a = np.array([[-1,2], [1,5], [6,7], [5,2], [3,4], [0, 0], [-1,-1]])
ranges = (0,4), (0,4)
def conditionRange(X : np.ndarray, ranges : list) -> np.ndarray:
idx = set()
for column, r in enumerate(ranges):
tmp = np.where(np.logical_and(X[:, column] >= r[0], X[:, column] <= r[1]))[0]
if idx:
idx = idx & set(tmp)
else:
idx = set(tmp)
idx = np.array(list(idx))
return X[idx, :]
b = conditionRange(a, ranges)
print(b)

s=[52, 33, 70, 39, 57, 59, 7, 2, 46, 69, 11, 74, 58, 60, 63, 43, 75, 92, 65, 19, 1, 79, 22, 38, 26, 3, 66, 88, 9, 15, 28, 44, 67, 87, 21, 49, 85, 32, 89, 77, 47, 93, 35, 12, 73, 76, 50, 45, 5, 29, 97, 94, 95, 56, 48, 71, 54, 55, 51, 23, 84, 80, 62, 30, 13, 34]
dic={}
for i in range(0,len(s),10):
dic[i,i+10]=list(filter(lambda x:((x>=i)&(x<i+10)),s))
print(dic)
for keys,values in dic.items():
print(keys)
print(values)
Output:
(0, 10)
[7, 2, 1, 3, 9, 5]
(20, 30)
[22, 26, 28, 21, 29, 23]
(30, 40)
[33, 39, 38, 32, 35, 30, 34]
(10, 20)
[11, 19, 15, 12, 13]
(40, 50)
[46, 43, 44, 49, 47, 45, 48]
(60, 70)
[69, 60, 63, 65, 66, 67, 62]
(50, 60)
[52, 57, 59, 58, 50, 56, 54, 55, 51]

You can use np.clip() to achieve the same:
a = [1, 3, 5, 6, 9, 10, 14, 15, 56]
np.clip(a,6,10)
However, it holds the values less than and greater than 6 and 10 respectively.

Related

How can I extract a set of 2D slices from a larger 2D numpy array?

If I have a large 2D numpy array and 2 arrays which correspond to the x and y indices I want to extract, It's easy enough:
h = np.arange(49).reshape(7,7)
# h = [[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]]
x_indices = np.array([1,3,4])
y_indices = np.array([2,3,5])
reduced_h = h[x_indices, y_indices]
#reduced_h = [ 9, 24, 33]
However, I would like to, for each x,y pair cut out a square (denoted by 'a' - the number of indices in each direction from the centre) surrounding this 'coordinate' and return an array of these little 2D arrays.
For example, for h, x,y_indices as above and a=1:
reduced_h = [[[1,2,3],[8,9,10],[15,16,17]], [[16,17,18],[23,24,25],[30,31,32]], [[25,26,27],[32,33,34],[39,40,41]]]
i.e one 3x3 array for each x-y index pair corresponding to the 3x3 square of elements centred on the x-y index. In general, this should return a numpy array which has shape (len(x_indices),2a+1, 2a+1)
By analogy to reduced_h[0] = h[x_indices[0]-1:x_indices[0]+1 , y_indices[0]-1:y_indices[0]+1] = h[1-1:1+1 , 2-1:2+1] = h[0:2, 1:3] my first try was the following:
h[x_indices-a : x_indices+a, y_indices-a : y_indices+a]
However, perhaps unsurprisingly, slicing between the arrays fails.
So the obvious next thing to try is to create this slice manually. np.arange seems to struggle with this but linspace works:
a=1
xrange = np.linspace(x_indices-a, x_indices+a, 2*a+1, dtype=int)
# xrange = [ [0, 2, 3], [1, 3, 4], [2, 4, 5] ]
yrange = np.linspace(y_indices-a, y_indices+a, 2*a+1, dtype=int)
Now can try h[xrange,yrange] but this unsurprisingly does this element-wise meaning I get only one (2a+1)x(2a+1) array (the same dimensions as xrange and yrange). It there a way to, for every index, take the right slices from these ranges (without loops)? Or is there a way to make the broadcast work initially without having to set up linspace explicitly? Thanks
You can index np.lib.stride_tricks.sliding_window_view using your x and y indices:
import numpy as np
h = np.arange(49).reshape(7,7)
x_indices = np.array([1,3,4])
y_indices = np.array([2,3,5])
a = 1
window = (2*a+1, 2*a+1)
out = np.lib.stride_tricks.sliding_window_view(h, window)[x_indices-a, y_indices-a]
out:
array([[[ 1, 2, 3],
[ 8, 9, 10],
[15, 16, 17]],
[[16, 17, 18],
[23, 24, 25],
[30, 31, 32]],
[[25, 26, 27],
[32, 33, 34],
[39, 40, 41]]])
Note that you may need to pad h first to handle windows around your coordinates that reach "outside" h.

Skipping rows in numpy ndarrays slicing

Suppose I have a numpy array img, with img.shape == (468,832,3). What does img[::2, ::2] do? It reduces the shape to (234,416,3) Can you please explain the logic?
Let's read documentation together (Source).
(Just read the bold part first)
The basic slice syntax is i:j:k where i is the starting index, j is the stopping index, and k is the step (k \neq 0). This selects the m elements (in the corresponding dimension) with index values i, i + k, ..., i + (m - 1) k where m = q + (r\neq0) and q and r are the quotient and remainder obtained by dividing j - i by k: j - i = q k + r, so that i + (m - 1) k < j.
...
Assume n is the number of elements in the dimension being sliced.
Then, if i is not given it defaults to 0 for k > 0 and n - 1 for k < 0
. If j is not given it defaults to n for k > 0 and -n-1 for k < 0 . If
k is not given it defaults to 1. Note that :: is the same as : and
means select all indices along this axis.
Now looking at your part.
[::2, ::2] will be translated to [0:468:2, 0:832:2] because you do not specify the first two or i and j in the documentation. (You only specify k here. Recall the i:j:k notation above.) You select elements on these axes at the step size 2 which means you select every other elements along the axes specified.
Because you did not specify for the 3rd dimension, all will be selected.
It slices every alternate row, and then every alternate column, from an array, returning an array of size (n // 2, n // 2, ...).
Here's an example of slicing with a 2D array -
>>> a = np.arange(16).reshape(4, 4)
>>> a
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
>>> a[::2, ::2]
array([[ 0, 2],
[ 8, 10]])
And, here's another example with a 3D array -
>>> a = np.arange(27).reshape(3, 3, 3)
>>> a
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]]])
>>> a[::2, ::2] # same as a[::2, ::2, :]
array([[[ 0, 1, 2],
[ 6, 7, 8]],
[[18, 19, 20],
[24, 25, 26]]])
Well, we have the RGB image as a 3D array of shape:
img.shape=(468,832,3)
Now, what does img[::2, ::2] do?
we're just downsampling the image (i.e. we're shrinking the image size by half by taking only every other pixel from the original image and we do this by using a step size of 2, which means to skip one pixel). This should be clear from the example below.
Let's take a simple grayscale image for easier understanding.
In [13]: arr
Out[13]:
array([[10, 11, 12, 13, 14, 15],
[20, 21, 22, 23, 24, 25],
[30, 31, 32, 33, 34, 35],
[40, 41, 42, 43, 44, 45],
[50, 51, 52, 53, 54, 55],
[60, 61, 62, 63, 64, 65]])
In [14]: arr.shape
Out[14]: (6, 6)
In [15]: arr[::2, ::2]
Out[15]:
array([[10, 12, 14],
[30, 32, 34],
[50, 52, 54]])
In [16]: arr[::2, ::2].shape
Out[16]: (3, 3)
Notice which pixels are in the sliced version. Also, observe how the array shape changes after slicing (i.e. it is reduced by half).
Now, this downsampling happens for all three channels in the image since there's no slicing happening in the third axis. Thus, you will get the shape reduced only for the first two axis in your example.
(468, 832, 3)
. . |
. . |
(234, 416, 3)

How to generate random numbers with if-statement in Python?

I would like to generate random numbers with a specific restriction using python. The code should do the following:
If an entered number is:
0, then generate 0 random non-recurrent numbers
<1, then generate 1 random non-recurrent numbers
<9, then generate 2 random non-recurrent numbers
<15, then generate 3 random non-recurrent numbers
<26, then generate 5 random non-recurrent numbers
<51, then generate 8 random non-recurrent numbers
<91, then generate 13 random non-recurrent numbers
<151, then generate 20 random non-recurrent numbers
<281, then generate 32 random non-recurrent numbers
The value of the random numbers should be limited by the value of the entered number. So if a 75 is entered, then the code should generate 13 random numbers with being 75 the highest value of the 13 numbers. 75 doesn't have to be the actual highest number, just in terms of max value.
My guess was to use numpy. Here is what I got until now (with an users help).
num_files=[0,1,9,...]
num_nums=[0,1,2,3,5,...]
for zipp in zip(num_files,num_nums)
if len(docx_files)<zipp[0]:
list_of_rands=np.random.choice(len(docx_files)+1,
zipp[1],replace=False)
Any ideas or more starting points?
Here's one way of doing it. Just zip the lists of numbers and the cutoffs, and check if the number input (the variable number in the code below) is above the cutoff. Note that this doesn't handle the case of numbers larger than 281, since I'm not sure what's supposed to happen there based on your description.
import numpy as np
number = 134
parameters = zip([9, 15, 26, 51, 91, 151], [3, 5, 8, 13, 20, 32])
nums = 2
for item in parameters:
if number > item[0]:
nums = item[1]
np.random.choice(number, nums)
You could define a function using a dictionary with ranges as keys and number of random numbers as values:
import random
def rand_nums(input_num):
d = {26: 5, 51: 8, 91: 13}
for k, v in d.items():
if input_num in range(k):
nums = random.sample(range(k+1), v)
return nums
print(rand_nums(20))
print(rand_nums(50))
print(rand_nums(88))
[14, 23, 11, 9, 5]
[9, 49, 23, 16, 8, 50, 47, 33]
[20, 16, 28, 77, 21, 87, 85, 82, 10, 47, 43, 90, 57]
>>>
You can avoid a many-branched if-elif-else using np.searchsorted:
import numpy as np
def generate(x):
boundaries = np.array([1, 2, 9, 15, 26, 51, 91, 151, 281])
numbers = np.array([0, 1, 2, 3, 5, 8, 13, 20, 32])
return [np.random.choice(j, n, False)+1 if j else np.array([], np.int64)
for j, n in np.broadcast(x, numbers[boundaries.searchsorted(x, 'right')])]
# demo
from pprint import pprint
# single value
pprint(generate(17))
# multiple values in one go
pprint(generate([19, 75, 3, 1, 2, 0, 8, 9]))
# interactive
i = int(input('Enter number: '))
pprint(generate(i))
Sample output:
[array([ 9, 1, 14, 4, 12])]
[array([ 8, 12, 6, 17, 4]),
array([17, 29, 2, 20, 16, 37, 36, 13, 34, 58, 49, 72, 41]),
array([1, 3]),
array([1]),
array([2, 1]),
array([], dtype=int64),
array([1, 8]),
array([3, 2, 6])]
Enter number: 280
[array([184, 73, 80, 280, 254, 164, 192, 145, 176, 29, 58, 251, 37,
107, 5, 51, 7, 128, 142, 125, 135, 87, 259, 83, 260, 10,
108, 210, 8, 36, 181, 64])]
How about:
def gen_rand_array(n):
mapping = np.array([[1,1],
[26,5],
[51,8],
[91,13]])
k = mapping[np.max(np.where(n > mapping[:,0])),1]
return np.random.choice(n+1,k)
Example:
>>> gen_rand_array(27)
array([ 0, 21, 26, 25, 23])
>>> gen_rand_array(27)
array([21, 5, 10, 3, 13])
>>> gen_rand_array(57)
array([30, 26, 50, 31, 44, 51, 39, 13])
>>> gen_rand_array(57)
array([21, 18, 35, 8, 13, 13, 20, 3])
Here's a screen shot putting it all together:
Explanation:
The line k = mapping[np.max(np.where(n > mapping[:,0])),1] is just finding the number of random values needed from the array mapping. n > mapping[:,0] return a boolean array whose values will be True for all the numbers smaller then n, False otherwise. np.where(...) will return the indexes of the elements of the array that are true. Since the values in the first column of mapping (i.e. mapping[:,0]) are ascending, we can find the index of the largest one that is less than n be calling np.max(...). Finally we want the corresponding value from the second column which is why we pass the result of that as an index to mapping again i.e. mapping[...,1] where the 1 is for the second column.
I don't know how to implement it in your code but with this code you then you get the randoms:
import random
x = 51
if x < 26:
ar_Random = [None]*5
for i in range(0, 6):
ar_Random[i] = random.randint(startNumOfRandom, stopNumOfRandom)
elif x < 51:
ar_Random = [None]*8
for i in range (0,9):
ar_Random[i] = random.randint(startNumOfRandom, stopNumOfRandom)
...
I'm not sure how you're mapping the length to the input but this is how you generate N random numbers with a maximum using Numpy.
import numpy as np
//set entered_num and desired_length to whatever you want
random_nums = np.random.randint(entered_num, size = desired_length)
import random
Starting_Number = int(input())
if Starting_Number < 26:
print(random.sample(range(1, 26), 5))
elif Starting_Number < 51:
print(random.sample(range(1, 51), 8))
elif Starting_Number < 91:
print(random.sample(range(1, 91), 13))
Here you go!!!
random.sample is the module you are looking for.
Have a good one!

Multiple indices for numpy array: IndexError: failed to coerce slice entry of type numpy.ndarray to integer

Is there a way to do multiple indexing in a numpy array as described below?
arr=np.array([55, 2, 3, 4, 5, 6, 7, 8, 9])
arr[np.arange(0,2):np.arange(5,7)]
output:
IndexError: too many indices for array
Desired output:
array([55,2,3,4,5],[2,3,4,5,6])
This problem might be similar to calculating a moving average over an array (but I want to do it without any function that is provided).
Here's an approach using strides -
start_index = np.arange(0,2)
L = 5 # Interval length
n = arr.strides[0]
strided = np.lib.stride_tricks.as_strided
out = strided(arr[start_index[0]:],shape=(len(start_index),L),strides=(n,n))
Sample run -
In [976]: arr
Out[976]: array([55, 52, 13, 64, 25, 76, 47, 18, 69, 88])
In [977]: start_index
Out[977]: array([2, 3, 4])
In [978]: L = 5
In [979]: out
Out[979]:
array([[13, 64, 25, 76, 47],
[64, 25, 76, 47, 18],
[25, 76, 47, 18, 69]])

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]?
it means 'nothing for the first argument, nothing for the second, and jump by three'. It gets every third item of the sequence sliced.
Extended slices is what you want. New in Python 2.3
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.
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]
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 3rd item from s:
>>> s[::3]
[0, 3, 6, 9, 12, 15, 18]
Let's take every 3rd 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 3rd 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 3rd item from s[:10]:
>>> s[:10]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> s[:10:3]
[0, 3, 6, 9]
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!
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'
remember that the foundations is what a[start:end:step] means. From there you can get a[1::2] get every odd index, a[::2] get every even, a[2::2] get every even starting at 2, a[2:4:2] get every even starting at 2 and ending at 4. Inspired by https://stackoverflow.com/a/3453102/1601580
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?
The third parameter is the step. So [::3] would return every 3rd element of the list/string.
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 uses the :: to separate the End, the Start, and the Step value.

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