NumPy arrays iterate over the left-most axis first. Thus if B has shape
(2,3,4), then B[0] has shape (3,4) and B[1] has shape (3,4). In this sense,
you could think of B as 2 arrays of shape (3,4). You can sort of see the two
arrays in the repr of B:
In [233]: B = np.arange(2*3*4).reshape((2,3,4))
array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7], <-- first (3,4) array
[ 8, 9, 10, 11]],
[[12, 13, 14, 15],
[16, 17, 18, 19], <-- second (3,4) array
[20, 21, 22, 23]]])
You can also think of B as containing four 2x3 arrays by iterating over the last index first:
for i in range(4):
print(B[:,:,i])
# [[ 0 4 8]
# [12 16 20]]
# [[ 1 5 9]
# [13 17 21]]
# [[ 2 6 10]
# [14 18 22]]
# [[ 3 7 11]
# [15 19 23]]
but you could just as easily think of B as three 2x4 arrays:
for i in range(3):
print(B[:,i,:])
# [[ 0 1 2 3]
# [12 13 14 15]]
# [[ 4 5 6 7]
# [16 17 18 19]]
# [[ 8 9 10 11]
# [20 21 22 23]]
NumPy arrays are completely flexible this way. But as far as the repr of B is concerned, what you see corresponds to two (3x4) arrays since B iterates over the left-most axis first.
for arr in B:
print(arr)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]]
# [[12 13 14 15]
# [16 17 18 19]
# [20 21 22 23]]
NumPy arrays iterate over the left-most axis first. Thus if B has shape
(2,3,4), then B[0] has shape (3,4) and B[1] has shape (3,4). In this sense,
you could think of B as 2 arrays of shape (3,4). You can sort of see the two
arrays in the repr of B:
In [233]: B = np.arange(2*3*4).reshape((2,3,4))
array([[[ 0, 1, 2, 3],
[ 4, 5, 6, 7], <-- first (3,4) array
[ 8, 9, 10, 11]],
[[12, 13, 14, 15],
[16, 17, 18, 19], <-- second (3,4) array
[20, 21, 22, 23]]])
You can also think of B as containing four 2x3 arrays by iterating over the last index first:
for i in range(4):
print(B[:,:,i])
# [[ 0 4 8]
# [12 16 20]]
# [[ 1 5 9]
# [13 17 21]]
# [[ 2 6 10]
# [14 18 22]]
# [[ 3 7 11]
# [15 19 23]]
but you could just as easily think of B as three 2x4 arrays:
for i in range(3):
print(B[:,i,:])
# [[ 0 1 2 3]
# [12 13 14 15]]
# [[ 4 5 6 7]
# [16 17 18 19]]
# [[ 8 9 10 11]
# [20 21 22 23]]
NumPy arrays are completely flexible this way. But as far as the repr of B is concerned, what you see corresponds to two (3x4) arrays since B iterates over the left-most axis first.
for arr in B:
print(arr)
# [[ 0 1 2 3]
# [ 4 5 6 7]
# [ 8 9 10 11]]
# [[12 13 14 15]
# [16 17 18 19]
# [20 21 22 23]]
I hope the below example would clarify the second part of your question where you have asked about getting a 2X3 matrics when you type print(B[:,:,1])
import numpy as np
B = [[[1,2,3,4],
[5,6,7,8],
[9,10,11,12]],
[[13,14,15,16],
[17,18,19,20],
[21,22,23,24]]]
B = np.array(B)
print(B)
print()
print(B[:,:,1])
[[[ 1 2 3 4]
[ 5 6 7 8]
[ 9 10 11 12]]
[[13 14 15 16]
[17 18 19 20]
[21 22 23 24]]]
[[ 2 6 10]
[14 18 22]]
Since the dimension of B is 2X3X4, It means you have two matrices of size 3X4 as far as repr of B is concerned
Now in B[:,:,1] we are passing : , : and 1. First : indicates that we are selecting both the 3X4 matrices. The second : indicates that we are selecting all the rows from both the 3X4 matrices. The third parameter 1 indicates that we are selecting only the second column values of all the rows from both the 3X4 matrices. Hence we get
[[ 2 6 10]
[14 18 22]]
python - Accessing slice of 3D numpy array - Stack Overflow
python - How does slicing in three dimensional numpy array work? - Stack Overflow
python - Slice a 3D numpy array by a list of indices - Stack Overflow
Numpy slicing behaving weirdly when slicing three dimensional array. Any explanation?
arr[:][123][:] is processed piece by piece, not as a whole.
arr[:] # just a copy of `arr`; it is still 3d
arr[123] # select the 123 item along the first dimension
# oops, 1st dim is only 31, hence the error.
arr[:, 123, :] is processed as whole expression. Select one 'item' along the middle axis, and return everything along the other two.
arr[12][123] would work, because that first select one 2d array from the 1st axis. Now [123] works on that 285 length dimension, returning a 1d array. Repeated indexing [][].. works just often enough to confuse new programmrs, but usually it is not the right expression.
I think you might just want to do arr[:,123,:]. That gives you a 2D array with a shape of (31, 286), with the contents of the 124th place along that axis.
arr = np.array([[[1,2,3], [4,5,6]], [[7,8,9],[10,11,12]]])
# array([[[ 1, 2, 3],
# [ 4, 5, 6]],
# [[ 7, 8, 9],
# [10, 11, 12]]])
arr.shape # ------> (2,2,3)
# Think of them as axis
# lets create the first 2 axis of (`2`, ...)
# |(1)
# |
# |
# |
# |---------(0)
# now lets create second 2 axis of (2, `2`, ..)
# (1,1)
# |
# |---(1,0)
# |
# |
# |
# | |(0,1)
# |---------|---(0,0)
# now lets create the last 3 axis of (2, 2, `3`)
# /``(1,1,0) = 10
# |-- (1,1,1) = 11
# | \__(1,1,2) = 12
# |
# | /``(1,0,0) = 7
# |--|--(1,0,1) = 8
# | \__(1,0,2) = 9
# |
# |
# | /``(0,1,0) = 4
# | |--(0,1,1) = 5
# | \__(0,1,2) = 6
# | |
# | |
# |---------|---/``(0,0,0) = 1
# |--(0,0,1) = 2
# \__(0,0,2) = 3
# now suppose you ask
arr[0, :, :] # give me the first axis alon with all it's component
# |
# | /``(0,1,0) = 4
# | |--(0,1,1) = 5
# | \__(0,1,2) = 6
# | |
# | |
# |---------|---/``(0,0,0) = 1
# |--(0,0,1) = 2
# \__(0,0,2) = 3
# So it will print
# array([[1, 2, 3],
# [4, 5, 6]])
arr[:, 0, :] # you ask take all the first axis 1ut give me only the first axis of the first axis and all its components
#
#
#
#
# | /``(1,0,0) = 7
# |--|--(1,0,1) = 8
# | \__(1,0,2) = 9
# |
# |
# |
# |
# |
# |
# |
# |---------|---/``(0,0,0) = 1
# |--(0,0,1) = 2
# \__(0,0,2) = 3
# so you get the output
# array([[1, 2, 3],
# [7, 8, 9]])
# like wise you ask
print(arr[0:2, : , 2])
# so you are saying give (0,1) first axis, all of its children and only 3rd (index starts at 0 so 2 means 3) children
# 0:2 means 0 to 2 `excluding` 2; 0:5 means 0,1,2,3,4
#
# |
# | \__(1,1,2) = 12
# |
# |
# |--
# | \__(1,0,2) = 9
# |
# |
# |
# |
# | \__(0,1,2) = 6
# | |
# | |
# |---------|---/
# |
# \__(0,0,2) = 3
# so you get
# array([[ 3, 6],
# [ 9, 12]])
Going In
Given the array:
arr = np.array([[[1,2,3], [4,5,6]], [[7,8,9],[10,11,12]]])
The shape of arr, given by arr.shape is (2,2,3).
That basically means that if we start from outside (and ignore the first square bracket), we have 2 arrays in that scope. If we enter to one of those we can count 2 arrays and if we enter either we find 3 elements. I think having this view is quite helpful in understanding the following.
Specifying arr[1,1,2] selects the second array(index 1) in the outermost scope, and then selects the second array in the following scope and then selects the third element. The output is a single number:
12
specifying arr[:,1,2] first simultaneously selects all arrays at the outermost scope and then for each of these selects the second array (index 1). When we then enter into the next scope, we pick out the 3rd elements. This outputs two numbers:
array([ 6, 12])
specifying arr[:, : , 2] outputs 4 numbers since 1. at the first scope we selected all arrays (2) 2. at the next scope we selected all array (2 for each in the first)
array([[ 3, 6],
[ 9, 12]])
Coming out
Instinctually, the reason why they appear as 2x2 array can be viewed as receding from the lowermost scope. Elements are getting enclosed in square brackets because that share a scope. 3 and 6 will be in one array while 9 and 12 will be in another array.
Hi, guys!
I have a three dimensional array x=np.random.randn(10, 5, 5)
When I use: x2 = x[:, [1, 2], [1, 2]] I should get an array of size [10, 2, 2] but I don't. Instead I get an array of shape [10, 2].
To get what I want, I have to use two lines of code: x2 = x[:, [1, 2], :]; x2=x2[:, :, [1, 2]].
I don't understand why is numpy behaving in this manner. Any explanation is much appreciated
Thank you!
In [145]: arr[np.arange(3), np.arange(3),:]
Out[145]:
array([[ 0, 1, 2],
[12, 13, 14],
[24, 25, 26]])
You can try with np.diagonal:
arr = np.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]]])
np.diagonal(arr).T
array([[ 0, 1, 2],
[12, 13, 14],
[24, 25, 26]])