I have two matrices of the same size, A, B. I want to use the columns of B to acsses the columns of A, on a per column basis. For example,
A = np.array([[1, 4, 7],
[2, 5, 8],
[3, 6, 9]])
and
B = np.array([[0, 0, 2],
[1, 2, 1],
[2, 1, 0]])
I want something like:
A[B] = [[1, 4, 9],
[2, 6, 8],
[3, 5, 7]]
I.e., I've used the j'th column of B as indices to the j'th column of A.
Is there any effiecnt way of doing so?
Thanks!
You can use advanced indexing:
A[B, np.arange(A.shape[0])]
array([[1, 4, 9],
[2, 6, 8],
[3, 5, 7]])
Or with np.take_along_axis:
np.take_along_axis(A, B, axis=0)
array([[1, 4, 9],
[2, 6, 8],
[3, 5, 7]])
Related
Let's say I have data structured in a 2D array like this:
[[1, 3, 4, 6],
[1, 4, 8, 2],
[1, 3, 2, 9],
[2, 2, 4, 8],
[2, 4, 9, 1],
[2, 2, 9, 3]]
The first column denotes a third dimension, so I want to convert this to the following 3D array:
[[[3, 4, 6],
[4, 8, 2],
[3, 2, 9]],
[[2, 4, 8],
[4, 9, 1],
[2, 9, 3]]]
Is there a built-in numpy function to do this?
You can try code below:
import numpy as np
array = np.array([[1, 3, 4, 6],
[1, 4, 8, 2],
[1, 3, 2, 9],
[2, 2, 4, 8],
[2, 4, 9, 1],
[2, 2, 9, 3]])
array = np.delete(array, 0, 1)
array.reshape(2,3,-1)
Output
array([[[3, 4, 6],
[4, 8, 2],
[3, 2, 9]],
[[2, 4, 8],
[4, 9, 1],
[2, 9, 3]]])
However, this code can be used when you are aware of the array's shape. But if you are sure that the number of columns in the array is a multiple of 3, you can simply use code below to show the array in the desired format.
array.reshape(array.shape[0]//3,3,-3)
Use numpy array slicing with reshape function.
import numpy as np
arr = [[1, 3, 4, 6],
[1, 4, 8, 2],
[1, 3, 2, 9],
[2, 2, 4, 8],
[2, 4, 9, 1],
[2, 2, 9, 3]]
# convert the list to numpy array
arr = np.array(arr)
# remove first column from numpy array
arr = arr[:,1:]
# reshape the remaining array to desired shape
arr = arr.reshape(len(arr)//3,3,-1)
print(arr)
Output:
[[[3 4 6]
[4 8 2]
[3 2 9]]
[[2 4 8]
[4 9 1]
[2 9 3]]]
You list a non numpy array. I am unsure if you are just suggesting numpy as a means to get a non numpy result, or you are actually looking for a numpy array as result. If you don't actually need numpy, you could do something like this:
arr = [[1, 3, 4, 6],
[1, 4, 8, 2],
[1, 3, 2, 9],
[2, 2, 4, 8],
[2, 4, 9, 1],
[2, 2, 9, 3]]
# Length of the 3rd and 2nd dimension.
nz = arr[-1][0] + (arr[0][0]==0)
ny = int(len(arr)/nz)
res = [[arr[ny*z_idx+y_idx][1:] for y_idx in range(ny)] for z_idx in range(nz)]
OUTPUT:
[[[3, 4, 6], [4, 8, 2], [3, 2, 9]], [[2, 4, 8], [4, 9, 1], [2, 9, 3]]]
Note that the calculation of nz takes into account that the 3rd dimension index in your array is either 0-based (as python is per default) or 1-based (as you show in your example).
I have two numpy 2d arrays, and I would like to stack them but into specific row on where I'm going to put them.
a = ([[4, 2],
[7, 3],
[1, 8]])
b = ([[10, 6], (put in 3rd row)
[9, 5]]) (put in 5th row)
Expected output = ([[4, 2],
[7, 3],
[10, 6],
[1, 8],
[9, 5]])
What is the fastest way to do this in python?
For your particular example:
import numpy as np
a = np.array([[1, 2],[3, 4],[7, 8]])
b = np.array([[5, 6],[9, 10]])
np.insert(a,[2,3],b,axis=0)
output:
array([[ 1, 2],
[ 3, 4],
[ 5, 6],
[ 7, 8],
[ 9, 10]])
Let's say I have this numpy matrix:
>>> mat = np.matrix([[3,4,5,2,1], [1,2,7,6,5], [8,9,4,5,2]])
>>> mat
matrix([[3, 4, 5, 2, 1],
[1, 2, 7, 6, 5],
[8, 9, 4, 5, 2]])
Now let's say I have some indexes in this form:
>>> ind = np.matrix([[0,2,3], [0,4,2], [3,1,2]])
>>> ind
matrix([[0, 2, 3],
[0, 4, 2],
[3, 1, 2]])
What I would like to do is to get three values from each row of the matrix, specifically values at columns 0, 2, and 3 for the first row, values at columns 0, 4 and 2 for the second row, etc. This is the expected output:
matrix([[3, 5, 2],
[1, 5, 7],
[5, 9, 4]])
I've tried using np.take but it doesn't seem to work. Any suggestion?
This is take_along_axis.
>>> np.take_along_axis(mat, ind, axis=1)
matrix([[3, 5, 2],
[1, 5, 7],
[5, 9, 4]])
This will do it: mat[np.arange(3).reshape(-1, 1), ind]
In [245]: mat[np.arange(3).reshape(-1, 1), ind]
Out[245]:
matrix([[3, 5, 2],
[1, 5, 7],
[5, 9, 4]])
(but take_along_axis in #user3483203's answer is simpler).
How can I sort 2d array in NumPy by the 1st column ascending and 2nd column descending?
For example,
a = array([[9, 2, 3],
[4, 5, 6],
[7, 0, 5],
[7, 1, 6]])
Result :
array([[4, 5, 6],
[7, 1, 6],
[7, 0, 5],
[9, 2, 3]])
You can use the np.lexsort function for this
import numpy as np
a = np.asarray([[9, 2, 3],
[4, 5, 6],
[7, 0, 5],
[7, 1, 6]])
a[np.lexsort((-a[:, 1], a[:, 0]))]
Output
array([[4, 5, 6],
[7, 1, 6],
[7, 0, 5],
[9, 2, 3]])
let's say I have the following 3x4 array
master_array = [[1, 3, 4, 5],
[6, 5, 4, 1],
[7, 8, 4, 1]]
Then, I want to delete number 4 from each of the 3 1x4 subarrays. Would I use the following?
for i in range(master_array.shape[0]):
np.delete(master_array[i], 3)
Then, when I print the master_array, would I get?
[[1, 3, 5],
[6, 5, 1],
[7, 8, 1]]
In case master_array is a list of lists, like in your example, you could do
master_array = [[1, 3, 4, 5],
[6, 5, 4, 1],
[7, 8, 4, 1]]
for row in master_array:
del row[2]
In case master_array is indeed a numpy array, you would simply do
master_array = np.array([[1, 3, 4, 5],
[6, 5, 4, 1],
[7, 8, 4, 1]])
np.delete(master_array, 2, axis=1)