I have a matrix 1500X2, and I have to create 10 submatrices of 150 rows. How can i do this without for loop. I need a function, because with the [:] is too slow and complicated
You could use the numpy.take function to select a range of rows of your matrix. You can pass the indices you want to select and the axis over which you want to select your items.
import numpy as np
indices = list(range(0,3))
array = np.random.rand(5,2)
print(array)
res = np.take(array, indices, 0)
print(res)
You get something like this:
[[0.63680493 0.27066094]
[0.71182288 0.48258969]
[0.61321531 0.02215374]
[0.98148503 0.5669895 ]
[0.42720908 0.57326236]]
[[0.63680493 0.27066094]
[0.71182288 0.48258969]
[0.61321531 0.02215374]]
Related
I have a numpy array with n row and p columns.
I want to check if a given row is in my array and find the index.
For exemple I have a numpy array like this :
[[1,0,8,7,2,2],[1,3,7,0,3,0],[1,7,1,0,1,0],[1,9,1,0,6,0],[1,8,1,7,9,0],....]
I want to check if this array [6,0,5,8,2,1] is in my numpy array or and where.
Is there a numpy function for that ?
I'm sorry for asking naive question but I'm quite confuse right now.
You can use == and .all(axis=1) to match entire rows, then use numpy.where() to get the index:
import numpy as np
a = np.array([[1,0,8,7,2,2],[1,3,7,0,3,0],[1,7,1,0,1,0],[1,9,1,0,6,0],[1,8,1,7,9,0], [6,0,5,8,2,1]])
b = np.array([6,0,5,8,2,1])
print(np.where((a==b).all(axis=1)))
Output:
(array([5], dtype=int32),)
I have a numpy multidimensional array with shape = (12,2,3,3)
import numpy as np
arr = np.arange(12*2*3*3).reshape((12,2,3,3))
I need to select those elements based on the 2nd dimension where the dindices are stored in another list
indices = [0,1,0,0,1,1,0,1,1,0,1,1]
in one array, and the rest in another array.
the output in either case should be in another array of shape (12,3,3)
arr2 = np.empty((arr.shape[0],*arr.shape[-2:]))
I could do it using a for loop
for i, ii in enumerate(indices):
arr2[i] = arr[i, indices[ii],...]
However, I am searching for a one liner.
When I try indexing using the list as indices
test = arr[:,indices,...]
I get test of shape (12,12,3,3) instead of (12,3,3). Could you help me please?
You can use np.arange for indexing the first dimension:
test = arr[np.arange(arr.shape[0]),indices,...]
or just the python range function:
test = arr[range(arr.shape[0]),indices,...]
I'm unsure of how to access an element in an array (of arrays?). Basically, I need to be able to assign random numbers to a series of arrays but I'm not sure how indexing works.
array_20 = np.zeros((5,10))
a = [[array_20]]*10
#This gives me 10 arrays of 5x10. I'd like to be able to then assign random
#numbers to all of the elements.
You could use numpy.random.rand like so:
import numpy as np
a = np.random.rand(10, 5, 10)
You can then index a like a python list. (i.e. a[1][2][0])
I have a numpy array of indexes e.g. [1,3,12]. I want to create another array from this such that at these indexes, I get a non-zero elements e.g. 1. So in this case, with input [1,3,12], I should get [0,1,0,1,0,0,0,0,0,0,0,0,1]. I can do it in a for loop, is there a short numpy function to achieve this?
With numpy you can index with lists directly:
a = [1,3,12]
vector = numpy.zeros(shape=max(a) + 1)
vector[a] = 1
I have numpy array called data of dimensions 150x4
I want to create a new numpy array called mean of dimensions 3x4 by choosing random elements from data.
My current implementation is:
cols = (data.shape[1])
K=3
mean = np.zeros((K,cols))
for row in range(K):
index = np.random.randint(data.shape[0])
for col in range(cols):
mean[row][col] = data[index][col]
Is there a faster way to do the same?
You can specify the number of random integers in numpy.randint (third argument). Also, you should be familiar with numpy.array's index notations. Here, you can access all the elements in one row by : specifier.
mean = data[np.random.randint(0,len(data),3),:]