If I have a 2D numpy array A:
[[6 9 6]
[1 1 2]
[8 7 3]]
And I have access to array [1 1 2]. Clearly, [1 1 2] belongs to index 1 of array A. But how do I do this?
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4 Answers
Access the second row using the following operator:
import numpy as np
a = np.array([[6, 9, 6],
[1, 1, 2],
[8, 7, 3]])
row = [1, 1, 2]
i = np.where(np.all(a==row, axis=1))
print(i[0][0])
np.where will return a tuple of indices (lists), which is why you need to use the operators [0][0] consecutively in order to obtain an int.
1 Comment
Black Beard 53
Sorry if I wasn't clear but my question was how do I code it so that the program returns 1 given that I know that array [1 1 2] is a member of the matrix A.
One option:
a = np.array([[6, 9, 6],
[1, 1, 2],
[8, 7, 3]])
b = np.array([1, 1, 2])
np.nonzero((a == b).all(1))[0]
output: [1]
Comments
arr1 = [[6,9,6],[1,1,2],[8,7,3]]
ind = arr1.index([1,1,2])
Output:
ind = 1
EDIT for 2D np.array:
arr1 = np.array([[6,9,6],[1,1,2],[8,7,3]])
ind = [l for l in range(len(arr1)) if (arr1[l,:] == np.array([1,1,2])).all()]
2 Comments
Black Beard 53
I don't think it works if my array is a numpy araay. I need to convert it to list instead.
imburningbabe
I edited for 2D np.array
import numpy as np
a = np.array([[6, 9, 6],
[1, 1, 2],
[8, 7, 3]])
b = np.array([1, 1, 2])
[x for x,y in enumerate(a) if (y==b).all()] # here enumerate will keep the track of index
#output
[1]
