Given m x n numpy array
X = np.array([
[1, 2],
[10, 20],
[100, 200]
])
how to find index of a row, i.e. [10, 20] -> 1?
n could any - 2, 3, ..., so I can have n x 3 arrays
Y = np.array([
[1, 2, 3],
[10, 20, 30],
[100, 200, 300]
])
so I need to pass a vector of size n, in this case n=3, i.e a vector [10, 20, 30] to get its row index 1? Again, n could be of any value, like 100 or 1000.
Numpy arrays could be big, so I don't want to convert them to lists to use .index()
Just in case that the query array contains duplicate rows that you are looking for, the function below returns multiple indices in such case.
def find_rows(source, target):
return np.where((source == target).all(axis=1))[0]
looking = [10, 20, 30]
Y = np.array([[1, 2, 3],
[10, 20, 30],
[100, 200, 300],
[10, 20, 30]])
print(find_rows(source=Y, target=looking)) # [1, 3]
You can use numpy.equal, which will broadcast and compare row vector against each row of the original array, and if all elements of a row are equal to the target, the row is identical to the target:
import numpy as np
np.flatnonzero(np.equal(X, [10, 20]).all(1))
# [1]
np.flatnonzero(np.equal(Y, [10, 20, 30]).all(1))
# [1]
You can make a function as follow:
def get_index(seq, *arrays):
for array in arrays:
try:
return np.where(array==seq)[0][0]
except IndexError:
pass
then:
>>>get_index([10,20,30],Y)
1
Or with just indexing:
>>>np.where((Y==[10,20,30]).all(axis=1))[0]
1
Y = np.array([ [10, 2, 3], [10, 20, 30], [100, 20, 30] ]) so np.where(Y[:,:]==[10, 20, 30]) gives (array([0, 1, 1, 1, 2, 2]), array([0, 0, 1, 2, 1, 2]))
[10, 20]then you refer to a vector[10, 20, 30]: is the query tensor you are using to find the index of variable size?np.where((X[:,0] == 10) & (X[:,1] == 20))do the job, but I don't know how to make a condition to handle arbitrary number n of elements in a vector, i.e. how given a vector[10, 20, 30]automatically get a condition like(Y[:,0] == 10) & (Y[:,1] == 20) & (Y[:,2] == 30)