删除 numpy 数组中第一次出现的元素

Question

I have a numpy array:

a = np.array([-1,2,3,-1,5,-2,2,9])

I want to only keep values in the array which occurs more than 2 times, so the result should be:

a = np.array([-1,2,-1,2])

Is there a way to do this only using numpy? I have a solution using a dictionary and dictionary filtering, but this is kind of slow, and I was wondering if there was a faster solution only using numpy.

Thanks !

Answer 1

import numpy as np
a = np.array([-1, 2, 3, -1, 5, -2, 2, 9])
values, counts = np.unique(a, return_counts=True)
values_filtered = values[counts >= 2]
result = a[np.isin(a, values_filtered)]
print(result)  # return [-1 2 -1 2]

Answer 2

import numpy as np

arr = np.array([1, 2, 3,4,4,4,1])

filter_arr = [np.count_nonzero(arr == i)>1 for i in arr]

newarr = arr[filter_arr]

print(filter_arr)
print(np.unique(newarr))

Answer 3

Here's one way using broadcasting and some advanced indexing techniques:

In [24]: a[(a[:, None] == a).sum(0) > 1]
Out[24]: array([-1,  2, -1,  2])

Answer 4

thanks a lot!

All answers solved the problem, but the solution from Matvey_coder3 was the fastest.

KR