计算 2D numpy 数组每行的重复数

Question

Is there a simple way in python to check the amount of duplicates in different rows. For example:

Row1: 12  13  20  25  45  46  
Row2: 14  24  30  38  39  47  
Row3:  1   9  15  21  29  39  
Row4:  2   6  14  19  26  45  
Row5:  5  23  25  27  32  40  
Row6:  6   8  25  26  27  45  

I want to compare the Row6 to previous "n" rows. If n=5, then the output should be something like this: [2 0 0 3 2]

Of course, I can compare each value in Row6 to each value from other row in the loop, and increase the counter for each row.

But do you know any already existing function in python?

Answer 1

If you're working with numpy arrays, use broadcasted comparison,

>>> n = 5
>>> v = df.values 
>>> v
array([[12, 13, 20, 25, 45, 46],
       [14, 24, 30, 38, 39, 47],
       [ 1,  9, 15, 21, 29, 39],
       [ 2,  6, 14, 19, 26, 45],
       [ 5, 23, 25, 27, 32, 40],
       [ 6,  8, 25, 26, 27, 45]])
>>> (v[None, -(n+1):-1, None] == v[-1, :, None]).sum(-1).sum(-1).squeeze()
array([2, 0, 0, 3, 2])

Answer 2

You could use unique from numpy

>>> import numpy as np 
>>> np.unique([1, 1, 2, 2, 3, 3])
array([1, 2, 3])
>>> a = np.array([[1, 1], [2, 3]])
>>> np.unique(a)
array([1, 2, 3])

You would still need to loop over the n rows and then checking the length of the resulting array. Maybe you find something more suitable still using numpy.