Python Forum

I have two numpy-matrices ("A" and "B"). I want to compare every row-vector in "A" matrix with every row-vector in "B" matrix.

Here is my (not working) code. The expected output is 5.

import numpy as np
a=np.matrix([[1,2,3],[42,68,69],[1,2,3],[85,89,95]])
b=np.matrix([[42,68,69],[1,2,3],[85,89,95], [42,68,69]])


found_one=0
for i in range(3):
    for j in range(3):
        if a[i,]==b[j,]]:
            found_one=found_one+1
            
print(found_one)

Any suggestion?

I see a solution using collections.Counter

from collections import Counter
import numpy as np
a=np.matrix([[1,2,3],[42,68,69],[1,2,3],[85,89,95]])
b=np.matrix([[42,68,69],[1,2,3],[85,89,95], [42,68,69]])

ca, cb = (Counter(tuple(row) for row in m.A) for m in (a, b))
found = sum( v * cb.get(k, 0) for k, v in ca.items())
print(found)

If I correctly understand the objective (which in doubt) in 'pure' Python it could be done this way:

>>> a = [[1,2,3],[42,68,69],[1,2,3],[85,89,95]]
>>> b= [[42,68,69],[1,2,3],[85,89,95], [42,68,69]]
>>> len([(x, y) for x in a for y in b if x == y])
5

To 'translate' this into numpy array it needs little adjustments:

>>> import numpy as np
>>> a = np.matrix([[1,2,3],[42,68,69],[1,2,3],[85,89,95]])
>>> b = np.matrix([[42,68,69],[1,2,3],[85,89,95], [42,68,69]])
>>> len([(x, y) for x in a for y in b if np.array_equal(x, y)])
5

Thank you for the answers!

Use itertools.product, to avoid nested loops:

[x for x,y in itertools.product(a,b) if np.array_equal(x,y)]

There is also a hint on StackOverflow how to make a Cartesian product with numpy:

• https://stackoverflow.com/questions/2868...ls-product
  
• https://stackoverflow.com/questions/1208...two-arrays
  
• https://stackoverflow.com/questions/4709...t-speed-up
  

(Apr-18-2019, 12:19 PM)DeaD_EyE Wrote: [ -> ]Use itertools.product, to avoid nested loops:

[x for x,y in itertools.product(a,b) if np.array_equal(x,y)]

This is very good point.

To get answer OP looking for:

>>> sum(1 for x, y in product(a, b) if np.array_equal(x, y))
5

Note that the product method has complexity O(A*B) while the hashtable method has O(A+B) complexity. For large data, the Counter code should be faster, while the product method is probably faster for small data due to the current implementation.

Hi all!

Thank you for the solutions. I applied the Counter-code, because the matrices have lots of rows (much more than in my question).

Hi, after a long pause, I started to play with Python again. I have a question:

How can I print which row-vektors were identical, if I am using solution given by Gribouillis? (I want to print the ordinal number 'ordinal number' of the identical rows in both matrices, and then the coordinates of the identical vectors.

The only idea I have:
Transforming the matrices into two lists, and a nested for cycle...

a_lista=a.tolist()
b_lista=b.tolist()
#"a" and "b" were my previously defined np.matrices
for i in range(4):
   for j in range(4):
      if a_lista[i]==b_lista[j]:
         print(i,j,a_lista[i])

Unfortunately, my "solution" is realy crap, because I want to use this on far bigger matrices :-(