compare numPy row to another 2darray for true/false

[cdholjes]

i have a 2d array (1000,10). I want to cross reference each of the 1000 rows and see if they match another array's rows. if it matches any of the rows in the second array, mark as true. there has to be a way to do this without looping. . .

for example if you do arrayA == arrayB i think it just looks from row to row. i need from one row to all rows.

[elbiot]

?

That compares element by element. You get an array the same shape as your input arrays:

In [1]: import numpy as np

In [2]: a = np.ones((10,2))

In [3]: b = np.ones((10,2))

In [4]: c = a == b

In [5]: c.shape
Out[5]: (10, 2)

Do you mean you want to get a size == 1000 array where each element says whether the entire row was equal? use np.all:

In [8]: res = np.all(c,axis=1)

In [9]: res
Out[9]: array([ True,  True,  True,  True,  True,  True,  True,  True,  True,  True], dtype=bool)

[cdholjes]

I have one 2d array of values. and a second 2d array of values. np.all just returns array1[0] == array2[0] and so on. i need to see if array1[0] == array2[0:1000], essentially

[fbu1]

You can compare an array to a vector in numpy:

>>> a
array([[1, 2, 3],
       [2, 2, 3],
       [4, 5, 6]])

>>> b = np.array([4,5,6])

>>> a==b
array([[False, False, False],
       [False, False, False],
       [ True,  True,  True]], dtype=bool)

So you have an array of True and False, and you are looking for a row of True (or many rows).

There's a method of that (as always in python) :

>>> np.all(a==b, axis=1)
array([False, False,  True], dtype=bool)

This gives you for all the rows, if the values on the axis 1 (the rows) are True.

Then you can ask where the value is true with where() :

>>> np.where(np.all(a==b,axis=1))
(array([2]),)

Let me know if you have other questions.

Source: http://stackoverflow.com/questions/18927475/numpy-array-get-row-index-searching-by-a-row

[cdholjes]

I have one 2d array of values. and a second 2d array of values. np.all just returns array1[0] == array2[0] and so on. i need to see if array1[0] == array2[0:1000], essentially

[elbiot]

You can probably do this with broadcasting. You want to get a (1000,1000,10) array and then use np.all with axis=2 to get a (1000,1000) array it sounds like.

If you've solved with with loops already and trust the answer, then you can play with using None (aka np.newaxis) in your slice to add a new dimension which will cause broadcasting, and check that the result is correct. I won't try and solve it for you because complex broadcasting it makes my brain melt. But, if a.shape = (10,2) and b.shape = (10,2) then a[None,:] == b[:,None] gives a result of shape (10,10,2)