Pythonic way to test if a row is in an array

Question

This seems like a simple question, but I haven't been able to find a good answer.

I'm looking for a pythonic way to test whether a 2d numpy array contains a given row. For example:

myarray = numpy.array([[0,1],
                       [2,3],
                       [4,5]])

myrow1 = numpy.array([2,3])
myrow2 = numpy.array([2,5])
myrow3 = numpy.array([0,3])
myrow4 = numpy.array([6,7])

Given myarray, I want to write a function that returns True if I test myrow1, and False if I test myrow2, myrow3 and myrow4.

I tried the "in" keyword, and it didn't give me the results I expected:

>>> myrow1 in myarray
True
>>> myrow2 in myarray
True
>>> myrow3 in myarray
True
>>> myrow4 in myarray
False

It seems to only check if one or more of the elements are the same, not if all elements are the same. Can someone explain why that's happening?

I can do this test element by element, something like this:

def test_for_row(array,row):
    numpy.any(numpy.logical_and(array[:,0]==row[0],array[:,1]==row[1]))

But that's not very pythonic, and becomes problematic if the rows have many elements. There must be a more elegant solution. Any help is appreciated!

Answer 1

The SO question below should help you out, but basically you can use:

any((myrow1 == x).all() for x in myarray)

Numpy.Array in Python list?