Matching array with elements in rows of matrix

Question

I have a matrix x_faces with 3 columns and N elements (in this example 4). Per row I want to know if it contains any of the elements from array matches

x_faces = [[   0,  43, 446],
           [   8,  43, 446],
           [   0,  10, 446],
           [   0,   5, 425]
          ]
matches = [8, 10, 44, 425, 440]

Which should return this:

results = [
        False,
        True,
        True,
        True
    ]

I can think of a for loop that does this, but is there a neat way to do this in python?

I found a faster way: np.any(np.in1d(x_faces, matches[i]).reshape(-1,3), axis=1) — bl3nd3d
– bl3nd3d, Commented Apr 5, 2016 at 14:39

xiº · Accepted Answer · 2016-04-04 15:50:52Z

2

You could use any() function for that purpose:

result = [any(x in items for x in matches) for items in x_faces]

Output:

[False, True, True, True]

answered Apr 4, 2016 at 15:50

xiº

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3 Comments

Ilja Over a year ago

Is this internally really different from a for-loop? Does list comprehension do some magis to be more effective additionaly to more good-looking? Edit: ah, the thing inside the any is a generator expression, right?

Iron Fist Over a year ago

I think you should link to any in Python3 version instead

bl3nd3d Over a year ago

Thanks, the any() function was what I was looking for

Ilja · Accepted Answer · 2016-04-04 15:59:08Z

0

You can use numpy and convert both arrays to 3D and compare them. Then I use sum to determine, whether any of the values in the last two axis is True:

x_faces = np.array([[   0,  43, 446],
           [   8,  43, 446],
           [   0,  10, 446],
           [   0,   5, 425]
          ])
matches = np.array([8, 10, 44, 425, 440])
shape1, shape2 = x_faces.shape, matches.shape
x_faces = x_faces.reshape((shape1 + (1, )))
mathes = matches.reshape((1, 1) + shape2)
equals = (x_faces == matches)
print np.sum(np.sum(equals, axis=-1), axis=-1, dtype=bool)

answered Apr 4, 2016 at 15:59

Ilja

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1 Comment

Ilja Over a year ago

maybe :) depends on the size, doesn't it. I thought, that what you do is in effect a for loop, and he asked for something else :)

Giuseppe Cammarota · Accepted Answer · 2016-04-04 16:00:57Z

0

I would do something like:

result = [any([n in row for n in matches]) for row in x_faces]

answered Apr 4, 2016 at 16:00

Giuseppe Cammarota

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1 Comment

Sнаđошƒаӽ Over a year ago

Generator expressions are best suited in cases like this. Take a look at this SO post, and this post any()+generator expression. Also, in this I tried to explain why generator expression is better in cases like this than list comprehensions.

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