CODEWITHSUNDEEP
pythonarraysnumpy
Trung Tran
Trung Tran

Reputation: 13721

Python - return intersection of two arrays

I have two arrays and I am trying to return a new array that equals the intersection of my original two arrays. The two original arrays should be of the same length. For example, if I have:

arr1 = np.array([(255, 255, 255), (255, 255, 255)])

arr2 = np.array([(255, 255, 255), (255, 255, 255)])

I should get:

intersectedArr = ([(255, 255, 255), (255, 255, 255])

However, if I have:

arr1 = np.array([(100, 100, 100), (255, 255, 255)])

arr2 = np.array([(255, 255, 255), (255, 255, 255)])

I should get

([(255, 255, 255)])

So far i've tried:

intersectedArr = np.intersect1d(arr1, arr2)

but this returns [255] instead of the expected ([(255, 255, 255)])

Can someone help? Thanks in advance!

Upvotes: 2

Views: 25496

Answers (6)

active_VTA
active_VTA

Reputation: 11

In your case, you want to compare against rows instead of elements, so it`s a matter of 2D array. I would recommend an improvement of intersect1d which is intersection of 2D numpy arrays. I found a good solution here Intersection of 2D numpy ndarrays.

def multidim_intersect(arr1, arr2):
    arr1_view = arr1.view([('',arr1.dtype)]*arr1.shape[1])
    arr2_view = arr2.view([('',arr2.dtype)]*arr2.shape[1])
    intersected = numpy.intersect1d(arr1_view, arr2_view)
    return intersected.view(arr1.dtype).reshape(-1, arr1.shape[1])

The code above convert the shape of the original array, combine them row-wise, and then convert it back to 2-dim shape.

Upvotes: 0

Matt Rundle
Matt Rundle

Reputation: 81

If you want to keep duplicates, like in your examples, you can use a list comprehension:

def intersection(list_a, list_b):
    return [ e for e in list_a if e in list_b ]

which produces:

in:
    [(255, 255, 255), (255, 255, 255)]
    [(255, 255, 255), (255, 255, 255)]
out:
    [(255, 255, 255), (255, 255, 255)]

in:
    [(100, 100, 100), (255, 255, 255)]
    [(255, 255, 255), (255, 255, 255)]
out:
    [(255, 255, 255)]

If you want uniquie combinations between the lists (sets) though:

def intersection(a, b):
    return list(set(a).intersection(b))

which produces:

in:
    [(255, 255, 255), (255, 255, 255)]
    [(255, 255, 255), (255, 255, 255)]
out:
    [(255, 255, 255)]

in:
    [(100, 100, 100), (255, 255, 255)]
    [(255, 255, 255), (255, 255, 255)]
out:
    [(255, 255, 255)]

Cheers!

Upvotes: 7

Andy Hayden
Andy Hayden

Reputation: 375415

For larger arrays it might help to use pandas' groupby and cumcount:

In [11]: df1 = pd.DataFrame(arr1)

In [12]: df1["cumcount"] = df1.groupby([0, 1, 2]).cumcount()

In [13]: df1
Out[13]:
     0    1    2  cumcount
0  100  100  100         0
1  255  255  255         0

In [14]: df2 = pd.DataFrame(arr2)

In [15]: df2["cumcount"] = df2.groupby([0, 1, 2]).cumcount()

In [16]: df2
Out[16]:
     0    1    2  cumcount
0  255  255  255         0
1  255  255  255         1

Now a merge gets you the array you desire:

In [21]: df1.merge(df1).iloc[:, :3].values
Out[21]:
array([[100, 100, 100],
       [255, 255, 255]])

In [22]: df1.merge(df2).iloc[:, :3].values
Out[22]: array([[255, 255, 255]])

In [23]: df2.merge(df2).iloc[:, :3].values
Out[23]:
array([[255, 255, 255],
       [255, 255, 255]])

Upvotes: 0

f5r5e5d
f5r5e5d

Reputation: 3706

how about a numpy answer? 

import numpy as np


arr1 = np.array([(255, 255, 255), (255, 255, 25)])  # changed some to 25
arr2 = np.array([(255, 25, 255), (255, 255, 25)])

arr1[np.where(arr1==arr2)]

array([255, 255, 255, 255,  25])

2nd example 

arr1 = np.array([(100, 100, 100), (255, 255, 255)])
arr2 = np.array([(255, 255, 255), (255, 255, 255)])

arr1[np.where(arr1==arr2)]

array([255, 255, 255])

Upvotes: 3

Ken Y-N
Ken Y-N

Reputation: 15009

NOTE: This assumes [a, b, c] and [b, c, a] gives [a, b, c], that is the order of elements is ignored.

OK, I've done a little experimenting and this might be what you are after. Given:

arr1a = np.array([(255, 255, 255), (255, 255, 255)])
arr1b = np.array([(100, 100, 100), (255, 255, 255)])
arr2 = np.array([(255, 255, 255), (255, 255, 255)])

Then we can find an intersection with:

np.array([item in arr2 for item in arr1a])

ie, for each element in arr1a, check to see it appears in arr2 also. This gives a result of:

>>> array([ True,  True], dtype=bool)

Similarly:

np.array([item in arr2 for item in arr1b])
>>> array([False,  True], dtype=bool)

Now, we can use this result to pick the common values from the original lists:

mask = np.array([item in arr2 for item in arr1a])
arr1a[mask]
>>> array([[255, 255, 255],
           [255, 255, 255]])

And:

mask = np.array([item in arr2 for item in arr1b])
arr1b[mask]
>>> array([[255, 255, 255]])

Upvotes: 0

AC1009
AC1009

Reputation: 64

Not sure how big your arrays will get, but if they remain fairly small, this could work:

import numpy as np

arr1 = np.array([(255, 255, 255), (255, 255, 255)])
arr2 = np.array([(255, 255, 255), (255, 255, 255)])
intersectedArr = []

for a1, a2 in zip(arr1, arr2):
    if np.array_equal(a1, a2):
        intersectedArr.append(a1)
print(np.array(intersectedArr))

arr1 = np.array([(100, 100, 100), (255, 255, 255)])
arr2 = np.array([(255, 255, 255), (255, 255, 255)])
intersectedArr = []

for a1, a2 in zip(arr1, arr2):
    if np.array_equal(a1, a2):
        intersectedArr.append(a1)
print(np.array(intersectedArr))

Upvotes: 3

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