How to get 1:1 corresponding matches using sklearn KNearest neighbors

Question

I’m writing an algorithm to match each person from setA with someone from setB, based on interest similarity, using NearestNeighbors(n_neighbors = 1).

This is what I have so far:

dfA = pd.DataFrame(np.array([[1, 1, 1, 1], [1,1,2,2], [4, 5, 2, 0], [8, 8, 8, 8]]),
                   columns=['interest0', 'interest2', 'interest3','interest4'],
                  index=['personA0','personA1','personA2','personA3'])


dfB = pd.DataFrame(np.array([[1, 1, 1, 1], [1, 1, 1, 2], [2,3,2,2], [8, 6, 8, 8]]),
                   columns=['interest0', 'interest2', 'interest3','interest4'],
                  index=['personB0','personB1','personB2','personB3'])


knn = NearestNeighbors(n_neighbors = 1, metric = my_dist).fit(dfA)
distances, indices = knn.kneighbors(dfB)


>>> dfA
          drink  interest2  interest3  interest4
personA0      1          1          1          1
personA1      1          1          2          2
personA2      4          5          2          0
personA3      8          8          8          8



>>> dfB
          drink  interest2  interest3  interest4
personB0      1          1          1          1
personB1      1          1          1          2
personB2      2          3          2          2
personB3      8          6          8          8

>>> print("Distancesnn", distances, "nnIndicesnn", indices)

Distances

 [[0.   ]
 [0.125]
 [1.125]
 [0.5  ]] 

Indices

 [[0]
 [0]
 [1]
 [3]]

Looking at the output, it suggests personB0’s top match is personA0 (distance = 0). However, personB1’s top match is also personA0(distance = 0.125)!

I want to somehow match personB0 with personA0 (as their distances are smallest), move them to another table, then re-run the K-Neighbors algorithm, which will hopefully now suggest personB1’s top match is personA1 (as A0 is now removed). I’ve started to write a for loop to iterate through this, however, it’s quite complicated for me (having to iterate through multiple different arrays, dataframes etc) so I’m wondering what is the best way? I want a final dataframe like below, which has 1:1 correspondence:

  SetA         SetB
personA0     personB0
personA1     personB1
personA2     personB3
personA3     personB2

Asked By: chillingfox

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Answer 1

You could use a list to check whether a person has been matched or not. Besides, you need to get a list of neighbours ordered by their distance rather than the nearest neighbour by changing tha value passed to parameter n_neighbors.

knn = NearestNeighbors(n_neighbors=len(dfB)).fit(dfB)
distances, indices = knn.kneighbors(dfA)

matched = []
pairs = []
for indexA, candidatesB in enumerate(indices):
    personA = dfA.index[indexA]
    for indexB in candidatesB:
        if indexB not in matched:
            matched.append(indexB)
            personB = dfB.index[indexB]
            pairs.append([personA, personB])
            break

matches = pd.DataFrame(pairs, columns=['SetA', 'SetB'])

The resulting dataframe looks like this:

       SetA      SetB
0  personA0  personB0
1  personA1  personB1
2  personA2  personB2
3  personA3  personB3

Please notice that I have used the default metric (minkowski with p=2). Results may vary if you pass metric=my_dist to NearestNeighbors.

Answered By: Tonechas

How to get 1:1 corresponding matches using sklearn KNearest neighbors

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