How to left join numpy array python

Question

What’s the numpy "pythonic" way to left join arrays? Let’s say I have two 2-D arrays that share a key:

a.shape # (20, 2)
b.shape # (200, 3)

Both arrays share a common key in their first dimension:
a[:, 0] # values from 0..19
b[:, 0] # values from 0..19

How are I left join the values from a[:, 1] to b?

Asked By: Ruslan

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Source

Answer 1

It’s trivial with pandas:

import numpy as np
import pandas as pd

a = np.array([[4, 9], [5, 8], [6, 7]])
b = np.array([[5, 4, 1], [4, 6, 8], [5, 4, 8], [3, 8, 9]])

dfa = pd.DataFrame(a).add_prefix('a')
dfb = pd.DataFrame(b).add_prefix('b')
out = dfa.merge(dfb, left_on='a0', right_on='b0', how='left')

Output:

>>> a
array([[4, 9],
       [5, 8],
       [6, 7]])

>>> b
array([[5, 4, 1],
       [4, 6, 8],
       [5, 4, 8],
       [3, 8, 9]])

>>> out.values
array([[ 4.,  9.,  4.,  6.,  8.],
       [ 5.,  8.,  5.,  4.,  1.],
       [ 5.,  8.,  5.,  4.,  8.],
       [ 6.,  7., nan, nan, nan]])

It’s more complicated with numpy to do a left join:

import numpy as np

a = np.array([[4, 9], [5, 8], [6, 7]])
b = np.array([[5, 4, 1], [4, 6, 8], [5, 4, 8], [3, 8, 9]])

i, j = np.where(a[:, 0, None] == b[:, 0])
k = np.setdiff1d(np.arange(len(a)), i)
c = np.pad(a[k], [(0, 0), (0, b.shape[1])], constant_values=-1)
out = np.vstack([np.hstack([a[i], b[j]]), c])

Output:

>>> out
array([[ 4,  9,  4,  6,  8],
       [ 5,  8,  5,  4,  1],
       [ 5,  8,  5,  4,  8],
       [ 6,  7, -1, -1, -1]])

Answered By: Corralien

How to left join numpy array python

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