How to get 2d array of indices from 1d array?

Question

I'm looking for an efficient way to return indices for a 2d array based on values in a 1d array. I currently have a nested for loop set up that is painfully slow.

Here is some example data and what I want to get:

data2d = np.array( [  [1,2] , [1,3] ,[3,4], [1,2] , [7,9] ])

data1d = np.array([1,2,3,4,5,6,7,8,9])

I would like to return the indices where data2d is equal to data1d. My desired output would be this 2d array:

locs = np.array([[0, 1], [0, 2], [2, 3], [0, 1], [6, 8]])

The only thing I've come up with is the nested for loop:

locs = np.full((np.shape(data2d)), np.nan)

for i in range(0, 5):
    for j in range(0, 2):
        loc_val = np.where(data1d == data2d[i, j])
        loc_val = loc_val[0]
        locs[i, j] = loc_val

This would be fine for a small set of data but I have 87,600 2d grids that are each 428x614 grid points.

Answer 1

Use np.searchsorted :

np.searchsorted(data1d, data2d.ravel()).reshape(data2d.shape)

array([[0, 1],
       [0, 2],
       [2, 3],
       [0, 1],
       [6, 8]])

searchsorted performs binary search with the ravelled data2d . The result is then reshaped.

---

Another option is to build an index and query it in constant time. You can do this with pandas' Index API.

import pandas as pd

idx = pd.Index([1,2,3,4,5,6,7,8,9])
idx
#  Int64Index([1, 2, 3, 4, 5, 6, 7, 8, 9], dtype='int64')

idx.get_indexer(data2d.ravel()).reshape(data2d.shape)

array([[0, 1],
       [0, 2],
       [2, 3],
       [0, 1],
       [6, 8]])

Answer 2

This should be fast also

import numpy as np
data2d = np.array( [  [1,2] , [1,3] ,[3,4], [1,2] , [7,9] ])
data1d = np.array([1,2,3,4,5,6,7,8,9])
idxdict = dict(zip(data1d,range(len(data1d))))
locs = data2d
for i in range(len(locs)):
    for j in range(len(locs[i])):
        locs[i][j] = idxdict[locs[i][j]]