I'm trying to sort the rows of one array by the values of another. For example:
import numpy as np
arr1 = np.random.normal(1, 1, 80)
arr2 = np.random.normal(1,1, (80,100))
I want to sort arr1 in descending order, and to have the current relationship between arr1 and arr2 to be maintained (ie, after sorting both, the rows of arr1[0] and arr2[0, :] are the same).
Use argsort as follows:
arr1inds = arr1.argsort()
sorted_arr1 = arr1[arr1inds[::-1]]
sorted_arr2 = arr2[arr1inds[::-1]]
This example sorts in descending order.
len(arr1) has to match the rows of arr2. For example, I had a (5,) vector and a (2, 5) array, which I wanted to sort column-wise by the first vector. I did this using two transposes like arr2.T[arr1.argsort()].T, although there is probably a more elegant solution that doesn't require two transposes.
arr = np.array([4, 1, 5, 3, 2]); arr[np.argsort(arr)] gives me the output [1, 2, 3, 4, 5].
argsort sorts in ascending order, but this solution accesses the indices in reverse order (with [::-1]) to give an answer in descending order.
argsort(axis=N), you'll have to use numpy's take_along_axis instead of advanced slicing to sort.Use the zip function: zip( *sorted( zip(arr1, arr2) ) ) This will do what you need.
Now the explanation:
zip(arr1, arr2) will combine the two lists, so you've got [(0, [...list 0...]), (1, [...list 1...]), ...]
Next we run sorted(...), which by default sorts based on the first field in the tuple.
Then we run zip(...) again, which takes the tuples from sorted, and creates two lists, from the first element in the tuple (from arr1) and the second element (from arr2).
