Lets assume I have the follwing arrays:
distance = np.array([2, 3, 5, 4, 8, 2, 3])
idx = np.array([0, 0, 1, 1, 1, 2, 2 ])
Now I want the smallest distance within one index. So my goald would be:
result = [2, 4, 2]
My only idea right now would be something like this:
for i in idx_unique:
result.append(np.amin(distances[np.argwhere(idx = i)]))
But is there a faster way without a loop??
You can convert idx to a boolean vector to use indexing within the distance vector:
distance = np.array([2, 3, 5, 4, 8])
idx = np.array([0, 0, 1, 1, 1]).astype(np.bool)
result = [np.min(distance[~idx]), np.min(distance[idx])]
Although not truly free from loops, here is one way to do that:
import numpy as np
distance = np.array([2, 3, 5, 4, 8, 2, 3])
idx = np.array([0, 0, 1, 1, 1, 2, 2 ])
t = np.split(distance, np.where(idx[:-1] != idx[1:])[0] + 1)
print([np.min(x) for x in t])
Actually, this provides no improvement as both the OP's solution and this one has the same runtime:
res1 = []
def soln1():
for i in idx_unique:
res1.append(np.amin(distances[np.argwhere(idx = i)]))
def soln2():
t = np.split(distance, np.where(idx[:-1] != idx[1:])[0] + 1)
res2 = [np.min(x) for x in t]
Timeit gives:
%timeit soln1
#10000000 loops, best of 5: 24.3 ns per loop
%timeit soln2
#10000000 loops, best of 5: 24.3 ns per loop
