CODEWITHSUNDEEP
pythonnumpy
Oroshimaru
Oroshimaru

Reputation: 338

Numpy: Get the smallest value within indices without loop?

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??

Upvotes: 2

Views: 175

Answers (2)

anurag
anurag

Reputation: 1932

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

Upvotes: 1

Josemi
Josemi

Reputation: 346

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])]

Upvotes: 1

Related Questions