Claudiu - 6 months ago 48

Python Question

What's an efficient way, given a numpy matrix (2-d array), to return the min/max

`n`

`def n_max(arr, n):`

res = [(0,(0,0))]*n

for y in xrange(len(arr)):

for x in xrange(len(arr[y])):

val = float(arr[y,x])

el = (val,(y,x))

i = bisect.bisect(res, el)

if i > 0:

res.insert(i, el)

del res[0]

return res

This takes 3x longer than the image template matching algorithm that

`pyopencv`

Answer

Since there is no heap implementation in NumPy, probably your best guess is to sort the whole array and take the last `n`

elements:

```
def n_max(arr, n):
indices = arr.ravel().argsort()[-n:]
indices = (numpy.unravel_index(i, arr.shape) for i in indices)
return [(arr[i], i) for i in indices]
```

(This will probably return the list in reverse order compared to your implementation - did not check.)

**Edit**: A more efficient solution that works with newer versions of Numpy is given in this answer