Tricky numpy argmax on last dimension of 3-dimensional ndarray

Question

if have an array of shape (9,1,3).

array([[[  6,  12, 108]],

   [[122, 112,  38]],

   [[ 57, 101,  62]],

   [[119,  76, 177]],

   [[ 46,  62,   2]],

   [[127,  61, 155]],

   [[  5,   6, 151]],

   [[  5,   8, 185]],

   [[109, 167,  33]]])

I want to find the argmax index of the third dimension, in this case it would be 185, so index 7.

I guess the solution is linked to reshaping but I can't wrap my head around it. Thanks for any help!

It doesn't really make much sense to say "the argmax index of the third dimension". You could ask for the third coordinate of the argmax (which would be 2, not 7) or the first coordinate, which would be 7, or the argmax of the sliced array after taking [:, 0, 2] (as PP suggested, which is indeed 7). — DSM, Jan 20 '18 at 19:30
I don't get why every item is a singleton list, seems like a bad modeling. — Willem Van Onsem, Jan 20 '18 at 19:41
@kmario23 or `np.unravel_index(np.argmax(arr), arr.shape)`? Anything that results in something with a seven in it. — Paul Panzer, Jan 20 '18 at 20:06

kmario23 · Answer 1 · 2018-01-20T20:58:01.180

2

I'm not sure what's tricky about it. But, one way to get the index of the greatest element along the last axis would be by using np.max and np.argmax like:

# find `max` element along last axis 
# and get the index using `argmax` where `arr` is your array
In [53]: np.argmax(np.max(arr, axis=2))
Out[53]: 7

Alternatively, as @PaulPanzer suggested in his comments, you could use:

In [63]: np.unravel_index(np.argmax(arr), arr.shape)
Out[63]: (7, 0, 2)

In [64]: arr[(7, 0, 2)]
Out[64]: 185

edited Jan 20 '18 at 20:58

answered Jan 20 '18 at 19:57

kmario23

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Oh, please don't promote my throw-away comments ;-) Or, if you must use the other one; I think it's better. – Paul Panzer Jan 20 '18 at 20:11
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Thank you, trickiness is relative and I haven't spent that much time with np as of yet, although I will in the future. – Tshosh Harris Jan 21 '18 at 09:11
@TshoshHarris Sure! but, You don't have to thank explicitly. Upvoting answers, if it was helpful to you, is more than enough in SO. – kmario23 Jan 21 '18 at 15:35
1

@kmario23 You're right, and I do, but you have to get up to 15 Rep until it's visible ;) – Tshosh Harris Jan 21 '18 at 16:11

score 1 · Accepted Answer · edited Jan 20 '18 at 22:29

1

You may have to do it like this:

data = np.array([[[  6,  12, 108]],

   [[122, 112,  38]],

   [[ 57, 101,  62]],

   [[119,  76, 177]],

   [[ 46,  62,   2]],

   [[127,  61, 155]],

   [[  5,   6, 151]],

   [[  5,   8, 185]],

   [[109, 167,  33]]])

np.argmax(data[:,0][:,2])
 7

edited Jan 20 '18 at 22:29

Xenolion

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answered Jan 20 '18 at 19:25

jrjames83

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What's `data[:, 0][:, 2]`? Please just do `data[:, 0, 2]` instead of creating an unnecessary intermediate. – Paul Panzer Jan 20 '18 at 19:30
Yeah, very good. But you are not helping by spreading this kind sloppiness. – Paul Panzer Jan 20 '18 at 19:35

Tricky numpy argmax on last dimension of 3-dimensional ndarray

2 Answers2