PyTorch torch.max over multiple dimensions

Question

Have tensor like :x.shape = [3, 2, 2].

import torch

x = torch.tensor([
    [[-0.3000, -0.2926],[-0.2705, -0.2632]],
    [[-0.1821, -0.1747],[-0.1526, -0.1453]],
    [[-0.0642, -0.0568],[-0.0347, -0.0274]]
])

I need to take .max() over the 2nd and 3rd dimensions. I expect some like this [-0.2632, -0.1453, -0.0274] as output. I tried to use: x.max(dim=(1,2)), but this causes an error.

Answer 1

Now, you can do this. The PR was merged (Aug 28 2020) and it is now available in the nightly release.

Simply use torch.amax():

import torch

x = torch.tensor([
    [[-0.3000, -0.2926],[-0.2705, -0.2632]],
    [[-0.1821, -0.1747],[-0.1526, -0.1453]],
    [[-0.0642, -0.0568],[-0.0347, -0.0274]]
])

print(torch.amax(x, dim=(1, 2)))

# Output:
# >>> tensor([-0.2632, -0.1453, -0.0274])

---

Original Answer

As of today (April 11, 2020), there is no way to do .min() or .max() over multiple dimensions in PyTorch. There is an open issue about it that you can follow and see if it ever gets implemented. A workaround in your case would be:

import torch

x = torch.tensor([
    [[-0.3000, -0.2926],[-0.2705, -0.2632]],
    [[-0.1821, -0.1747],[-0.1526, -0.1453]],
    [[-0.0642, -0.0568],[-0.0347, -0.0274]]
])

print(x.view(x.size(0), -1).max(dim=-1))

# output:
# >>> values=tensor([-0.2632, -0.1453, -0.0274]),
# >>> indices=tensor([3, 3, 3]))

So, if you need only the values: x.view(x.size(0), -1).max(dim=-1).values.

If x is not a contiguous tensor, then .view() will fail. In this case, you should use .reshape() instead.

---

Update August 26, 2020

This feature is being implemented in PR#43092 and the functions will be called amin and amax. They will return only the values. This is probably being merged soon, so you might be able to access these functions on the nightly build by the time you're reading this :) Have fun.

Answer 2

Although the solution of Berriel solves this specific question, I thought adding some explanation might help everyone to shed some light on the trick that's employed here, so that it can be adapted for (m)any other dimensions.

Let's start by inspecting the shape of the input tensor x:

In [58]: x.shape   
Out[58]: torch.Size([3, 2, 2])

So, we have a 3D tensor of shape (3, 2, 2). Now, as per OP's question, we need to compute maximum of the values in the tensor along both 1^st and 2^nd dimensions. As of this writing, the torch.max()'s dim argument supports only int. So, we can't use a tuple. Hence, we will use the following trick, which I will call as,

The Flatten & Max Trick: since we want to compute max over both 1^st and 2^nd dimensions, we will flatten both of these dimensions to a single dimension and leave the 0^th dimension untouched. This is exactly what is happening by doing:

In [61]: x.flatten().reshape(x.shape[0], -1).shape   
Out[61]: torch.Size([3, 4])   # 2*2 = 4

So, now we have shrinked the 3D tensor to a 2D tensor (i.e. matrix).

In [62]: x.flatten().reshape(x.shape[0], -1) 
Out[62]:
tensor([[-0.3000, -0.2926, -0.2705, -0.2632],
        [-0.1821, -0.1747, -0.1526, -0.1453],
        [-0.0642, -0.0568, -0.0347, -0.0274]])

Now, we can simply apply max over the 1^st dimension (i.e. in this case, first dimension is also the last dimension), since the flattened dimensions resides in that dimension.

In [65]: x.flatten().reshape(x.shape[0], -1).max(dim=1)    # or: `dim = -1`
Out[65]: 
torch.return_types.max(
values=tensor([-0.2632, -0.1453, -0.0274]),
indices=tensor([3, 3, 3]))

We got 3 values in the resultant tensor since we had 3 rows in the matrix.

---

Now, on the other hand if you want to compute max over 0^th and 1^st dimensions, you'd do:

In [80]: x.flatten().reshape(-1, x.shape[-1]).shape 
Out[80]: torch.Size([6, 2])    # 3*2 = 6

In [79]: x.flatten().reshape(-1, x.shape[-1]) 
Out[79]: 
tensor([[-0.3000, -0.2926],
        [-0.2705, -0.2632],
        [-0.1821, -0.1747],
        [-0.1526, -0.1453],
        [-0.0642, -0.0568],
        [-0.0347, -0.0274]])

Now, we can simply apply max over the 0^th dimension since that is the result of our flattening. ((also, from our original shape of (3, 2, 2), after taking max over first 2 dimensions, we should get two values as result.)

In [82]: x.flatten().reshape(-1, x.shape[-1]).max(dim=0) 
Out[82]: 
torch.return_types.max(
values=tensor([-0.0347, -0.0274]),
indices=tensor([5, 5]))

In a similar vein, you can adapt this approach to multiple dimensions and other reduction functions such as min.

---

Note: I'm following the terminology of 0-based dimensions (0, 1, 2, 3, ...) just to be consistent with PyTorch usage and the code.

Answer 3

If you only want to use the torch.max() function to get the indices of the max entry in a 2D tensor, you can do:

max_i_vals, max_i_indices = torch.max(x, 0)
print('max_i_vals, max_i_indices: ', max_i_vals, max_i_indices)
max_j_index = torch.max(max_i_vals, 0)[1]
print('max_j_index: ', max_j_index)
max_index = [max_i_indices[max_j_index], max_j_index]
print('max_index: ', max_index)

In testing, the above printed out for me:

max_i_vals: tensor([0.7930, 0.7144, 0.6985, 0.7349, 0.9162, 0.5584, 1.4777, 0.8047, 0.9008, 1.0169, 0.6705, 0.9034, 1.1159, 0.8852, 1.0353], grad_fn=\<MaxBackward0>)   
max_i_indices: tensor([ 5,  8, 10,  6, 13, 14,  5,  6,  6,  6, 13,  4, 13, 13, 11])  
max_j_index:  tensor(6)  
max_index:  [tensor(5), tensor(6)]

This approach can be extended for 3 dimensions. While not as visually pleasing as other answers in this post, this answer shows that the problem can be solved using only the torch.max() function (though I do agree built-in support for torch.max() over multiple dimensions would be a boon).

FOLLOW UP
I stumbled upon a similar question in the PyTorch forums and the poster ptrblck offered this line of code as a solution for getting the indices of the maximal entry in the tensor x:

x = (x==torch.max(x)).nonzero()

Not only does this one-liner work with N-dimensional tensors without needing adjustments to the code, but it is also much faster than the approach I wrote of above (at least 2:1 ratio) and faster than the accepted answer (about 3:2 ratio) according to my benchmarks.

Answer 4

If you use torch <= 1.10, please use this custom max function

def torch_max(x,dim):
    s1 = [i for i in range(len(x.shape)) if i not in dim] 
    s2 = [i for i in range(len(x.shape)) if i in dim]
    x2 = x.permute(tuple(s1+s2))
    s = [d for (i,d) in enumerate(x.shape) if i not in dim] + [-1]
    x2 = torch.reshape(x2, tuple(s))
    max,_ = x2.max(-1)
    return max 

Then run like

print(torch_max(x, dim=(1, 2)))