Pairwise cosine distance

I want to find cosine distance between each pair of 2 tensors.
That is given [a,b] and [p,q], I want a 2x2 matrix which finds
[ cosDist(a,p), cosDist(a,q)
cosDist(b,p), cosDist(b,q) ]
I want to be able to use this matrix for triplet loss with hard mining.
What is the best way to do this?


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I am not sure about cosine distance. You might do something similar to below code.

For L2 loss, you might use this:

Thanks @InnovArul, I had been referring to this code. I was wondering if it is possible to use nn.CosineSimilarity() instead of computing the cosine similarity manually, just to be sure that there are no errors by me.

@learnpytorch: Have you find the solution?

Something like:

import torch

def cosine_distance_torch(x1, x2=None, eps=1e-8):
    x2 = x1 if x2 is None else x2
    w1 = x1.norm(p=2, dim=1, keepdim=True)
    w2 = w1 if x2 is x1 else x2.norm(p=2, dim=1, keepdim=True)
    return 1 -, x2.t()) / (w1 * w2.t()).clamp(min=eps)

def cosine_similarity_n_space(m1=None, m2=None, dist_batch_size=100):
    NoneType = type(None)
    device = torch.device('cuda' if torch.cuda.is_available() else 'cpu')
    if type(m1) != torch.Tensor: # only numpy conversion supported
        m1 = torch.from_numpy(m1).float()
    if type(m2) != torch.Tensor and type(m2)!=NoneType:
        m2 = torch.from_numpy(m2).float() # m2 could be None
    m2 = m1 if m2 is None else m2
    assert m1.shape[1] == m2.shape[1]
    result = torch.zeros([1, m2.shape[0]])
    for row_i in range(0, int(m1.shape[0] / dist_batch_size) + 1):
        start = row_i * dist_batch_size
        end = min([(row_i + 1) * dist_batch_size, m1.shape[0]])
        if end <= start:
            break # cause I'm too lazy to elegantly handle edge cases
        rows = m1[start: end] 
        # sim = cosine_similarity(rows, m2) # rows is O(1) size        
        sim = cosine_distance_torch(,
        result = (result, sim.cpu()), 0)
    result = result[1:, :] # deleting the first row, as it was used for setting the size only
    del sim
    return result.numpy() # return 1 - ret # should be used with sklearn cosine_similarity

Thanks but it too complex and time consuming

I worked on this during the summer. Cosine worked well for me :slight_smile:

Have a look at this file

@Oli: You just used original cosine sim function. It only returns the triangle value of matrix. I want to return a paired sim, so it will be full matrix

@John1231983; it isn’t time consuming if the GPU is enabled.

Usage example:

input1 = torch.randn(5, 7)
input2 = torch.randn(5, 7)

dist_mat  = cosine_distance_torch(input1, input2, 2.4)
print(dist_mat )

Actually, I have input of 5D. so I sure that your way is time-comsuming

In fact, it does not work for a 5D input.

@Deeply: I think torch.bmm is more simple and faster your way. But in my case. bmm is memory error

You can use this snippet. Cosine similarity is the same as the scalar product of the normalized inputs and you can get the pw scalar product through matrix multiplication.
Cosine distance in turn is just 1-cosine_similarity.

def pw_cosine_distance(input_a, input_b):
   normalized_input_a = torch.nn.functional.normalize(input_a)  
   normalized_input_b = torch.nn.functional.normalize(input_b)
   res =, normalized_input_b.T)
   res *= -1 # 1-res without copy
   res += 1
   return res

This github issue gives a nice way to achieve this using the built in cosine_similarity function.

th.nn.functional.cosine_similarity(x[:,:,None], x.t()[None,:,:])  
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