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Create a PyTorch tensor of sequences which excludes specified value

I have a 1d PyTorch tensor containing integers between 0 and n-1. Now I need to create a 2d PyTorch tensor with n-1 columns, where each row is a sequence from 0 to n-1 excluding the value in the first tensor. How can I achieve this efficiently?
Ex:
n = 3
a = torch.Tensor([0, 1, 2, 1, 2, 0])
# desired output
b = [
[1, 2],
[0, 2],
[0, 1],
[0, 2],
[0, 1],
[1, 2]
]
Typically, the a.numel() >> n.
Detailed Explanation:
The first element of a is 0, hence it has to map to the sequence [0, 1, 2] excluding 0, which is [1, 2].
Similarly, the second element of a is 1, hence it has to map to [0, 2] and so on.
PS: I actually have an additional batch dimension, which I've excluded here for simplicity. Hence, I need the solution to be easily extendable to one additional dimension.

We can construct a tensor with the desired sequences and index with tensor a.
import torch
n = 3
a = torch.Tensor([0, 1, 2, 1, 2, 0]) # using torch.tensor is recommended
def exclude_gather(a, n):
sequences = torch.nonzero(torch.arange(n) != torch.arange(n)[:,None], as_tuple=True)[1].reshape(-1, n-1)
return sequences[a.long()]
exclude_gather(a, n)
Output
tensor([[1, 2],
[0, 2],
[0, 1],
[0, 2],
[0, 1],
[1, 2]])
We can add a batch dimension with functorch.vmap
from functorch import vmap
n = 4
b = torch.Tensor([[0, 1, 2, 1, 3, 0],[0, 3, 1, 0, 2, 1]])
vmap(exclude_gather, in_dims=(0, None))(b, n)
Output
tensor([[[1, 2, 3],
[0, 2, 3],
[0, 1, 3],
[0, 2, 3],
[0, 1, 2],
[1, 2, 3]],
[[1, 2, 3],
[0, 1, 2],
[0, 2, 3],
[1, 2, 3],
[0, 1, 3],
[0, 2, 3]]])

All you have to do is initialize a multi-dimension array with all possible indices using torch.arange(). After that, purge indices that you don't want from each tensor using a boolean mask.
import torch
a = torch.Tensor([0, 1, 2, 1, 2, 0])
n = 3
b = [torch.arange(n) for i in range(len(a))]
c = [b[i]!=a[i] for i in range(len(b))]
# use the boolean array as a mask to apply on b
d = [[b[i][c[i]] for i in range(len(b))]]
print(d) # this can be converted to a list of numbers or torch tensor
This prints the output - [[tensor([1, 2]), tensor([0, 2]), tensor([0, 1]), tensor([0, 2]), tensor([0, 1]), tensor([1, 2])]] which you can convert to int/numpy/torch array/tensor easily.
This can be extended to multiple dimensions as well.

The following does the trick
b = []
for i in range(n-1):
b.append(i * torch.ones_like(a) + (a <= i))
b = torch.stack(b, dim=1)
Since n << size(a), the for loop should not be very costly.

Batch dot product with numpy?

I need to get the dot product of many vectors with one vector. Example code:
a = np.array([0, 1, 2])
b = np.array([
[0, 1, 2],
[4, 5, 6],
[-1, 0, 1],
[-3, -2, 1]
])
I would like to get the dot product of each row of b against a. I can iterate:
result = []
for row in b:
result.append(np.dot(row, a))
print(result)
which gives:
[5, 17, 2, 0]
How can I get this without iterating? Thanks!

Use numpy.dot or numpy.matmul without for loop:
import numpy as np
np.matmul(b, a)
# or
np.dot(b, a)
Output:
array([ 5, 17, 2, 0])

I will just do #
b#a
Out[108]: array([ 5, 17, 2, 0])

Numpy - Apply a custom function on all combination of rows in matrix to get a new matrix?

I have the following function, that applies the histogram intersection kernel for 2 arrays:
def histogram_intersection_kernel(X, Y):
k = np.array([])
for x_i,y_i in zip(X,Y):
k = np.append(k,np.minimum(x_i,y_i))
return np.sum(k)
now, lets say I have the following matrix "mat":
[[1,0,0,2,3],
[2,3,4,0,1],
[3,3,5,0,1]]
I would like to find an efficient way to get the matrix that is the result of applying "histogram_intersection_kernel" to all of the combinations of rows in mat. In this example it would be:
[[6,2,2],
[6,10,10],
[2,10,12]]

Extend dimensions to 3D and leverage broadcasting -
np.minimum(a[:,None,:],a[None,:,:]).sum(axis=2)
Or simply -
np.minimum(a[:,None],a).sum(2)
Sample run -
In [248]: a
Out[248]:
array([[1, 0, 0, 2, 3],
[2, 3, 4, 0, 1],
[3, 3, 5, 0, 1]])
In [249]: np.minimum(a[:,None],a).sum(2)
Out[249]:
array([[ 6, 2, 2],
[ 2, 10, 10],
[ 2, 10, 12]])

Changing a number within a matrix

I randomly generate a matrix. Let's assume for simplicity that it is in the following form np.shape(A) = (2,4):
import numpy as np
A:
matrix([[ 1, 2, 3, 4],
[ 3, 4, 10, 8]])
Then, I estimate the following expression:
import numpy as np
K = 3
I = 4
C0 = np.sum(np.maximum(A[-1] - K, 0)) / I
The question is how do I input the following restriction: if any number of a column in the matrix A is less than or equal to (<=) K (3), then change the last number of that column to zero? So basically, my matrix should transform to this:
A:
matrix([[ 1, 2, 3, 4],
[ 0, 0, 0, 8]])

This is one way.
A[-1][np.any(A <= 3, axis=0)] = 0
# matrix([[1, 2, 3, 4],
# [0, 0, 0, 8]])
A[-1][np.any((A > 2) & (A <= 3), axis=0)] = 0
# matrix([[1, 2, 3, 4],
# [0, 4, 0, 8]])

How to get a value from every column in a Numpy matrix

I'd like to get the index of a value for every column in a matrix M. For example:
M = matrix([[0, 1, 0],
[4, 2, 4],
[3, 4, 1],
[1, 3, 2],
[2, 0, 3]])
In pseudocode, I'd like to do something like this:
for col in M:
idx = numpy.where(M[col]==0) # Only for columns!
and have idx be 0, 4, 0 for each column.
I have tried to use where, but I don't understand the return value, which is a tuple of matrices.

The tuple of matrices is a collection of items suited for indexing. The output will have the shape of the indexing matrices (or arrays), and each item in the output will be selected from the original array using the first array as the index of the first dimension, the second as the index of the second dimension, and so on. In other words, this:
>>> numpy.where(M == 0)
(matrix([[0, 0, 4]]), matrix([[0, 2, 1]]))
>>> row, col = numpy.where(M == 0)
>>> M[row, col]
matrix([[0, 0, 0]])
>>> M[numpy.where(M == 0)] = 1000
>>> M
matrix([[1000, 1, 1000],
[ 4, 2, 4],
[ 3, 4, 1],
[ 1, 3, 2],
[ 2, 1000, 3]])
The sequence may be what's confusing you. It proceeds in flattened order -- so M[0,2] appears second, not third. If you need to reorder them, you could do this:
>>> row[0,col.argsort()]
matrix([[0, 4, 0]])
You also might be better off using arrays instead of matrices. That way you can manipulate the shape of the arrays, which is often useful! Also note ajcr's transpose-based trick, which is probably preferable to using argsort.
Finally, there is also a nonzero method that does the same thing as where in this case. Using the transpose trick now:
>>> (M == 0).T.nonzero()
(matrix([[0, 1, 2]]), matrix([[0, 4, 0]]))

As an alternative to np.where, you could perhaps use np.argwhere to return an array of indexes where the array meets the condition:
>>> np.argwhere(M == 0)
array([[[0, 0]],
[[0, 2]],
[[4, 1]]])
This tells you each the indexes in the format [row, column] where the condition was met.
If you'd prefer the format of this output array to be grouped by column rather than row, (that is, [column, row]), just use the method on the transpose of the array:
>>> np.argwhere(M.T == 0).squeeze()
array([[0, 0],
[1, 4],
[2, 0]])
I also used np.squeeze here to get rid of axis 1, so that we are left with a 2D array. The sequence you want is the second column, i.e. np.argwhere(M.T == 0).squeeze()[:, 1].

The result of where(M == 0) would look something like this
(matrix([[0, 0, 4]]), matrix([[0, 2, 1]])) First matrix tells you the rows where 0s are and second matrix tells you the columns where 0s are.
Out[4]:
matrix([[0, 1, 0],
[4, 2, 4],
[3, 4, 1],
[1, 3, 2],
[2, 0, 3]])
In [5]: np.where(M == 0)
Out[5]: (matrix([[0, 0, 4]]), matrix([[0, 2, 1]]))
In [6]: M[0,0]
Out[6]: 0
In [7]: M[0,2] #0th row 2nd column
Out[7]: 0
In [8]: M[4,1] #4th row 1st column
Out[8]: 0

This isn't anything new on what's been already suggested, but a one-line solution is:
>>> np.where(np.array(M.T)==0)[-1]
array([0, 4, 0])
(I agree that NumPy matrix objects are more trouble than they're worth).

>>> M = np.array([[0, 1, 0],
... [4, 2, 4],
... [3, 4, 1],
... [1, 3, 2],
... [2, 0, 3]])
>>> [np.where(M[:,i]==0)[0][0] for i in range(M.shape[1])]
[0, 4, 0]