I have rather simple question but still couldn´t make it work.
I want a block diagonal n^2*n^2 matrix. The blocks are sparse n*n matrices with just the diagonal, first off diagonals and forth off diag. For the simple case of n=4 this can easily be done
datanew = ones((5,n1))
datanew[2] = -2*datanew[2]
diagsn = [-4,-1,0,1,4]
DD2 = sparse.spdiags(datanew,diagsn,n,n)
new = sparse.block_diag([DD2,DD2,DD2,DD2])
Since this only useful for small n's, is there a way better way to use block_diag? Thinking of n -> 1000
A simple way of constructing a long list of DD2 matrices, is with a list comprehension:
In [128]: sparse.block_diag([DD2 for _ in range(20)]).A
Out[128]:
array([[-2, 1, 0, ..., 0, 0, 0],
[ 1, -2, 1, ..., 0, 0, 0],
[ 0, 1, -2, ..., 0, 0, 0],
...,
[ 0, 0, 0, ..., -2, 1, 0],
[ 0, 0, 0, ..., 1, -2, 1],
[ 0, 0, 0, ..., 0, 1, -2]])
In [129]: _.shape
Out[129]: (80, 80)
At least in my version, block_diag wants a list of arrays, not *args:
In [133]: sparse.block_diag(DD2,DD2,DD2,DD2)
...
TypeError: block_diag() takes at most 3 arguments (4 given)
In [134]: sparse.block_diag([DD2,DD2,DD2,DD2])
Out[134]:
<16x16 sparse matrix of type '<type 'numpy.int32'>'
with 40 stored elements in COOrdinate format>
This probably isn't the fastest way to construct such a block diagonal array, but it's a start.
================
Looking at the code for sparse.block_mat I deduce that it does:
In [145]: rows=[]
In [146]: for i in range(4):
arow=[None]*4
arow[i]=DD2
rows.append(arow)
.....:
In [147]: rows
Out[147]:
[[<4x4 sparse matrix of type '<type 'numpy.int32'>'
with 10 stored elements (5 diagonals) in DIAgonal format>,
None,
None,
None],
[None,
<4x4 sparse matrix of type '<type 'numpy.int32'>'
...
None,
<4x4 sparse matrix of type '<type 'numpy.int32'>'
with 10 stored elements (5 diagonals) in DIAgonal format>]]
In other words, rows is a 'matrix' of None with DD2 along the diagonals. It then passes these to sparse.bmat.
In [148]: sparse.bmat(rows)
Out[148]:
<16x16 sparse matrix of type '<type 'numpy.int32'>'
with 40 stored elements in COOrdinate format>
bmat in turn collects the data,rows,cols from the coo format of all the input matricies, joins them into master arrays, and builds a new coo matrix from them.
So an alternative is to construct those 3 arrays directly.
Related
I'm trying to calculate the alpha values as explained here.
I have as argument a tensor with shape (1, 512, 14, 14). To calculate alpha values I need to calculate the average of all dimensions except the channel dimension, so the output will have the shape (1, k, 1, 1) which is essentialy (k,).
How can I do this in PyTorch?
Thanks!
You could permute the first and second axis to keep the channel dimension on dim=0, then flatten all other dimensions, and lastly, take the mean on that new axis:
x.permute(1, 0, 2, 3).flatten(start_dim=1).mean(dim=1)
Here are the shapes, step by step:
>>> x.permute(1, 0, 2, 3).shape
(512, 1, 14, 14)
>>> x.permute(1, 0, 2, 3).flatten(start_dim=1).shape
(512, 1, 196)
>>> x.permute(1, 0, 2, 3).flatten(start_dim=1).mean(dim=1).shape
(512,)
in the pyrr.Matrix docs it states:
Matrices are laid out in row-major format and can be loaded directly into OpenGL. To convert to column-major format, transpose the array using the numpy.array.T method.
creating a transformation matrix gives me:
Matrix44.from_translation( np.array([1,2,3]))
Matrix44([[1, 0, 0, 0],
[0, 1, 0, 0],
[0, 0, 1, 0],
[1, 2, 3, 1]])
If the layout is row-major, I would expect the output to be the transpose:
Matrix44([[1, 0, 0, 1],
[0, 1, 0, 2],
[0, 0, 1, 3],
[0, 0, 0, 1]])
I'm most likely confused (I come from C/OpenGL background), but could anyone please enlighten me?
Jonathan
I was writing down a great answer. But I found this really interesting link I invite you to read it !
This is a small resume :
If it's row-major matrix, then the translation is stored in the 3, 7, and 11th indices.
If it's column-major, then the translation is stored in the 12, 13, and 14th indices.
The difference behind the scene is the way to store the data. As it is 16 float in memory, those floats are contiguous in the memory. So you have to define if you either store them in 4 float x 4 columns or 4 float x 4 rows. And then it change the way you access and use it.
You can look at this link too.
I am using scipy and its cdist function to compute a distance matrix from an array of vectors.
import numpy as np
from scipy.spatial import distance
vectorList = [(0, 10), (4, 8), (9.0, 11.0), (14, 14), (16, 19), (25.5, 17.5), (35, 16)]
#Convert to numpy array
arr = np.array(vectorList)
#Computes distances matrix and set self-comparisons to NaN
d = distance.cdist(arr, arr)
np.fill_diagonal(d, None)
Let's say I want to return all the distances that are below a specific threshold (6 for example)
#Find pairs of vectors whose separation distance is < 6
id1, id2 = np.nonzero(d<6)
#id1 --> array([0, 1, 1, 2, 2, 3, 3, 4])
#id2 --> array([1, 0, 2, 1, 3, 2, 4, 3])
I now have 2 arrays of indices.
Question: how can I return the distances between these pairs of vectors as an array / list ?
4.47213595499958 #d[0][1]
4.47213595499958 #d[1][0]
5.830951894845301 #d[1][2]
5.830951894845301 #d[2][1]
5.830951894845301 #d[2][2]
5.830951894845301 #d[3][2]
5.385164807134504 #d[3][4]
5.385164807134504 #d[4][3]
d[id1][id2] returns a matrix, not a list, and the only way I found so far is to iterate over the distance matrix again which doesn't make sense.
np.array([d[i1][i2] for i1, i2 in zip(id1, id2)])
Use
d[id1, id2]
This is the form that numpy.nonzero example shows (i.e. a[np.nonzero(a > 3)]) which is different from the d[id1][id2] you are using.
See arrays.indexing for more details on numpy indexing.
I have a 2D binary array where the value can take 0 and 1 only.
I have a set of indices to check whether the entries of the binary matrix for those indices are duplicate or not. I want to get the matrix with duplicate rows removed and the set of duplicate indices.
For example,
>>>> a
array([[1, 0, 1, 0],
[0, 0, 1, 1],
[1, 0, 1, 0],
[0, 0, 1, 1],
[1, 1, 1, 0],
[1, 1, 1, 0],
[1, 1, 1, 0],
[1, 1, 1, 0],
])
I am given set of indices (0,2,3,4,6,7). From the set, the rows corresponding to (0,2) and (4,6,7) are duplicates. I want the resulting matrix with the duplicates removed (as shown below)
>>>> b
array([[1, 0, 1, 0],
[0, 0, 1, 1],
[0, 0, 1, 1],
[1, 1, 1, 0],
[1, 1, 1, 0],
])
and a method for reconstruction of the matrix 'a' from 'b'
If the order in the output array is not relevant, then you can probably just use Eelco Hoogendoorn's answer. However, if you want to keep the same relative order as in the original array, here is another possible approach.
import numpy as np
a = np.array([
[1, 0, 1, 0],
[0, 0, 1, 1],
[1, 0, 1, 0],
[0, 0, 1, 1],
[1, 1, 1, 0],
[1, 1, 1, 0],
[1, 1, 1, 0],
[1, 1, 1, 0],
])
idx = np.array([0, 2, 3, 4, 6, 7])
# Make an array of row numbers
r = np.arange(len(a))
# Replace row numbers in idx with -1
# (use assume_unique only if indices in idx are unique)
r[np.isin(r, idx, assume_unique=True)] = -1
# Add the column to the array
a2 = np.concatenate([a, r[:, np.newaxis]], axis=-1)
# Find unique indices and inverse indices
_, uniq_idx, inv_idx = np.unique(a2, return_index=True, return_inverse=True, axis=0)
# Sort indices to make output array and inverse indices
s = np.argsort(uniq_idx)
a_uniq = a[uniq_idx[s]]
inv_idx = s[inv_idx]
print(a_uniq)
# [[1 0 1 0]
# [0 0 1 1]
# [0 0 1 1]
# [1 1 1 0]
# [1 1 1 0]]
print(np.all(a_uniq[inv_idx] == a))
# True
EDIT: Some further explanation.
The idea in the solution above is to apply np.unique, but in a way that the rows that are not included in idx are not affected by it. In order to do that, you can just add a new number to each row. For the rows included in idx, this number will always be -1, and for the rest of rows it will be a different number for each. That way, it is impossible that rows that are not in idx get removed by np.unique. In order to do that, I build r, first with np.arange(len(a)), which gives you a number per row:
[0 1 2 3 4 5 6 7]
Then I check which of those are in idx with np.isin(r, idx, assume_unique=True) (assume_unique can only be used if elements in idx are guaranteed to be unique), so r[np.isin(r, idx, assume_unique=True)] = -1 will turn all indices idx into -1:
[-1 1 -1 -1 -1 5 -1 -1]
That is added as new column to a into a2:
[[ 1 0 1 0 -1]
[ 0 0 1 1 1]
[ 1 0 1 0 -1]
[ 0 0 1 1 -1]
[ 1 1 1 0 -1]
[ 1 1 1 0 5]
[ 1 1 1 0 -1]
[ 1 1 1 0 -1]]
Now it's just a matter of applying np.unique to a2. As expected, only rows in idx may be eliminated. However, since we want to keep the original relative order, we cannot use the output of np.unique, because it is sorted. We use return_index and return_inverse to get the indices that make the array of unique rows and the indices that get you back to the original array, and actually discard the new array.
To form the final array, you need to sort uniq_idx to keep the relative order, and then inv_idx accordingly. np.argsort gives you the indices that sort uniq_idx into s. uniq_idx[s] is just the array of unique row indices sorted, and s[inv_idx] will map every inverse index in inv_idx to the corresponding one in the resorted array. So, finally, a[uniq_idx[s]] gives you the output array, and the new inv_idx takes you back to the original one.
It feels like you could phrase your question at a higher level to get a more elegant solution; but this seems to solve the literal problem as stated.
idx = [0,2,3,4,6,7]
b = np.concatenate([np.unique(a[idx], axis=0), np.delete(a, idx, axis=0)], axis=0)
I have two square matrices A and B
I must convert B to CSR Format and determine the product C
A * B_csr = C
I have found a lot of information online regarding CSR Matrix - Vector multiplication. The algorithm is:
for (k = 0; k < N; k = k + 1)
result[i] = 0;
for (i = 0; i < N; i = i + 1)
{
for (k = RowPtr[i]; k < RowPtr[i+1]; k = k + 1)
{
result[i] = result[i] + Val[k]*d[Col[k]];
}
}
However, I require Matrix - Matrix multiplication.
Further, it seems that most algorithms apply A_csr - vector multiplication where I require A * B_csr. My solution is to transpose the two matrices before converting then transpose the final product.
Can someone explain how to compute a Matrix - CSR Matrix product and/or a CSR Matrix - Matrix product?
Here is a simple solution in Python for the Dense Matrix X CSR Matrix. It should be self-explanatory.
def main():
# 4 x 4 csr matrix
# [1, 0, 0, 0],
# [2, 0, 3, 0],
# [0, 0, 0, 0],
# [0, 4, 0, 0],
csr_values = [1, 2, 3, 4]
col_idx = [0, 0, 2, 1]
row_ptr = [0, 1, 3, 3, 4]
csr_matrix = [
csr_values,
col_idx,
row_ptr
]
dense_matrix = [
[1, 3, 3, 4],
[1, 2, 3, 4],
[1, 4, 3, 4],
[1, 2, 3, 5],
]
res = [
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
]
# matrix order, assumes both matrices are square
n = len(dense_matrix)
# res = dense X csr
csr_row = 0 # Current row in CSR matrix
for i in range(n):
start, end = row_ptr[i], row_ptr[i + 1]
for j in range(start, end):
col, csr_value = col_idx[j], csr_values[j]
for k in range(n):
dense_value = dense_matrix[k][csr_row]
res[k][col] += csr_value * dense_value
csr_row += 1
print res
if __name__ == '__main__':
main()
CSR Matrix X Dense Matrix is really just a sequence of CSR Matrix X Vector product for each row of the dense matrix right? So it should be really easy to extend the code you show above to do this.
Moving forward, I suggest you don't code these routines yourself. If you are using C++ (based on the tag), then you could have a look at Boost ublas for example, or Eigen. The APIs may seem a bit cryptic at first but it's really worth it in the long term. First, you gain access to a lot more functionality, which you will probably require in the future. Second these implementations will be better optimised.