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I have the following segment of for loop in my code. The nested loop is slowing down my complete execution.
for q in range(batchSize):
temp=torch.where((composition_matrix == pred[q]).all(dim=1))[0]
if len(temp)==0:
output[q]=0
else:
output[q]=int(temp[0])
Here, composition_matrix is [14000,2] dimensional pytorch tensor with only positive integers as cell values. pred and output both are a [batchSize,2] dimensional torch tensor.
As this for loop is slowing my code a lot and I am unable to get the equivalent broadcasting solution to this code segment.
Does a broadcasting solution exists to eleminate this for loop?
I shall be grateful for any help.
A minimum reproducible example is
import torch
composition_matrix=torch.randint(3, 10, (14000,2))
batchSize=64
pred=torch.randint(3, 10, (batchSize,2))
output=torch.zeros([batchSize])
for q in range(batchSize):
temp=torch.where((composition_matrix == pred[q]).all(dim=1))[0]
if len(temp)==0:
output[q]=0
else:
output[q]=int(temp[0])
To make it simple, you first need to understand what the operation is essentially doing. You've got two tensors. Tensor A is of shape (14000, 2) and tensor B is of shape (64, 2). The operation you want to do is:
For each row B[i] in B, compare that B[i] (of shape (2,) with A (of
shape (14000, 2)). If B[i] occurs within A, set output[i] = index of
first occurrence.
This can actually be done in two lines of code (maybe even one line):
comp = (composition_matrix[:, None, :] == pred).all(dim=-1)
output = torch.argmax(comp.float(), axis=0)
The first line creates comp, the broadcasted comparison of composition_matrix and pred, a boolean tensor of shape (14000, 64).
The second line needs to find the "index of the first match". This can be done quite simply with argmax: it will return the index of the first "1" (or if all the values are "0", will return the first index, ie, 0).
(Note that torch does not support argmax for "bool" tensors, and so comp needed to be cast to another data type.)
Sorry for the short and probably over-simplified example. I fear a bigger one would be much more difficult to visualize. But I hope this suits your purpose.
My solution may seem a little complicated but it's fully vectorized and includes no explicit loops.
Here's what I would do:
import torch
torch.manual_seed(0)
batchSize = 8
pred = torch.randint(0, 10, (batchSize, 2))
output = torch.zeros((batchSize, 2))
composition_matrix = torch.randint(0, 10, (14, 2))
# compair all vectors in composition_matrix to all vectors in pred
comparisons = (composition_matrix.unsqueeze(0) == pred.unsqueeze(1))
comparisons = comparisons.all(2)
# form an index array the shape of the comparisons array
comparison_idxs = torch.arange(comparisons.shape[1])
comparison_idxs = comparison_idxs.repeat(batchSize).reshape(*comparisons.shape)
# multipy the comparisons array by the index array
where_result = (comparison_idxs*comparisons)
# replace invalind zeros with the maximal value in each sample
batch_idxs = torch.arange(comparisons.shape[0])
batch_idxs = batch_idxs.repeat(comparisons.shape[1])
batch_idxs = batch_idxs.reshape(comparisons.shape[1], comparisons.shape[0]).T
maxima = where_result.max(1).values[batch_idxs]
maxima_vecor = maxima[(1-comparisons.int()).bool()]
where_result[(1-comparisons.int()).bool()] = maxima_vecor
vectorized_output = where_result.min(1)[0]
output = torch.zeros([batchSize])
for q in range(batchSize):
temp=torch.where((composition_matrix == pred[q]).all(dim=1))[0]
if len(temp)==0:
output[q]=0
else:
output[q]=int(temp[0])
output:
composition_matrix =
tensor([[6, 8],
[4, 3],
[6, 9],
[1, 4],
[4, 1],
[9, 9],
[9, 0],
[1, 2],
[3, 0],
[5, 5],
[2, 9],
[1, 8],
[8, 3],
[6, 9]])
pred =
tensor([[4, 9],
[3, 0],
[3, 9],
[7, 3],
[7, 3],
[1, 6],
[6, 9],
[8, 6]])
output =
tensor([0., 8., 0., 0., 0., 0., 2., 0.])
vectorized_output =
tensor([0, 8, 0, 0, 0, 0, 2, 0])
Some timing results:
torch.manual_seed(0)
batchSize = 8
pred = torch.randint(0, 10, (batchSize, 2))
composition_matrix = torch.randint(0, 10, (14000, 2))
print('timing the vectorized_solution:')
%timeit -n 1000 vectorized_solution(composition_matrix, pred,)
print('timing the loop_solution:')
%timeit -n 1000 loop_solution(composition_matrix, pred,)
output:
timing the vectorized_solution:
1000 loops, best of 5: 137 µs per loop
timing the loop_solution:
1000 loops, best of 5: 1.89 ms per loop
I am trying to replicate how zip works by using a simple example and i want the output to be an array.
I have the following data
s = (2, 2)
array = np.zeros(s)
x = np.array([1, 0, 1, 0, 1, 1, 1, 1])
y = np.array([1, 0, 0, 0, 1, 0, 1, 1])
What i want to do is have a 2x2 matrix as output, which works like this:
for i, j in zip(x, y):
array[i][j] += 1
This outputs
[[2 0]
[2 4]]
I tried obtaining the same results without using the zip for lists but i get a (1,1) tuple
for i in range(len(x)):
array = x[i], y[i]
will output: (1, 1)
for i in range(len(x)):
array[x[i]][y[i]] += 1
This will do the same as
for i, j in zip(x, y):
array[i][j] += 1
This makes the transformation clear for you
for idx in range(len(x)):
i = x[idx]
j = y[idx]
array[i][j] += 1
print(array)
Output:
[[2 0]
[2 4]]
import numpy as np
array = np.zeros(2, dtype=int)
x = np.array([1, 0, 1, 0, 1, 1, 1, 1])
y = np.array([1, 0, 0, 0, 1, 0, 1, 1])
# empirically what zip does
[(x[i], y[i]) for i in range(sorted([len(x), len(y)])[0])]
#proof
print( list(zip(x, y)) == [(x[i], y[i]) for i in range(sorted([len(x), len(y)])[0])] )
True
Why use sorted?
Since zip stops at the index of the smaller list we need to iterate by the range of the smallest list; therefore sorted([1,2])[0] is 1 (the smaller of the 2)
so even if i had y = np.array([1, 0, 0, 0, 1, 0, 1, 1, 1]) that would still return the correct zip of the 2
In order to do it in a vectorized way, try to work on array.ravel() instead. This command performs dynamic changes of array:
np.add.at(array.ravel(),
np.ravel_multi_index(np.array([x,y]), s),
np.repeat(1, len(x)))
So you can check out after running it that array has changed:
>>> array
array([[2, 0],
[2, 4]])
Signature: np.argwhere(a)
Docstring:
Find the indices of array elements that are non-zero, grouped by element.
Examples
>>> x = np.arange(6).reshape(2,3)
>>> x
array([[0, 1, 2],
[3, 4, 5]])
>>> np.argwhere(x>1)
array([[0, 2],
[1, 0],
[1, 1],
[1, 2]])
What does it mean by 'non-zero' and 'grouped by element'? and what is "x>1"?
In each row the first entry is the row index and the second entry is the column index of the entries of x that satisfy the condition.
For example:
2 is greater than 1
so the first row of argwhere gives you [0, 2]
pointing to the position of 2 in x.
Find the indices (positions) of array elements that are non-zero (true), grouped by element (each index is its own row).
Basically, if you pass a boolean array, you will find the indices where that array is true, but transposed so that the indices in the form [[x1, x2, ...], [y1, y2, ...]] become in the form [[x1, y1], [x2, y2], ...].
x > 1 is a boolean array which is True wherever x > 1 and False wherever x <= 1. In your example, it looks loke
[[False, False, True],
[True, True, True]]
I have two dense matrices, A [200000,10], B [10,100000]. I need to multiply them to get matrix C. I can't do that directly, since the resulting matrix won't fit into the memory. Moreover, I need only a few elements from the resulting matrix, like 1-2% of the total number of elements. I have a third matrix W [200000,100000] which is sparse and has non-zero elements on exactly those places which are interesting to me in the matrix C.
Is there a way to use W as a "mask" so that the resulting matrix C will be sparse and will contain only the needed elements?
Since a matrix multiplication is just a table of dot products, we can just perform the specific dot products we need, in a vectorized fashion.
import numpy as np
import scipy as sp
iX, iY = sp.nonzero(W)
values = np.sum(A[iX]*B[:, iY].T, axis=-1) #batched dot product
C = sp.sparse.coo_matrix(values, np.asarray([iX,iY]).T)
First, get the indexes of the non zero places in W, and then you can just get the (i,j) element of the result matrix by multiplying the i-th row in A with the j-th column in B, and save the result as a tuple (i,j,res) instead of saving it as a matrix (this is the right way to save sparse matrices).
Here's one approach using np.einsum for a vectorized solution -
from scipy import sparse
from scipy.sparse import coo_matrix
# Get row, col for the output array
r,c,_= sparse.find(W)
# Get the sum-reduction using valid rows and corresponding cols from A, B
out = np.einsum('ij,ji->i',A[r],B[:,c])
# Store as sparse matrix
out_sparse = coo_matrix((out, (r, c)), shape=W.shape)
Sample run -
1) Inputs :
In [168]: A
Out[168]:
array([[4, 6, 1, 1, 1],
[0, 8, 1, 3, 7],
[2, 8, 3, 2, 2],
[3, 4, 1, 6, 3]])
In [169]: B
Out[169]:
array([[5, 2, 4],
[2, 1, 3],
[7, 7, 2],
[5, 7, 5],
[8, 5, 0]])
In [176]: W
Out[176]:
<4x3 sparse matrix of type '<type 'numpy.bool_'>'
with 5 stored elements in Compressed Sparse Row format>
In [177]: W.toarray()
Out[177]:
array([[ True, False, False],
[False, False, False],
[ True, True, False],
[ True, False, True]], dtype=bool)
2) Using dense array to perform direct calculations and verify results later on :
In [171]: (A.dot(B))*W.toarray()
Out[171]:
array([[52, 0, 0],
[ 0, 0, 0],
[73, 57, 0],
[84, 0, 56]])
3) Use the proposed codes and get sparse matrix output :
In [172]: # Using proposed codes
...: r,c,_= sparse.find(W)
...: out = np.einsum('ij,ji->i',A[r],B[:,c])
...: out_sparse = coo_matrix((out, (r, c)), shape=W.shape)
...:
4) Finally verify results by converting to dense/array version and checking against direct version -
In [173]: out_sparse.toarray()
Out[173]:
array([[52, 0, 0],
[ 0, 0, 0],
[73, 57, 0],
[84, 0, 56]])
I have a numpy array A of size ((s1,...sm)) with integer entries and a dictionary D with integers as keys and numpy arrays of size ((t)) as values. I would like to evaluate the dictionary on every entry of the array A to get a new array B of size ((s1,...sm,t)).
For example
D={1:[0,1],2:[1,0]}
A=np.array([1,2,1])
The output shout be
array([[0,1],[1,0],[0,1]])
Motivation: I have an array with indexes of unit vectors as entries and I need to transform it into an array with the vectors as entries.
If you can rename your keys to be 0-indexed, you might use direct array querying on your unit vectors:
>>> units = np.array([D[1], D[2]])
>>> B = units[A - 1] # -1 because 0 indexed: 1 -> 0, 2 -> 1
>>> B
array([[0, 1],
[1, 0],
[0, 1]])
And similarly for any shape:
>>> A = np.random.random_integers(0, 1, (10, 11, 12))
>>> A.shape
(10, 11, 12)
>>> B = units[A]
>>> B.shape
(10, 11, 12, 2)
You can learn more about advanced indexing on the numpy doc
>>> np.asarray([D[key] for key in A])
array([[0, 1],
[1, 0],
[0, 1]])
Here's an approach using np.searchsorted to locate those row indices to index into the values of the dictionary and then simply indexing it to get the desired output, like so -
idx = np.searchsorted(D.keys(),A)
out = np.asarray(D.values())[idx]
Sample run -
In [45]: A
Out[45]: array([1, 2, 1])
In [46]: D
Out[46]: {1: [0, 1], 2: [1, 0]}
In [47]: idx = np.searchsorted(D.keys(),A)
...: out = np.asarray(D.values())[idx]
...:
In [48]: out
Out[48]:
array([[0, 1],
[1, 0],
[0, 1]])