I am trying to sum specific indices per row in a numpy matrix, based on values in a second numpy vector. For example, in the image, there is the matrix A and the vector of indices inds. Here I want to sum:
A[0, inds[0]] + A[1, inds[1]] + A[2, inds[2]] + A[3, inds[3]]
I am currently using a python for loop, making the code quite slow. Is there a way to do this using vectorisation? Thanks!
Yes, numpy's magic indexing can do this. Just generate a range for the 1st dimension and use your coords for the second:
import numpy as np
x1 = np.array( [[1,2,3,4],[5,6,7,8],[9,10,11,12],[13,14,15,16]] )
print(x1[ [0,1,2,3],[2,0,3,1] ].sum())
Related
Say I have one 2d numpy array X with shape (3,3) and one numpy array Y with shape (3,) where
X = np.array([[0,1,2],
[3,4,5],
[1,9,2]])
Y = np.array([[1,0,1]])
How can I create a numpy array, Z for example, from multiplying X,Y element-wise and then summation row-wise?
multiplying element-wise would yield: 0,0,2, 3,0,5, 1,0,2
then, adding each row would yield:
Z = np.array([2,8,3])
I have tried variations of
Z = np.sum(X * Y) --> adds all elements of entire array, not row-wise.
I know I can use a forloop but the dataset is very large and so I am trying to find a more efficient numpy-specific way to perform the operation. Is this possible?
You can do the following:
sum_row = np.sum(X*Y, axis=1) # axis=0 for columnwise
for example, I have a matrix of dimensions (a,b,c,d). I want to calculate L2 norm of all d matrices of dimensions (a,b,c). Is there any way to use numpy.linalg.norm with out any looping structure?
I mean, the resultant array should be 1 x d
How about this?
import numpy as np
mat = np.arange(2*3*4*5).reshape(2,3,4,5) # create 4d array
mat2 = np.moveaxis(mat,-1,0) # bring last axis to the front
*outarr, = map(np.linalg.norm,mat2) # use map
I need the code in python
for example i have a numpy array sized (x,y,z)
i want to sum it into an array of (x,y), sum z only
z was an array of number, after sum he become a number to finaly get a 2d matrix
You can specify the axis on which the sum will be performed for the numpy function sum:
import numpy as np
res = np.sum(arr, axis=2)
# np.sum(arr, axis=-1) is equivalent in this case
How can I take an inner product of 2 column vectors in python's numpy
Below code does not work
import numpy as np
x = np.array([[1], [2]])
np.inner(x, x)
It returned
array([[1, 2],
[2, 4]])`
instead of 5
The inner product of a vector with dimensions 2x1 (2 rows, 1 column) with another vector of dimension 2x1 (2 rows, 1 column) is a matrix with dimensions 2x2 (2 rows, 2 columns). When you take the inner product of any tensor the inner most dimensions must match (which is 1 in this case) and the result is a tensor with the dimensions matching the outter, i.e.; a 2x1 * 1x2 = 2x2.
What you want to do is transpose both such that when you multiply the dimensions are 1x2 * 2x1 = 1x1.
More generally, multiplying anything with dimensions NxM by something with dimensionsMxK, yields something with dimensions NxK. Note the inner dimensions must both be M. For more, review your matrix multiplication rules
The np.inner function will automatically transpose the second argument, thus when you pass in two 2x1, you get a 2x2, but if you pass in two 1x2 you will get a 1x1.
Try this:
import numpy as np
x = np.array([[1], [2]])
np.inner(np.transpose(x), np.transpose(x))
or simply define your x as row vectors initially.
import numpy as np
x = np.array([1,2])
np.inner(x, x)
i think you mean to have:
x= np.array([1,2])
in order to get 5 as output, your vector needs to be 1xN not Nx1 if you want to apply np.inner on it
Try the following it will work
np.dot(np.transpose(a),a))
make sure col_vector has shape (N,1) where N is the number of elements
then simply sum one to one multiplication result
np.sum(col_vector*col_vector)
This question has two parts (maybe one solution?):
Sample vectors from a sparse matrix: Is there an easy way to sample vectors from a sparse matrix?
When I'm trying to sample lines using random.sample I get an TypeError: sparse matrix length is ambiguous.
from random import sample
import numpy as np
from scipy.sparse import lil_matrix
K = 2
m = [[1,2],[0,4],[5,0],[0,8]]
sample(m,K) #works OK
mm = np.array(m)
sample(m,K) #works OK
sm = lil_matrix(m)
sample(sm,K) #throws exception TypeError: sparse matrix length is ambiguous.
My current solution is to sample from the number of rows in the matrix, then use getrow(),, something like:
indxSampls = sample(range(sm.shape[0]), k)
sampledRows = []
for i in indxSampls:
sampledRows+=[sm.getrow(i)]
Any other efficient/elegant ideas? the dense matrix size is 1000x30000 and could be larger.
Constructing a sparse matrix from a list of sparse vectors: Now imagine I have the list of sampled vectors sampledRows, how can I convert it to a sparse matrix without densify it, convert it to list of lists and then convet it to lil_matrix?
Try
sm[np.random.sample(sm.shape[0], K, replace=False), :]
This gets you out an LIL-format matrix with just K of the rows (in the order determined by the random.sample). I'm not sure it's super-fast, but it can't really be worse than manually accessing row by row like you're currently doing, and probably preallocates the results.
The accepted answer to this question is outdated and no longer works. With newer versions of numpy, you should use np.random.choice in place of np.random.sample, e.g.:
sm[np.random.choice(sm.shape[0], K, replace=False), :]
as opposed to:
sm[np.random.sample(sm.shape[0], K, replace=False), :]