In numpy operation, I have two vectors, let's say vector A is 4X1, vector B is 1X5, if I do AXB, it should result a matrix of size 4X5.
But I tried lot of times, doing many kinds of reshape and transpose, they all either raise error saying not aligned or return a single value.
How should I get the output product of matrix I want?
Normal matrix multiplication works as long as the vectors have the right shape. Remember that * in Numpy is elementwise multiplication, and matrix multiplication is available with numpy.dot() (or with the # operator, in Python 3.5)
>>> numpy.dot(numpy.array([[1], [2]]), numpy.array([[3, 4]]))
array([[3, 4],
[6, 8]])
This is called an "outer product." You can get it using plain vectors using numpy.outer():
>>> numpy.outer(numpy.array([1, 2]), numpy.array([3, 4]))
array([[3, 4],
[6, 8]])
If you are using numpy.
First, make sure you have two vectors. For example, vec1.shape = (10, ) and vec2.shape = (26, ); in numpy, row vector and column vector are the same thing.
Second, you do res_matrix = vec1.reshape(10, 1) # vec2.reshape(1, 26) ;.
Finally, you should have: res_matrix.shape = (10, 26).
numpy documentation says it will deprecate np.matrix(), so better not use it.
Function matmul (since numpy 1.10.1) works fine:
import numpy as np
a = np.array([[1],[2],[3],[4]])
b = np.array([[1,1,1,1,1],])
ab = np.matmul(a, b)
print (ab)
print(ab.shape)
You have to declare your vectors right. The first has to be a list of lists of one number (this vector has to have columns in one row), and the second - a list of list (this vector has to have rows in one column) like in above example.
Output:
[[1 1 1 1 1]
[2 2 2 2 2]
[3 3 3 3 3]
[4 4 4 4 4]]
(4, 5)
Related
Slice a 3d numpy array using a 1d lookup between indices
import numpy as np
a = np.arange(12).reshape(2, 3, 2)
b = np.array([2, 0])
b maps i to j where i and j are the first 2 indexes of a, so a[i,j,k]
Desired result after applying b to a is:
[[4 5]
[6 7]]
Naive solution:
c = np.empty(shape=(2, 2), dtype=int)
for i in range(2):
j = b[i]
c[i, :] = a[i, j, :]
Question: Is there a way to do this using a numpy or scipy routine or routines or fancy indexing?
Application: Reinforcement Learning finite MDPs where b is a deterministic policy vector pi(a|s), a is the state transition probabilities p(s'|s,a) and c is the state transition matrix for that policy vector p(s'|s). The arrays will be large and this operation will be repeated a large number of times so needs to be scaleable and fast.
What I have tried:
Compiling using numba but line profiler suggests my code is slower compared to a similarly sized numpy routine. Also numpy is more widely understood and used.
Maintaining pi(a|s) as a sparse matrix (all zero except one 1 per row) b_as_a_matrix and then using einsum but this involves storing and updating the matrix and creates more work (an extra loop over j and sum operation).
c = np.einsum('ij,ijk->ik', b_as_a_matrix, a)
Numpy arrays can be indexed using other arrays as indices. See also: NumPy selecting specific column index per row by using a list of indexes.
With that in mind, we can vectorize your loop to simply use b for indexing:
>>> import numpy as np
>>> a = np.arange(12).reshape(2, 3, 2)
>>> b = np.array([2, 0])
>>> i = np.arange(len(b))
>>> i
array([0, 1])
>>> a[i, b, :]
array([[4, 5],
[6, 7]])
This question already has answers here:
Transposing a 1D NumPy array
(15 answers)
Closed 3 years ago.
Let's consider a as an 1D row/horizontal array:
import numpy as np
N = 10
a = np.arange(N) # array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
a.shape # (10,)
now I want to have b a 1D column/vertical array transposed of a:
b = a.transpose() # array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
b.shape # (10,)
but the .transpose() method returns an identical ndarray whith the exact same shape!
What I expected to see was
np.array([[0], [1], [2], [3], [4], [5], [6], [7], [8], [9]])
which can be achieved by
c = a.reshape(a.shape[0], 1) # or c = a; c.shape = (c.shape[0], 1)
c.shape # (10, 1)
and to my surprise, it has a shape of (10, 1) instead of (1, 10).
In Octave/Scilab I could do:
N = 10
b = 0:(N-1)
a = b'
size(b) % ans = 1 10
size(a) % ans = 10 1
I understand that numpy ndarrays are not matrices (as discussed here), but the behavior of the numpy's transpose function just doesn't make sense to me! I would appreciate it if you could help me understand how this behavior makes sense and what am I missing here.
P.S. So what I have understood so far is that b = a.transpose() is the equivalent of b = a; b.shape = b.shape[::-1] which if you had a "2D array" of (N, 1) would return a (1, N) shaped array, as you would expect from a transpose operator. However, numpy seems to treat the "1D array" of (N,) as a 0D scalar. I think they should have named this method something else, as this is very misleading/confusing IMHO.
To understand the numpy array better, you should take a look at this review paper: The NumPy array: a structure for efficient numerical computation
In short, numpy ndarrays have this attribute called the stride, which is
the number of bytes to skip in memory to proceed to the next element.
For a (10, 10) array of bytes, for example, the strides may be (10,
1), in other words: proceed one byte to get to the next column and ten
bytes to locate the next row.
For your ndarray a, a.stride = (8,), which shows that it is only 1 dimensional, and that to get to the next element on this single dimension, you need to advance 8 bytes in memory (each int is 64-bit).
Strides are useful for representing transposes:
By modifying strides, for example, an array can be transposed or
reshaped at zero cost (no memory needs to be copied).
So if there was a 2-dimensional ndarray, say b = np.ones((3,5)) for example, then b.strides = (40, 8), while b.transpose().strides = (8, 40). So as you see a transposed 2D-ndarray is simply the exact same array, whose strides have been reordered. And since your 1D ndarray has only 1 dimension, swapping the the values of its strides (i.e. taking its transpose), doesn't do anything.
As you already mentioned that numpy array are not matrix. The defination of transpose function is like below
Permute the dimensions of an array.
Which means that numpy's transpose method will move data from one dimension to another. As 1D array has only one dimension there is no other dimension to move the data t0. So you need add a dimension before transpose has any effect. This behavior make sense also to be consistent with higher dimensional array (3D, 4D ...) array.
There is a clean way to achive what you want
N = 10
a = np.arange(N)
a[ :, np.newaxis]
I have an nd-array A
A.shape
(2, 500, 3)
What's the difference between A[:] and A[:,2]
Coming from Python, the ',' in the array access is confusing me a lot.
The commas separate the subscripts for each dimension. So, for example, if the matrix M is defined as
M = np.array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
then M[2, 1] would be 8 (third row, second column).
The subscript for each dimension can also be a slice, where : represents a full slice, like a slice in normal Python sequences. For example, M[:, 2] would select from every row the third column, which would be [3, 6, 9].
Any additional dimensions for which a subscript is not provided are implicitly full slices. In your example, A[:,2] is equivalent to A[:, 2, :]. If you consider the (2, 500, 3) shaped array to be two stacked matrices with 500 rows and 3 columns, then A[:, 2, :] would select from both matrices the third row (and every column of the third row), which should have a shape of (2, 3).
When you have multidimensional NumPy arrays, the slicing operation [] can work if you provide tuple of slice() objects. If the number of tuples does not match your number of dimensions, this is equivalent to having a slice(None) (which abbreviates to :) in all the remaining dimensions. Note also that NumPy also accepts ... which means "fill the rest of the dimensions with :" - which is especially useful if you want to "fill" the initial dimensions.
So to recapitulate the following expression give identical results on your A array of A.ndim == 3:
A[:, 2]
A[:, 2, :]
A[:, 2, ...]
A[slice(None), 2]
A[slice(None), 2, slice(None)]
A[(slice(None), 2) + tuple(slice(None) for _ in range(A.ndim - 2))]
I have matrices that are 2 x 4 and 3 x 4. I want to find the euclidean distance across rows, and get a 2 x 3 matrix at the end. Here is the code with one for loop that computes the euclidean distance for every row vector in a against all b row vectors. How do I do the same without using for loops?
import numpy as np
a = np.array([[1,1,1,1],[2,2,2,2]])
b = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
dists = np.zeros((2, 3))
for i in range(2):
dists[i] = np.sqrt(np.sum(np.square(a[i] - b), axis=1))
Here are the original input variables:
A = np.array([[1,1,1,1],[2,2,2,2]])
B = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
A
# array([[1, 1, 1, 1],
# [2, 2, 2, 2]])
B
# array([[1, 2, 3, 4],
# [1, 1, 1, 1],
# [1, 2, 1, 9]])
A is a 2x4 array.
B is a 3x4 array.
We want to compute the Euclidean distance matrix operation in one entirely vectorized operation, where dist[i,j] contains the distance between the ith instance in A and jth instance in B. So dist is 2x3 in this example.
The distance
could ostensibly be written with numpy as
dist = np.sqrt(np.sum(np.square(A-B))) # DOES NOT WORK
# Traceback (most recent call last):
# File "<stdin>", line 1, in <module>
# ValueError: operands could not be broadcast together with shapes (2,4) (3,4)
However, as shown above, the problem is that the element-wise subtraction operation A-B involves incompatible array sizes, specifically the 2 and 3 in the first dimension.
A has dimensions 2 x 4
B has dimensions 3 x 4
In order to do element-wise subtraction, we have to pad either A or B to satisfy numpy's broadcast rules. I'll choose to pad A with an extra dimension so that it becomes 2 x 1 x 4, which allows the arrays' dimensions to line up for broadcasting. For more on numpy broadcasting, see the tutorial in the scipy manual and the final example in this tutorial.
You can perform the padding with either np.newaxis value or with the np.reshape command. I show both below:
# First approach is to add the extra dimension to A with np.newaxis
A[:,np.newaxis,:] has dimensions 2 x 1 x 4
B has dimensions 3 x 4
# Second approach is to reshape A with np.reshape
np.reshape(A, (2,1,4)) has dimensions 2 x 1 x 4
B has dimensions 3 x 4
As you can see, using either approach will allow the dimensions to line up. I'll use the first approach with np.newaxis. So now, this will work to create A-B, which is a 2x3x4 array:
diff = A[:,np.newaxis,:] - B
# Alternative approach:
# diff = np.reshape(A, (2,1,4)) - B
diff.shape
# (2, 3, 4)
Now we can put that difference expression into the dist equation statement to get the final result:
dist = np.sqrt(np.sum(np.square(A[:,np.newaxis,:] - B), axis=2))
dist
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
Note that the sum is over axis=2, which means take the sum over the 2x3x4 array's third axis (where the axis id starts with 0).
If your arrays are small, then the above command will work just fine. However, if you have large arrays, then you may run into memory issues. Note that in the above example, numpy internally created a 2x3x4 array to perform the broadcasting. If we generalize A to have dimensions a x z and B to have dimensions b x z, then numpy will internally create an a x b x z array for broadcasting.
We can avoid creating this intermediate array by doing some mathematical manipulation. Because you are computing the Euclidean distance as a sum-of-squared-differences, we can take advantage of the mathematical fact that sum-of-squared-differences can be rewritten.
Note that the middle term involves the sum over element-wise multiplication. This sum over multiplcations is better known as a dot product. Because A and B are each a matrix, then this operation is actually a matrix multiplication. We can thus rewrite the above as:
We can then write the following numpy code:
threeSums = np.sum(np.square(A)[:,np.newaxis,:], axis=2) - 2 * A.dot(B.T) + np.sum(np.square(B), axis=1)
dist = np.sqrt(threeSums)
dist
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
Note that the answer above is exactly the same as the previous implementation. Again, the advantage here is the we do not need to create the intermediate 2x3x4 array for broadcasting.
For completeness, let's double-check that the dimensions of each summand in threeSums allowed broadcasting.
np.sum(np.square(A)[:,np.newaxis,:], axis=2) has dimensions 2 x 1
2 * A.dot(B.T) has dimensions 2 x 3
np.sum(np.square(B), axis=1) has dimensions 1 x 3
So, as expected, the final dist array has dimensions 2x3.
This use of the dot product in lieu of sum of element-wise multiplication is also discussed in this tutorial.
I had the same problem recently working with deep learning(stanford cs231n,Assignment1),but when I used
np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))
There was a error
MemoryError
That means I ran out of memory(In fact,that produced a array of 500*5000*1024 in the middle.It's so huge!)
To prevent that error,we can use a formula to simplify:
code:
import numpy as np
aSumSquare = np.sum(np.square(a),axis=1);
bSumSquare = np.sum(np.square(b),axis=1);
mul = np.dot(a,b.T);
dists = np.sqrt(aSumSquare[:,np.newaxis]+bSumSquare-2*mul)
Simply use np.newaxis at the right place:
np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))
This functionality is already included in scipy's spatial module and I recommend using it as it will be vectorized and highly optimized under the hood. But, as evident by the other answer, there are ways you can do this yourself.
import numpy as np
a = np.array([[1,1,1,1],[2,2,2,2]])
b = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
np.sqrt((np.square(a[:,np.newaxis]-b).sum(axis=2)))
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
from scipy.spatial.distance import cdist
cdist(a,b)
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
Using numpy.linalg.norm also works well with broadcasting. Specifying an integer value for axis will use a vector norm, which defaults to Euclidean norm.
import numpy as np
a = np.array([[1,1,1,1],[2,2,2,2]])
b = np.array([[1,2,3,4],[1,1,1,1],[1,2,1,9]])
np.linalg.norm(a[:, np.newaxis] - b, axis = 2)
# array([[ 3.74165739, 0. , 8.06225775],
# [ 2.44948974, 2. , 7.14142843]])
I want to find a command in numpy for a column vector times a row vector equals to a matrix
[1,1,1,1 ] ^T * [ 2,3 ] = [[2,3],[2,3],[2,3],[2,3]]
First, let's define your 1-D numpy arrays:
In [5]: one = np.array([ 1,1,1,1 ]); two = np.array([ 2,3 ])
Now, lets multiply them:
In [6]: one[:, np.newaxis] * two[np.newaxis, :]
Out[6]:
array([[2, 3],
[2, 3],
[2, 3],
[2, 3]])
This used numpy's newaxis to add the appropriate axes to get a 4x2 output matrix.
The problem you are encountering is that both of your vectors are neither column nor row vectors - they're just vectors. If you look at len(vec.shape) it's 1.
What you can do is use numpy.reshape to turn your column vector into shape (m, 1) and your row vector into shape (1, n).
import numpy as np
colu = np.reshape(u, (u.shape[0], 1))
rowv = np.reshape(v, (1, v.shape[0]))
Now when you multiply colu and rowv you'll get a matrix with shape (m, n).
If you need a matrix - use matrices. This way you can use your expression nearly verbatim:
np.matrix([1,1,1,1]).T * np.matrix([2,3])
You might want to use numpy.kron(a,b) it takes the Kronecker product of two arrays. You can see the b vector as a block. The function puts this block, multiplied by the corresponding coefficient of the a vector, on the position of that coefficient. You can also use it for matrices.
For your example it would look like:
import numpy as np
vecA = np.array([[1],[1],[1],[1]])
vecB = np.array([2,3])
Out = np.kron(vecA,vecB)
this returns
>>> Out
array([[2, 3],
[2, 3],
[2, 3],
[2, 3]])
Hope this helps you.