If I have two single-dimensional arrays of length M and N what is the most efficient way to calculate the euclidean distance between all points with the resultant being an NxM array? I'm trying to figure this out with Numpy but am pretty new to it so I'm a little stuck.
Currently I am doing it this way:
def get_distances(x,y):
#compute distances between all points
distances = np.zeros((len(y),len(x)))
for i in range(len(y)):
for j in range(len(x)):
distances[i,j] = (x[j] - y[i])**2
return distances
Suppose you have 1-dimensional positions :
a = np.random.uniform(50,200,5)
b = np.random.uniform(50,200,3)
you can just use broadcasting:
result = np.abs(a[:, None] - b[None, :])
with the result being:
array([[ 44.37361012, 22.20152487, 89.04608885],
[ 42.83825434, 20.66616909, 87.51073307],
[ 0.19806059, 21.97402467, 44.87053932],
[ 8.42276237, 13.74932288, 53.0952411 ],
[ 8.12181467, 30.29389993, 36.55066406]])
So the i, j index is the distance between point i from array 1 and point j of array 2
if you want the result to be NxM in shape you need to exchange a and b:
result = np.abs(b[:, None] - a[None, :])
I was wondering if there's a way to compute multiple outer products and stack the results in a single operation.
Say I have an Nx1 vector and take the outer product with a 1xM vector, the result will be an NxM matrix.
What if I had an NxR matrix A, and an RxM matrix B. Is it possible to construct an NxMxR matrix where each layer of the output matrix is the outer product of the corresponding column of A and row of B?
I know it's really easy to do this in a single for loop over R, but I wanted to know if there's a faster way using numpy builtins (as there usually is when numpy is concerned).
I haven't been able to figure out a set of indices that work with einsum (and I'm not even sure if einsum is the right approach, since there is no summation involved here)
Yes, of course, using broadcasting or Einsum (the fact that there is no summation does not matter)
N, M, R = 8, 9, 16
A = numpy.random.rand(N)
B = numpy.random.rand(M)
C = A[:, None] * B[None, :]
D = numpy.einsum('a,b->ab', A, B)
numpy.allclose(C, D)
# True
C.shape
# (8, 9)
A = numpy.random.rand(N, R)
B = numpy.random.rand(M, R)
C = A[:, None, :] * B[None, :, :]
D = numpy.einsum('ar,br->abr', A, B)
numpy.allclose(C, D)
# True
C.shape
# (8, 9, 16)
I have two matrices, A of shape 512*3 and B of shape 1024*3
I want to calculate pairwise subtraction between their rows, so the result would be of shape 512*1024*3
(they are actually arrays of 3D point coordinates : x , y , z and I eventually want to find k nearest points from B to every point in A)
and I can't use for loops. is there any pythonic way to do this?
thank u.
From the reference I linked in my previous comment:
http://scipy.github.io/old-wiki/pages/EricsBroadcastingDoc
You are trying to do this.
Just follow the example, as in:
import numpy as np
np.random.seed(123)
a = np.random.uniform(size=(8,3)) # or (512,3)
b = np.random.uniform(size=(16,3)) # or (1024,3)
diff = a[np.newaxis,:,:]-b[:,np.newaxis,:]
dist = np.sqrt(np.sum(diff**2,axis=-1))
The difference:
diff = A[:, np.newaxis] - B[np.newaxis, :]
The closest k points in B for each point in A:
k = 5
dists = np.sum(np.square(A[:, np.newaxis] - B[np.newaxis, :]), axis=-1)
top_k = np.argpartition(dists, k, axis=1)[:, :k]
That top_k is not sorted by distance, though. You can sort it later or do instead:
top_k = np.argsort(dists, axis=1)[:, :k]
Which is less efficient but simpler.
I have two numpy arrays
import numpy as np
x = np.linspace(1e10, 1e12, num=50) # 50 values
y = np.linspace(1e5, 1e7, num=50) # 50 values
x.shape # output is (50,)
y.shape # output is (50,)
I would like to create a function which returns an array shaped (50,50) such that the first x value x0 is evaluated for all y values, etc.
The current function I am using is fairly complicated, so let's use an easier example. Let's say the function is
def func(x,y):
return x**2 + y**2
How do I shape this to be a (50,50) array? At the moment, it will output 50 values. Would you use a for loop inside an array?
Something like:
np.array([[func(x,y) for i in x] for j in y)
but without using two for loops. This takes forever to run.
EDIT: It has been requested I share my "complicated" function. Here it goes:
There is a data vector which is a 1D numpy array of 4000 measurements. There is also a "normalized_matrix", which is shaped (4000,4000)---it is nothing special, just a matrix with entry values of integers between 0 and 1, e.g. 0.5567878. These are the two "given" inputs.
My function returns the matrix multiplication product of transpose(datavector) * matrix * datavector, which is a single value.
Now, as you can see in the code, I have initialized two arrays, x and y, which pass through a series of "x parameters" and "y parameters". That is, what does func(x,y) return for value x1 and value y1, i.e. func(x1,y1)?
The shape of matrix1 is (50, 4000, 4000). The shape of matrix2 is (50, 4000, 4000). Ditto for total_matrix.
normalized_matrix is shape (4000,4000) and id_mat is shaped (4000,4000).
normalized_matrix
print normalized_matrix.shape #output (4000,4000)
data_vector = datarr
print datarr.shape #output (4000,)
def func(x, y):
matrix1 = x [:, None, None] * normalized_matrix[None, :, :]
matrix2 = y[:, None, None] * id_mat[None, :, :]
total_matrix = matrix1 + matrix2
# transpose(datavector) * matrix * datavector
# by matrix multiplication, equals single value
return np.array([ np.dot(datarr.T, np.dot(total_matrix, datarr) ) ])
If I try to use np.meshgrid(), that is, if I try
x = np.linspace(1e10, 1e12, num=50) # 50 values
y = np.linspace(1e5, 1e7, num=50) # 50 values
X, Y = np.meshgrid(x,y)
z = func(X, Y)
I get the following value error: ValueError: operands could not be broadcast together with shapes (50,1,1,50) (1,4000,4000).
reshape in numpy as different meaning. When you start with a (100,) and change it to (5,20) or (10,10) 2d arrays, that is 'reshape. There is anumpy` function to do that.
You want to take 2 1d array, and use those to generate a 2d array from a function. This is like taking an outer product of the 2, passing all combinations of their values through your function.
Some sort of double loop is one way of doing this, whether it is with an explicit loop, or list comprehension. But speeding this up depends on that function.
For at x**2+y**2 example, it can be 'vectorized' quite easily:
In [40]: x=np.linspace(1e10,1e12,num=10)
In [45]: y=np.linspace(1e5,1e7,num=5)
In [46]: z = x[:,None]**2 + y[None,:]**2
In [47]: z.shape
Out[47]: (10, 5)
This takes advantage of numpy broadcasting. With the None, x is reshaped to (10,1) and y to (1,5), and the + takes an outer sum.
X,Y=np.meshgrid(x,y,indexing='ij') produces two (10,5) arrays that can be used the same way. Look at is doc for other parameters.
So if your more complex function can be written in a way that takes 2d arrays like this, it is easy to 'vectorize'.
But if that function must take 2 scalars, and return another scalar, then you are stuck with some sort of double loop.
A list comprehension form of the double loop is:
np.array([[x1**2+y1**2 for y1 in y] for x1 in x])
Another is:
z=np.empty((10,5))
for i in range(10):
for j in range(5):
z[i,j] = x[i]**2 + y[j]**2
This double loop can be sped up somewhat by using np.vectorize. This takes a user defined function, and returns one that can take broadcastable arrays:
In [65]: vprod=np.vectorize(lambda x,y: x**2+y**2)
In [66]: vprod(x[:,None],y[None,:]).shape
Out[66]: (10, 5)
Test that I've done in the past show that vectorize can improve on the list comprehension route by something like 20%, but the improvement is nothing like writing your function to work with 2d arrays in the first place.
By the way, this sort of 'vectorization' question has been asked many times on SO numpy. Beyond these broad examples, we can't help you without knowning more about that more complicated function. As long as it is a black box that takes scalars, the best we can help you with is np.vectorize. And you still need to understand broadcasting (with or without meshgrid help).
I think there is a better way, it is right on the tip of my tongue, but as an interim measure:
You are operating on 1x2 windows of a meshgrid. You can use as_strided from numpy.lib.stride_tricks to rearrange the meshgrid into two-element windows, then apply your function to the resultant array. I like to use a generic nd solution, sliding_windows (http://www.johnvinyard.com/blog/?p=268) (Not mine) to transform the array.
import numpy as np
a = np.array([1,2,3])
b = np.array([.1, .2, .3])
z= np.array(np.meshgrid(a,b))
def foo((x,y)):
return x+y
>>> z.shape
(2, 3, 3)
>>> t = sliding_window(z, (2,1,1))
>>> t
array([[ 1. , 0.1],
[ 2. , 0.1],
[ 3. , 0.1],
[ 1. , 0.2],
[ 2. , 0.2],
[ 3. , 0.2],
[ 1. , 0.3],
[ 2. , 0.3],
[ 3. , 0.3]])
>>> v = np.apply_along_axis(foo, 1, t)
>>> v
array([ 1.1, 2.1, 3.1, 1.2, 2.2, 3.2, 1.3, 2.3, 3.3])
>>> v.reshape((len(a), len(b)))
array([[ 1.1, 2.1, 3.1],
[ 1.2, 2.2, 3.2],
[ 1.3, 2.3, 3.3]])
>>>
This should be somewhat faster.
You may need to modify your function's argument signature.
If the link to the johnvinyard.com blog breaks, I've posted the the sliding_window implementation in other SO answers - https://stackoverflow.com/a/22749434/2823755
Search around and you'll find many other tricky as_strided solutions.
In response to your edited question:
normalized_matrix
print normalized_matrix.shape #output (4000,4000)
data_vector = datarr
print datarr.shape #output (4000,)
def func(x, y):
matrix1 = x [:, None, None] * normalized_matrix[None, :, :]
matrix2 = y[:, None, None] * id_mat[None, :, :]
total_matrix = matrix1 + matrix2
# transpose(datavector) * matrix * datavector
# by matrix multiplication, equals single value
# return np.array([ np.dot(datarr.T, np.dot(total_matrix, datarr))])
return np.einsum('j,ijk,k->i',datarr,total_matrix,datarr)
Since datarr is shape (4000,), transpose does nothing. I believe you want the result of the 2 dots to be shape (50,). I'm suggesting using einsum. But it can be done with tensordot, or I think even np.dot(np.dot(total_matrix, datarr),datarr). Test the expression with smaller arrays, focusing on getting the shapes right.
x = np.linspace(1e10, 1e12, num=50) # 50 values
y = np.linspace(1e5, 1e7, num=50) # 50 values
z = func(x,y)
# X, Y = np.meshgrid(x,y)
# z = func(X, Y)
X,Y is wrong. func takes x and y that are 1d. Notice how you expand the dimensions with [:, None, None]. Also you aren't creating a 2d array from an outer combination of x and y. None of your arrays in func is (50,50) or (50,50,...). The higher dimensions are provided by nomalied_matrix and id_mat.
When showing us the ValueError you should also indicate where in your code that occurred. Otherwise we have to guess, or recreate the code ourselves.
In fact when I run my edited func(X,Y), I get this error:
----> 2 matrix1 = x [:, None, None] * normalized_matrix[None, :, :]
3 matrix2 = y[:, None, None] * id_mat[None, :, :]
4 total_matrix = matrix1 + matrix2
5 # transpose(datavector) * matrix * datavector
ValueError: operands could not be broadcast together with shapes (50,1,1,50) (1,400,400)
See, the error occurs right at the start. normalized_matrix is expanded to (1,400,400) [I'm using smaller examples]. The (50,50) X is expanded to (50,1,1,50). x expands to (50,1,1), which broadcasts just fine.
To address the edit and the broadcasting error in the edit:
Inside your function you are adding dimensions to arrays to try to get them to broadcast.
matrix1 = x [:, None, None] * normalized_matrix[None, :, :]
This expression looks like you want to broadcast a 1d array with a 2d array.
The results of your meshgrid are two 2d arrays:
X,Y = np.meshgrid(x,y)
>>> X.shape, Y.shape
((50, 50), (50, 50))
>>>
When you try to use X in in your broadcasting expression the dimensions don't line up, that is what causes the ValueError - refer to the General Broadcasting Rules:
>>> x1 = X[:, np.newaxis, np.newaxis]
>>> nm = normalized_matrix[np.newaxis, :, :]
>>> x1.shape
(50, 1, 1, 50)
>>> nm.shape
(1, 4000, 4000)
>>>
You're on the right track with your list comprehension, you just need to add in an extra level of iteration:
np.array([[func(i,j) for i in x] for j in y])
I have two arrays:
a = [[a11,a12],
[a21,a22]]
b = [[b11,b12],
[b21,b22]]
What I would like to do is build up a matrix as follows:
xx = np.mean(a[:,0]*b[:,0])
xy = np.mean(a[:,0]*b[:,1])
yx = np.mean(a[:,1]*b[:,0])
yy = np.mean(a[:,1]*b[:,1])
and return an array c such that
c = [[xx,xy],
yx,yy]]
Is there a nice pythonic way to do this in numpy? Because at the moment I have done it by hand, exactly as above, so the dimensions of the output array are coded in by hand, rather than determined as according to the size of the input arrays a and b.
Is there an error in your third element? If, as seems reasonable, you want yx = np.mean(a[:,1]*b[:,0]) instead of yx = np.mean(b[:,1]*a[:,0]), then you can try the following:
a = np.random.rand(2, 2)
b = np.random.rand(2, 2)
>>> c
array([[ 0.26951488, 0.19019219],
[ 0.31008754, 0.1793523 ]])
>>> np.mean(a.T[:, None, :]*b.T, axis=-1)
array([[ 0.26951488, 0.19019219],
[ 0.31008754, 0.1793523 ]])
It will actually be faster to avoid the intermediate array and express your result as a matrix multiplication:
>>> np.dot(a.T, b) / a.shape[0]
array([[ 0.26951488, 0.19019219],
[ 0.31008754, 0.1793523 ]])