Fast way to do consecutive one-to-all calculations on Numpy arrays without a for-loop?

Question

I'm working on an optimization problem, but to avoid getting into the details, I'm going to provide a simple example of a bug that's been giving me headaches for a few days.

Say I have a 2D numpy array with observed xy coordinates:

from scipy.optimize import distance
x = np.array([1,2], [2,3], [4,5], [5,6])

I also have a list of xy coordinates to compare to these points (y):

y = np.array([11,13], [12, 14])

I have a function that takes the sum of manhattan differences between a value of x and all of the values in y :

def find_sum(ref_row, comp_rows):
    modeled_counts = []
    y = ref_row * len(comp_rows)
    res = list(map(distance.cityblock, ref_row, comp_rows))
    modeled_counts.append(sum(res))

    return sum(modeled_counts)

Essentially, what I would like to do is find the sum of manhattan distances for every item in y with each item in x (so basically for each item in x, find the sum of the Manhattan distances between that (x,y) pair and every (x,y) pair in y ).

I've tried this out with the following line of code:

z = list(map(find_sum, x, y))

However, z is of length 2 (like y ), and not 4 like x . Is there a way to ensure that z is the result of consecutive one-to-all calculations? That is, I'd like to calculate the sum of all of the manhattan differences between x[0] and every set in y , and so on and so forth, so the length of z should be equal to the length of x .

Is there a simple way to do this without a for loop? My data is rather large (~ 4 million rows), so I'd really appreciate fast solutions. I'm fairly new to Python programming, so any explanations about why the solution works and is fast would be appreciated as well, but definitely isn't required!

Thanks!

Answer 1

This solution implements the distance in numpy , as I think it is a good example of broadcasting , which is a very useful thing to know if you need to use arrays and matrices.

By definition of Manhattan distance, you need to evaluate the sum of the absolute value of difference between each column. However, the first column of x , x[:, 0] , has shape (4,) and the first column of y , y[:, 0] , has shape (2,), so they are not compatible in the sense of applying subtraction: the broadcasting property says that each shape is compared starting with the trailing dimensions and two dimensions are compatible when they are equal or one of them is 1. Sadly, none of them are true for your columns.

However, you can add a new dimension of value 1 using np.newaxis , so

x[:, 0]

is array([1, 2, 4, 5]) , but

x[:, 0, np.newaxis]

is

array([[1],
       [2],
       [4],
       [5]])

and its shape is (4,1). Now, a matrix of shape (4, 1) subtracted by an array of shape 2 results in a matrix of shape (4, 2), by numpy 's broadcasting treatment:

   4 x 1
       2
=  4 x 2

You can obtain the differences for each column:

first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]

and evaluate the sum of their absolute values:

np.abs(first_column_difference) + np.abs(second_column_difference)

which results in a (4, 2) matrix. Now, you want to sum the values for each row, so that you have 4 values:

np.sum(np.abs(first_column_difference) + np.abs(second_column_difference), axis=1)

which results in array([73, 69, 61, 57]) . The rule is simple: the parameter axis will eliminate that dimension from the result, therefore using axis=1 for a (4, 2) matrix generates 4 values -- if you use axis=0 , it will generate 2 values.

So, this will solve your problem:

x = np.array([[1, 2], [2, 3], [4, 5], [5, 6]])
y = np.array([[11, 13], [12, 43]])

first_column_difference = x[:, 0, np.newaxis] - y[:, 0]
second_column_difference = x[:, 1, np.newaxis] - y[:, 1]
z = np.abs(first_column_difference) + np.abs(second_column_difference)
print(np.sum(z, axis=1))

You can also skip the intermediate steps for each column and evaluate everything at once (it is a little bit harder to understand, so I prefer the method described above to explain what is happening):

print(np.abs(x[:, np.newaxis] - y).sum(axis=(1, 2)))

It is a general case for an n-dimensional Manhattan distance: if x is (u, n) and y is (v, n), it generates u rows by broadcasting (u, 1, n) by (v, n) = (u, v, n) , then applying sum to eliminate the second and third axis.

Answer 2

Here is how you can do it using numpy broadcast with simplified explanation

Adjust Shape For Broadcasting

import numpy as np

start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])

## using np.newaxis as index add a new dimension at that position
## : give all the elements on that dimension
start_points = start_points[np.newaxis, :, :]
dest_points = dest_points[:, np.newaxis, :]

## Now lets check he shape of the point arrays
print('start_points.shape: ', start_points.shape) # (1, 4, 2)
print('dest_points.shape', dest_points.shape) # (2, 1, 2)

Lets try to understand

• last element of shape represent x and y of a point, size 2
• we can think of start_points as having 1 row and 4 columns of points
• we can think of dest_points as having 2 rows and 1 columns of points

We can think start_points and dest_points as matrix or a table of points of size (1X4) and (2X1) We clearly see that size are not compatible. What will happen if we perform arithmatic operation between them? Here is where a smart part of numpy comes, called broadcast.

• It will repeat rows of start_points to match that of dest_point making matrix of (2X4)
• It will repeat columns of dest_point to match that of start_points making matrix of (2X4)
• Result is arithmetic operation between every pair of elements on start_points and dest_points

Calculate the distance

diff_x_y = start_points - dest_points
print(diff_x_y.shape) # (2, 4, 2)
abs_diff_x_y = np.abs(start_points - dest_points)
man_distance = np.sum(abs_diff_x_y, axis=2)
print('man_distance:\n', man_distance)
sum_distance = np.sum(man_distance, axis=0)
print('sum_distance:\n', sum_distance)

Oneliner

start_points = np.array([[1,2], [2,3], [4,5], [5,6]])
dest_points = np.array([[11,13], [12, 14]])
np.sum(np.abs(start_points[np.newaxis, :, :] - dest_points[:, np.newaxis, :]), axis=(0,2))

Here is more detail explanation of broadcasting if you want to understand it more

Answer 3

With so many rows you can make substantial savings by using a smart algorithm. Let us for simplicity assume there is just one dimension; once we have established the algorithm, getting back to the general case is a simple matter of summing over coordinates.

The naive algorithm is O(mn) where m,n are the sizes of sets X,Y . Our algorithm is O((m+n)log(m+n)) so it scales much better.

We first have to sort the union of X and Y by coordinate and then form the cumsum over Y . Next, we find for each x in X the number YbefX of y in Y to its left and use it to look up the corresponding cumsum item YbefXval . The summed distances to all y to the left of x are YbefX times coordinate of x minus YbefXval , the distances to all y to the right are sum of all y coordinates minus YbefXval minus n - YbefX times coordinate of x .

Where does the saving come from? Sorting coordinates enables us to recycle the summations we have done before, instead of starting each time from scratch. This uses the fact that up to a sign we always sum the same y coordinates and going from left to right the signs flip one by one.

Code:

import numpy as np
from scipy.spatial.distance import cdist
from timeit import timeit

def pp(X,Y):
    (m,k),(n,k) = X.shape,Y.shape
    XY = np.concatenate([X.T,Y.T],1)
    idx = XY.argsort(1)
    Xmsk = idx<m
    Ymsk = ~Xmsk
    Xidx = np.arange(k)[:,None],idx[Xmsk].reshape(k,m)
    Yidx = np.arange(k)[:,None],idx[Ymsk].reshape(k,n)
    YbefX = Ymsk.cumsum(1)[Xmsk].reshape(k,m)
    YbefXval = XY[Yidx].cumsum(1)[np.arange(k)[:,None],YbefX-1]
    YbefXval[YbefX==0] = 0
    XY[Xidx] = ((2*YbefX-n)*XY[Xidx]) - 2*YbefXval + Y.sum(0)[:,None]
    return XY[:,:m].sum(0)

def summed_cdist(X,Y):
    return cdist(X,Y,"minkowski",p=1).sum(1)

# demo    
m,n,k = 1000,500,10
X,Y = np.random.randn(m,k),np.random.randn(n,k)
print("same result:",np.allclose(pp(X,Y),summed_cdist(X,Y)))
print("sort       :",timeit(lambda:pp(X,Y),number=1000),"ms")
print("scipy cdist:",timeit(lambda:summed_cdist(X,Y),number=100)*10,"ms")

Sample run, comparing smart algo "sort" to naive algo implemented using cdist library function:

same result: True
sort       : 1.4447695480193943 ms
scipy cdist: 36.41934019047767 ms