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 x-y coordinates:
from scipy.optimize import distance
x = np.array([1,2], [2,3], [4,5], [5,6])
I also have a list of x-y 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!
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.
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
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
I have some 3D numpy arrays that need to be transformed in various ways. E.g.:
x.shape = (4, 17, 17)
This array is 1 sample of 4 planes, each of size 17x17. What is the most efficient way to transform each plane: flipud, fliplr, and rot90? Is there a better way than using a for loop? Thanks!
for p in range(4):
x[p, :, :] = np.fliplr(x[p, :, :])
Look at the code of these functions:
def fliplr(...):
....
return m[:, ::-1]
In other words it returns a view with reverse slicing on the 2nd dimension
Your x[p, :, :] = np.fliplr(x[p, :, :] applies that reverse slicing to the last dimension, so the equivalent for the whole array should be
x[:, :, ::-1]
flipping the 2nd axis would be
x[:, ::-1, :]
etc.
np.rot90 has 4 case (k); for k=1 it is
return fliplr(m).swapaxes(0, 1)
in other words m[:, ::-1].swapaxes(0,1)
To work on your planes you would do something like
m[:, :,::-1].swapaxes(1,2)
or you could do the swapaxes/transpose first
m.transpose(0,2,1)[:, :, ::-1]
Does that give you enough tools to transform the plane's in what ever way you want?
As I discussed in another recent question, https://stackoverflow.com/a/41291462/901925, the flip... returns a view, but the rot90, with both flip and swap, will, most likely return a copy. Either way, numpy will be giving you the most efficient version.
I have a meshgrid in numpy. I make some calculations on the points. I want to filter out points that could not be calcutaled for some reason ( division by zero).
from numpy import arange, array
Xout = arange(-400, 400, 20)
Yout = arange(0, 400, 20)
Zout = arange(0, 400, 20)
Xout_3d, Yout_3d, Zout_3d = numpy.meshgrid(Xout,Yout,Zout)
#some calculations
# for example
b = z / ( y - x )
To perform z / ( y - x ) using those 3D mesh arrays, you can create a mask of the valid ones. Now, the valid ones would be the ones where any pair of combinations between y and x aren't identical. So, this mask would be of shape (M,N), where M and N are the lengths of the Y and X axes respectively. To get such a mask to span across all combinations between X and Y, we could use NumPy's broadcasting. Thus, we would have such a mask like so -
mask = Yout[:,None] != Xout
Finally, and again using broadcasting to broadcast the mask along the first two axes of the3D arrays, we could perform such a division and choose between an invalid specifier and the actual division result using np.where, like so -
invalid_spec = 0
out = np.where(mask[...,None],Zout_3d/(Yout_3d-Xout_3d),invalid_spec)
Alternatively, we can directly get to such an output using broadcasting and thus avoid using meshgrid and having those heavy 3D arrays in workspace. The idea is to simultaneously populate the 3D grids and perform the subtraction and division computations, both on the fly. So, the implementation would look something like this -
np.where(mask[...,None],Zout/(Yout[:,None,None] - Xout[:,None]),invalid_spec)
I need to normalize a numpy data cube say:
cube = np.random.random(100000).reshape(10,100,100)
and then normalise each of the 10 resulting planes by the median. So, e.g. for the first plane
cube[0, :, :] /= np.median(cube[0, :, :])
I just want to avoid a loop if possible 😊
thanks
You can pass a list of axes to np.median and then expand via None (np.newaxis):
>>> cube = np.random.random(100000).reshape(10,100,100)
>>> simple = cube / np.median(cube,axis=[1,2])[:,None,None]
>>>
>>> brute = cube.copy()
>>> for i in range(10):
... brute[i, :, :] /= np.median(cube[i, :, :])
...
>>> np.allclose(brute, simple)
True
but to be honest, looping over the shortest axis often isn't so bad performance-wise if the other axes are much longer.
I have a 2D numpy array input_array and two lists of indices (x_coords and y_coords). Id like to slice a 3x3 subarray for each x,y pair centered around the x,y coordinates. The end result will be an array of 3x3 subarrays where the number of subarrays is equal to the number of coordinate pairs I have.
Preferably by avoiding for loops. Currently I use a modification of game of life strides from the scipy cookbook:
http://wiki.scipy.org/Cookbook/GameOfLifeStrides
shape = (input_array.shape[0] - 2, input_array.shape[0] - 2, 3, 3)
strides = input_array.strides + input_array.strides
strided = np.lib.stride_trics.as_strided(input_array, shape=shape, strides=strides).\
reshape(shape[0]*shape[1], shape[2], shape[3])
This creates a view of the original array as a (flattened) array of all possible 3x3 subarrays. I then convert the x,y coordinate pairs to be able to select the subarrays I want from strided:
coords = x_coords - 1 + (y_coords - 1)*shape[1]
sub_arrays = strided[coords]
Although this works perfectly fine, I do feel it is a bit cumbersome. Is there a more direct approach to do this? Also, in the future I would like to extend this to the 3D case; slicing nx3x3 subarrays from a nxmxk array. It might also be possible using strides but so far I haven't been able to make it work in 3D
Here is a method that use array broadcast:
x = np.random.randint(1, 63, 10)
y = np.random.randint(1, 63, 10)
dy, dx = [grid.astype(int) for grid in np.mgrid[-1:1:3j, -1:1:3j]]
Y = dy[None, :, :] + y[:, None, None]
X = dx[None, :, :] + x[:, None, None]
then you can use a[Y, X] to select blocks from a. Here is an example code:
img = np.zeros((64, 64))
img[Y, X] = 1
Here is graph ploted by pyplot.imshow():
A very straight forward solution would be a list comprehension and itertools.product:
import itertools
sub_arrays = [input_array[x-1:x+2, y-1:y+2]
for x, y in itertools.product(x_coords, y_coords)]
This creates all possible tuples of coordinates and then slices the 3x3 arrays from the input_array.
But this is sort-of a for loop. And you will have to take care, that x_coords and y_coords are not on the border of the matrix.