Python: comparing cell arrays using minus function
I have 3 cell arrays with each cell array having different array sizes. How can I execute a minus function for each of the possible cell combinations?
For example:
import numpy as np
a=np.array([[np.array([[2,2,1,2]]),np.array([[1,3]])]])
b=np.array([[np.array([[4,2,1]])]])
c=np.array([[np.array([[1,2]]),np.array([[4,3]])]])
A possible combination here is ab, ac and bc.
Let's say a - b:
a=2,2,1,2 and 1,3
b=4,2,1
The desired result will come with switching windows due to the different size of the array:
(2,2,1)-(4,2,1) ----> -2,0,0
(2,1,2)-(4,2,1) ----> -2,-1,1
(1,3) -(4,2) ----> -3,1,1
(1,3) -(2,1) ----> 4,-1,2
I would like to know how to use python create shifting window that allow me to minus my cell arrays.
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I think this pair of functions does what you want. The former may require some tweaking to properly pair the differences.
import numpy as np
def diffs(a,b):
# collect sliding window differences
# length of window determined by the shorter array
# if a,b are not arrays, need to replace b[...]-a with
# a list comprehension
n,m=len(a),len(b)
if n>m:
# ensure s is the shorter
b,a=a,b # switch
n,m=len(a),len(b)
# may need to correct for sign switch
result=[]
for i in range(0,1+m-n):
result.append(b[i:i+n]-a)
return result
def alldiffs(a,b):
# collect all the differences for elements of a and b
# a,b could be lists or arrays of arrays, or 2d arrays
result=[]
for aa in a:
for bb in b:
result.append(diffs(aa,bb))
return result
# define the 3 arrays
# each is a list of 1d arrays
a=[np.array([2,2,1,2]),np.array([1,3])]
b=[np.array([4,2,1])]
c=[np.array([1,2]),np.array([4,3])]
# display the differences
print(alldiffs(a,b))
print(alldiffs(a,c))
print(alldiffs(b,c))
(with a pretty nice print):
1626:~/mypy$ python stack30678737.py
[[array([-2, 0, 0]), array([-2, -1, 1])],
[array([ 3, -1]), array([ 1, -2])]]
[[array([1, 0]), array([ 1, -1]), array([0, 0])],
[array([-2, -1]), array([-2, -2]), array([-3, -1])],
[array([ 0, -1])], [array([3, 0])]]
[[array([3, 0]), array([ 1, -1])],
[array([ 0, -1]), array([-2, -2])]]
Comparing my answer to your question, I wonder if you are adding your shorter arrays with 0, so the result always has 3 elements?
Change a
toa=[np.array([2,2,1,2]),np.array([0,1,3]),np.array([1,3,0])]
gives:
[[array([-2, 0, 0]), array([-2, -1, 1])],
[array([ 4, 1, -2])], [array([ 3, -1, 1])]]
I suppose you could do something more interesting with this inner loop:
for i in range(0,1+m-n):
result.append(b[i:i+n]-a)
But why? The first order of business is to clearly define the characteristics of the problem. Speed can wait. Besides sliding window code in image batches, np.lib.stride_tricks.as_strided
there is a neat trick striding
. But I doubt it will save time, especially not in small examples like this.
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You can use a function sliding_window()
from toolz library to make a switch window:
>>> import numpy as np
>>> import toolz
>>> a = np.array([2,2,1,2])
>>> b = np.array([4, 2, 1])
>>> for chunk in toolz.sliding_window(b.size, a):
...: print(chunk - b)
...:
[-2 0 0]
[-2 -1 1]
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