Issue
I have the following numpy array
u = np.array([a1,b1,a2,b2...,an,bn])
where I would like to subtract the a and b elements from each other and end up with a numpy array:
u_result = np.array([(a2-a1),(b2-b1),(a3-a2),(b3-b2),....,(an-a_(n-1)),(an-a_(n-1))])
How can I do this without too much array splitting and for loops? I'm using this in a larger loop so ideally, I would like to do this efficiently (and learn something new)
(I hope the indexing of the resulting array is clear)
Solution
you can use ravel torearrange as your original vector.
Short answer:
u_r = np.ravel([np.diff(u[::2]),
np.diff(u[1::2])], 'F')
Here a long and moore detailed explanation:
- separate
a
fromb
inu
this can be achieved indexing - differentiate
a
andb
you can use np.diff for easiness of code. - ravel again the differentiated values.
#------- Create u---------------
import numpy as np
a_aux = np.array([50,49,47,43,39,34,28])
b_aux = np.array([1,2,3,4,5,6,7])
u = np.ravel([a_aux,b_aux],'F')
print(u)
#-------------------------------
#1)
# get a as elements with index 0, 2, 4 ....
a = u[::2]
b = u[1::2] #get b as 1,3,5,....
#2)
#differentiate
ad = np.diff(a)
bd = np.diff(b)
#3)
#ravel putting one of everyone
u_result = np.ravel([ad,bd],'F')
print(u_result)
Answered By - Ulises Bussi
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