i have a set of 3 arrays which i would compare to eachother. array a contain a set of values and the values of array b and c should be partly the same.
it is so that if let say a[0] == b[0] than is c[0] always a wrong value
for better explanation i try to show what i mean.
import numpy as np
a = np.array([2, 2, 3, 4])
b = np.array([1, 3, 3, 4])
c = np.array([2, 2, 4, 3])
print(a == b)
# return: [False False True True]
print(a == c)
# return: [True True False False]
as you can see the from both sets i have two True and two False. so if one of both is true the total should be true. when i do the following i get a single True/ False for an array. and the answers are what i want...
print((a == b).all())
# return: False
print((a == c).all())
# return: False
print(a.all() == (b.all() or c.all()))
# return: True
When i make the array b and c so that i have one vallue that in both cases is wrong i should end with a False
import numpy as np
a = np.array([2, 2, 3, 4])
b = np.array([1, 3, 3, 4])
c = np.array([5, 2, 4, 3])
print(a == b)
# return: [False False True True]
print(a == c)
# return: [False True False False]
print((a == b).all())
# return: False
print((a == c).all())
# return: False
So far so Good
print(a.all() == (b.all() or c.all()))
# return: True
this part is not good this should be a False!! How do i get an or function like this so that i end with an True when for each value in a an same value exist in b or c??
EDIT: explanation about a[0] == b[0] than is c[0]: i have an python function where phase information comes in and have to do some actions. before that i want check if i have to do with an array of imaginary values or with an array with phase angles. I want to check this before i do something. The problem is the phase angle because the right side is the inverted phase +/- pi. so for every value i have two options. and yes most of the time it is an exclusive or but in the case +/- pi/2 it is not since both are true and that is also fine...
From your example and explanation, I guess what you want is:
For each position, exactly one of
borchas the same value asa
If that's the case, that can be done with the following code:
def is_exclusively_jointly_same(a, b, c):
return np.logical_xor(a==b, a==c).all() # or use ^ below
# return ((a==b)^(a==c)).all()
The ^ is exclusive or (XOR) operator, which returns True if and only if exactly one of its argument is True.
So the expression (a==b)^(a==c) means either a==b or a==c, but not both. Then the .all() checks whether this is true for all positions in the array.
Examples:
>>> a=np.array([1,2,3,4,5]) >>> b=np.array([1,2,0,0,5]) >>> c=np.array([0,0,3,4,0]) >>> is_exclusively_jointly_same(a, b, c) True >>> a=np.array([1,2,3,4,5]) >>> b=np.array([0,2,0,0,5]) # First value both not 1 >>> c=np.array([0,0,3,4,0]) >>> is_exclusively_jointly_same(a, b, c) False >>> a=np.array([1,2,3,4,5]) >>> b=np.array([1,2,0,0,5]) # First value both 1 >>> c=np.array([1,0,3,4,0]) >>> is_exclusively_jointly_same(a, b, c) False
If what you want is that:
For each position, at least one of
bandcshould have the same value asa
, then you need to change to OR instead of XOR, as follows:
def is_jointly_same(a, b, c):
return np.logical_or(a==b, a==c).all() # or use | below
# return ((a==b) | (a==c)).all()
Examples:
>>> a=np.array([1,2,3,4,5]) >>> b=np.array([1,2,0,0,5]) >>> c=np.array([0,0,3,4,0]) >>> is_jointly_same(a, b, c) True >>> a=np.array([1,2,3,4,5]) >>> b=np.array([0,2,0,0,5]) # First value both not 1 >>> c=np.array([0,0,3,4,0]) >>> is_jointly_same(a, b, c) False >>> a=np.array([1,2,3,4,5]) >>> b=np.array([1,2,0,0,5]) # First value both 1 >>> c=np.array([1,0,3,4,0]) >>> is_jointly_same(a, b, c) True
The key here is that .all() should be applied once as final aggregator, when each individual values has already been calculated. So when you see that you are applying .all() multiple times, you should be concerned.
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