I have the following numpy array which is basically a 3 channel image:
arr = np.zeros((6, 4, 3), dtype=np.float32)
# dictionary of values, key is array location
values_of_channel_0 = {
(0, 2) : 1,
(1, 0) : 1,
(1, 3) : 5,
(2, 1) : 2,
(2, 2) : 3,
(2, 3) : 1,
(3, 0) : 1,
(3, 2) : 2,
(4, 0) : 2,
(4, 2) : 20,
(5, 0) : 1,
(5, 2) : 10,
(5, 3) : 1
}
I am trying to find the most elegant way to set all the values of the 3rd channel according to the dictionary. Here is what I tried:
locations = list(values_of_channel_0.keys())
values = list(values_of_channel_0.values())
arr[lc,0] = values # trying to set the 3rd channel
But this fails.
Is there a way in which this can be done without looping over keys and values?
CodePudding user response:
What's wrong with a simple loop? Something will have to iterate over the key/value-pairs you provide in your dictionary in any case?
import numpy as np
arr = np.zeros((6, 4, 3), dtype=np.float32)
# dictionary of values, key is array location
values_of_channel_0 = {
(0, 2) : 1,
(1, 0) : 1,
(1, 3) : 5,
(2, 1) : 2,
(2, 2) : 3,
(2, 3) : 1,
(3, 0) : 1,
(3, 2) : 2,
(4, 0) : 2,
(4, 2) : 20,
(5, 0) : 1,
(5, 2) : 10,
(5, 3) : 1
}
for (a, b), v in values_of_channel_0.items():
arr[a, b, 0] = v
print(arr)
Result:
[[[ 0. 0. 0.]
[ 0. 0. 0.]
[ 1. 0. 0.]
[ 0. 0. 0.]]
[[ 1. 0. 0.]
[ 0. 0. 0.]
[ 0. 0. 0.]
[ 5. 0. 0.]]
[[ 0. 0. 0.]
[ 2. 0. 0.]
[ 3. 0. 0.]
[ 1. 0. 0.]]
[[ 1. 0. 0.]
[ 0. 0. 0.]
[ 2. 0. 0.]
[ 0. 0. 0.]]
[[ 2. 0. 0.]
[ 0. 0. 0.]
[20. 0. 0.]
[ 0. 0. 0.]]
[[ 1. 0. 0.]
[ 0. 0. 0.]
[10. 0. 0.]
[ 1. 0. 0.]]]
If you insist on not looping for the assignment, you can construct a data structure that can be assigned at once:
channel_0 = [[values_of_channel_0[b, a] if (b, a) in values_of_channel_0 else 0 for a in range(4)] for b in range(6)]
arr[..., 0] = channel_0
But this is clearly rather pointless and not even more efficient. If you have some control over how values_of_channel_0
is constructed, you could consider constructing it as a nested list or array of the right dimensions immediately, to allow for this type of assignment.
Users @mechanicpig and @michaelszczesny offer a very clean alternative (which will be more efficient since it relies on the efficient implementation of zip()
):
arr[(*zip(*values_of_channel_0), 0)] = list(values_of_channel_0.values())