Mixing
How to fill elements of a matrix with indices taken from another matrix in python
2
I have the following code, which constructs two matrices a (3*2 matrix of zeros) and b. I want to replace the 0s in matrix a with 1s depending on the index location stored in index b. I want to do it without a for loop.
import numpy as np
a = np.zeros((3, 2))
b = np.array([0, 1, 0])
The output should be
c = [[1, 0],
[0, 1],
[1, 0]]
python numpy matrix indexing
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
add a comment |
2
I have the following code, which constructs two matrices a (3*2 matrix of zeros) and b. I want to replace the 0s in matrix a with 1s depending on the index location stored in index b. I want to do it without a for loop.
import numpy as np
a = np.zeros((3, 2))
b = np.array([0, 1, 0])
The output should be
c = [[1, 0],
[0, 1],
[1, 0]]
python numpy matrix indexing
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
Can you showcas a proper 2D array?
– Mad Physicist
Jan 2 at 18:08
So are you saying that each element ofbcorresponds to a row ofa, and tells you which index to set?
– Mad Physicist
Jan 2 at 18:09
add a comment |
2
2
2
I have the following code, which constructs two matrices a (3*2 matrix of zeros) and b. I want to replace the 0s in matrix a with 1s depending on the index location stored in index b. I want to do it without a for loop.
import numpy as np
a = np.zeros((3, 2))
b = np.array([0, 1, 0])
The output should be
c = [[1, 0],
[0, 1],
[1, 0]]
python numpy matrix indexing
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
I have the following code, which constructs two matrices a (3*2 matrix of zeros) and b. I want to replace the 0s in matrix a with 1s depending on the index location stored in index b. I want to do it without a for loop.
import numpy as np
a = np.zeros((3, 2))
b = np.array([0, 1, 0])
The output should be
c = [[1, 0],
[0, 1],
[1, 0]]
python numpy matrix indexing
python numpy matrix indexing
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
edited Jan 2 at 18:14
Mad Physicist
38.4k1679112
38.4k1679112
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
asked Jan 2 at 18:03
Aziz Ur RehmanAziz Ur Rehman
136
136
Can you showcas a proper 2D array?
– Mad Physicist
Jan 2 at 18:08
So are you saying that each element ofbcorresponds to a row ofa, and tells you which index to set?
– Mad Physicist
Jan 2 at 18:09
add a comment |
Can you showcas a proper 2D array?
– Mad Physicist
Jan 2 at 18:08
So are you saying that each element ofbcorresponds to a row ofa, and tells you which index to set?
– Mad Physicist
Jan 2 at 18:09
Can you show
c as a proper 2D array?– Mad Physicist
Jan 2 at 18:08
Can you show
c as a proper 2D array?– Mad Physicist
Jan 2 at 18:08
So are you saying that each element of
b corresponds to a row of a, and tells you which index to set?– Mad Physicist
Jan 2 at 18:09
So are you saying that each element of
b corresponds to a row of a, and tells you which index to set?– Mad Physicist
Jan 2 at 18:09
add a comment |
1 Answer
1
active
oldest
votes
3
Numpy fancy indexing is your friend here. To make b work as the column index, you need an array of equal size to b which will tell you the row that each index applies to:
a[np.arange(b.size), b] = 1
This modifies a in-place. If that is not what you want, make a copy first:
c = a.copy()
c[np.arange(b.size), b] = 1
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
You can also usea[range(len(b)),b] = 1. Perhaps you can add this as an alternative
– Bazingaa
Jan 2 at 18:19
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,np.arangemakes sense
– Bazingaa
Jan 2 at 18:22
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
|
show 1 more comment
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
Numpy fancy indexing is your friend here. To make b work as the column index, you need an array of equal size to b which will tell you the row that each index applies to:
a[np.arange(b.size), b] = 1
This modifies a in-place. If that is not what you want, make a copy first:
c = a.copy()
c[np.arange(b.size), b] = 1
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
You can also usea[range(len(b)),b] = 1. Perhaps you can add this as an alternative
– Bazingaa
Jan 2 at 18:19
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,np.arangemakes sense
– Bazingaa
Jan 2 at 18:22
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
|
show 1 more comment
3
Numpy fancy indexing is your friend here. To make b work as the column index, you need an array of equal size to b which will tell you the row that each index applies to:
a[np.arange(b.size), b] = 1
This modifies a in-place. If that is not what you want, make a copy first:
c = a.copy()
c[np.arange(b.size), b] = 1
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
You can also usea[range(len(b)),b] = 1. Perhaps you can add this as an alternative
– Bazingaa
Jan 2 at 18:19
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,np.arangemakes sense
– Bazingaa
Jan 2 at 18:22
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
|
show 1 more comment
3
3
3
Numpy fancy indexing is your friend here. To make b work as the column index, you need an array of equal size to b which will tell you the row that each index applies to:
a[np.arange(b.size), b] = 1
This modifies a in-place. If that is not what you want, make a copy first:
c = a.copy()
c[np.arange(b.size), b] = 1
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
Numpy fancy indexing is your friend here. To make b work as the column index, you need an array of equal size to b which will tell you the row that each index applies to:
a[np.arange(b.size), b] = 1
This modifies a in-place. If that is not what you want, make a copy first:
c = a.copy()
c[np.arange(b.size), b] = 1
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
answered Jan 2 at 18:11
Mad PhysicistMad Physicist
38.4k1679112
38.4k1679112
You can also usea[range(len(b)),b] = 1. Perhaps you can add this as an alternative
– Bazingaa
Jan 2 at 18:19
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,np.arangemakes sense
– Bazingaa
Jan 2 at 18:22
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
|
show 1 more comment
You can also usea[range(len(b)),b] = 1. Perhaps you can add this as an alternative
– Bazingaa
Jan 2 at 18:19
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,np.arangemakes sense
– Bazingaa
Jan 2 at 18:22
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
You can also use
a[range(len(b)),b] = 1. Perhaps you can add this as an alternative– Bazingaa
Jan 2 at 18:19
You can also use
a[range(len(b)),b] = 1. Perhaps you can add this as an alternative– Bazingaa
Jan 2 at 18:19
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
@Bazingaa. If I'm not mistaken, that is a false memory saving device. Internally, the index will be expanded to a numpy array regardless of how you pass it in.
– Mad Physicist
Jan 2 at 18:20
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,
np.arange makes sense– Bazingaa
Jan 2 at 18:22
Anyway since NumPy is already used to create the matrix, it rules out the point of import overkill. So yeah,
np.arange makes sense– Bazingaa
Jan 2 at 18:22
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@Bazingaa. Exactly. If you're using the library already, theres no point in holding back.
– Mad Physicist
Jan 2 at 18:23
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
@MadPhysicist thank you so much this works perfect for me
– Aziz Ur Rehman
Jan 2 at 18:34
|
show 1 more comment
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Can you show
cas a proper 2D array?– Mad Physicist
Jan 2 at 18:08
So are you saying that each element of
bcorresponds to a row ofa, and tells you which index to set?– Mad Physicist
Jan 2 at 18:09