Subtract a column vector from matrix at specified vector of columns using only broadcast
I want to subtract a column vector from a numpy matrix using another vector which is index of columns where the first column vector needs to be subtracted from the main matrix. For eg.
M = array([[ 1, 2, 1, 1],
[ 2, 1, 1, 1],
[ 1, 1, 2, 1],
[ 2, 1, 1, 1],
[ 1, 1, 1, 2]]) # An example matrix
V = array([1, 1, 1, 1, 1]) # An example column vector
I = array([0, 3, 2, 3, 1, 3, 3]) # The index maxtrix
Now I want to subtract V from M at column numbers given in I.
For eg. I[0] is 0, so subtract V from first column (zero index) of matrix M.
Similarly I[1] = 3, subtract V from fourth column (three index) of matrix M.
At the end of operation, since 3 occurs 4 times in I, so V will be subtracted from third column i.e. last column of M- 4 times.
I need to do this using only broadcast, no loops.
I have tried the following:
M[:, I] - V[np.newaxis, :].T
but it ends up broadcasting resultant matrix to have more columns than there are in M.
python numpy matrix numpy-broadcasting
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
add a comment |
I want to subtract a column vector from a numpy matrix using another vector which is index of columns where the first column vector needs to be subtracted from the main matrix. For eg.
M = array([[ 1, 2, 1, 1],
[ 2, 1, 1, 1],
[ 1, 1, 2, 1],
[ 2, 1, 1, 1],
[ 1, 1, 1, 2]]) # An example matrix
V = array([1, 1, 1, 1, 1]) # An example column vector
I = array([0, 3, 2, 3, 1, 3, 3]) # The index maxtrix
Now I want to subtract V from M at column numbers given in I.
For eg. I[0] is 0, so subtract V from first column (zero index) of matrix M.
Similarly I[1] = 3, subtract V from fourth column (three index) of matrix M.
At the end of operation, since 3 occurs 4 times in I, so V will be subtracted from third column i.e. last column of M- 4 times.
I need to do this using only broadcast, no loops.
I have tried the following:
M[:, I] - V[np.newaxis, :].T
but it ends up broadcasting resultant matrix to have more columns than there are in M.
python numpy matrix numpy-broadcasting
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
add a comment |
I want to subtract a column vector from a numpy matrix using another vector which is index of columns where the first column vector needs to be subtracted from the main matrix. For eg.
M = array([[ 1, 2, 1, 1],
[ 2, 1, 1, 1],
[ 1, 1, 2, 1],
[ 2, 1, 1, 1],
[ 1, 1, 1, 2]]) # An example matrix
V = array([1, 1, 1, 1, 1]) # An example column vector
I = array([0, 3, 2, 3, 1, 3, 3]) # The index maxtrix
Now I want to subtract V from M at column numbers given in I.
For eg. I[0] is 0, so subtract V from first column (zero index) of matrix M.
Similarly I[1] = 3, subtract V from fourth column (three index) of matrix M.
At the end of operation, since 3 occurs 4 times in I, so V will be subtracted from third column i.e. last column of M- 4 times.
I need to do this using only broadcast, no loops.
I have tried the following:
M[:, I] - V[np.newaxis, :].T
but it ends up broadcasting resultant matrix to have more columns than there are in M.
python numpy matrix numpy-broadcasting
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
I want to subtract a column vector from a numpy matrix using another vector which is index of columns where the first column vector needs to be subtracted from the main matrix. For eg.
M = array([[ 1, 2, 1, 1],
[ 2, 1, 1, 1],
[ 1, 1, 2, 1],
[ 2, 1, 1, 1],
[ 1, 1, 1, 2]]) # An example matrix
V = array([1, 1, 1, 1, 1]) # An example column vector
I = array([0, 3, 2, 3, 1, 3, 3]) # The index maxtrix
Now I want to subtract V from M at column numbers given in I.
For eg. I[0] is 0, so subtract V from first column (zero index) of matrix M.
Similarly I[1] = 3, subtract V from fourth column (three index) of matrix M.
At the end of operation, since 3 occurs 4 times in I, so V will be subtracted from third column i.e. last column of M- 4 times.
I need to do this using only broadcast, no loops.
I have tried the following:
M[:, I] - V[np.newaxis, :].T
but it ends up broadcasting resultant matrix to have more columns than there are in M.
python numpy matrix numpy-broadcasting
python numpy matrix numpy-broadcasting
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
edited Nov 14 '18 at 7:47
Henrik
41.5k994109
41.5k994109
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
asked Nov 14 '18 at 7:14
Varun KuntalVarun Kuntal
234
234
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
One can use bincount and outer
>>> M - np.outer(V, np.bincount(I, None, M.shape[1]))
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
or subtract.at
>>> out = M.copy()
>>> np.subtract.at(out, (np.s_[:], I), V[:, None])
>>> out
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
bincountshould be more performant!
– Divakar
Nov 14 '18 at 7:43
add a comment |
We can use np.subtract.at on transposed view of M -
np.subtract.at(M.T,I,V)
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
extendingIseems unnecessary - still +1 for brevity
– Paul Panzer
Nov 14 '18 at 7:35
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
One can use bincount and outer
>>> M - np.outer(V, np.bincount(I, None, M.shape[1]))
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
or subtract.at
>>> out = M.copy()
>>> np.subtract.at(out, (np.s_[:], I), V[:, None])
>>> out
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
bincountshould be more performant!
– Divakar
Nov 14 '18 at 7:43
add a comment |
One can use bincount and outer
>>> M - np.outer(V, np.bincount(I, None, M.shape[1]))
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
or subtract.at
>>> out = M.copy()
>>> np.subtract.at(out, (np.s_[:], I), V[:, None])
>>> out
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
bincountshould be more performant!
– Divakar
Nov 14 '18 at 7:43
add a comment |
One can use bincount and outer
>>> M - np.outer(V, np.bincount(I, None, M.shape[1]))
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
or subtract.at
>>> out = M.copy()
>>> np.subtract.at(out, (np.s_[:], I), V[:, None])
>>> out
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
One can use bincount and outer
>>> M - np.outer(V, np.bincount(I, None, M.shape[1]))
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
or subtract.at
>>> out = M.copy()
>>> np.subtract.at(out, (np.s_[:], I), V[:, None])
>>> out
array([[ 0, 1, 0, -3],
[ 1, 0, 0, -3],
[ 0, 0, 1, -3],
[ 1, 0, 0, -3],
[ 0, 0, 0, -2]])
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
edited Nov 14 '18 at 7:32
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
answered Nov 14 '18 at 7:27
Paul PanzerPaul Panzer
30.1k21240
30.1k21240
bincountshould be more performant!
– Divakar
Nov 14 '18 at 7:43
add a comment |
bincountshould be more performant!
– Divakar
Nov 14 '18 at 7:43
bincount should be more performant!– Divakar
Nov 14 '18 at 7:43
bincount should be more performant!– Divakar
Nov 14 '18 at 7:43
add a comment |
We can use np.subtract.at on transposed view of M -
np.subtract.at(M.T,I,V)
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
extendingIseems unnecessary - still +1 for brevity
– Paul Panzer
Nov 14 '18 at 7:35
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
add a comment |
We can use np.subtract.at on transposed view of M -
np.subtract.at(M.T,I,V)
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
extendingIseems unnecessary - still +1 for brevity
– Paul Panzer
Nov 14 '18 at 7:35
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
add a comment |
We can use np.subtract.at on transposed view of M -
np.subtract.at(M.T,I,V)
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
We can use np.subtract.at on transposed view of M -
np.subtract.at(M.T,I,V)
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
edited Nov 14 '18 at 7:42
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
answered Nov 14 '18 at 7:23
DivakarDivakar
156k1485177
156k1485177
extendingIseems unnecessary - still +1 for brevity
– Paul Panzer
Nov 14 '18 at 7:35
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
add a comment |
extendingIseems unnecessary - still +1 for brevity
– Paul Panzer
Nov 14 '18 at 7:35
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
extending
I seems unnecessary - still +1 for brevity– Paul Panzer
Nov 14 '18 at 7:35
extending
I seems unnecessary - still +1 for brevity– Paul Panzer
Nov 14 '18 at 7:35
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
@PaulPanzer Yup, you are right there!
– Divakar
Nov 14 '18 at 7:42
add a comment |
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