Multiply numpy int and float arrays: Cannot cast ufunc multiply output from dtype
10
I'd like to multiply an int16 array but a float array, with auto rounding, but this fails :
import numpy
A = numpy.array([1, 2, 3, 4], dtype=numpy.int16)
B = numpy.array([0.5, 2.1, 3, 4], dtype=numpy.float64)
A *= B
I get:
TypeError: Cannot cast ufunc multiply output from dtype('float64') to dtype('int16') with casting rule 'same_kind'
python arrays numpy floating-point int
edited Nov 23 '18 at 22:37
seralouk
5,71522339
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
add a comment |
10
I'd like to multiply an int16 array but a float array, with auto rounding, but this fails :
import numpy
A = numpy.array([1, 2, 3, 4], dtype=numpy.int16)
B = numpy.array([0.5, 2.1, 3, 4], dtype=numpy.float64)
A *= B
I get:
TypeError: Cannot cast ufunc multiply output from dtype('float64') to dtype('int16') with casting rule 'same_kind'
python arrays numpy floating-point int
edited Nov 23 '18 at 22:37
seralouk
5,71522339
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
It seems it's possible withnumpy.multiply(A, B, out=A, casting='unsafe')but that's way much longer syntax! Is there a way to set casting='unsafe' in numpy by default?
– Basj
Jul 30 '16 at 12:10
See github.com/numpy/numpy/pull/6499/files
– Basj
Jul 30 '16 at 12:11
add a comment |
10
10
10
1
I'd like to multiply an int16 array but a float array, with auto rounding, but this fails :
import numpy
A = numpy.array([1, 2, 3, 4], dtype=numpy.int16)
B = numpy.array([0.5, 2.1, 3, 4], dtype=numpy.float64)
A *= B
I get:
TypeError: Cannot cast ufunc multiply output from dtype('float64') to dtype('int16') with casting rule 'same_kind'
python arrays numpy floating-point int
edited Nov 23 '18 at 22:37
seralouk
5,71522339
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
I'd like to multiply an int16 array but a float array, with auto rounding, but this fails :
import numpy
A = numpy.array([1, 2, 3, 4], dtype=numpy.int16)
B = numpy.array([0.5, 2.1, 3, 4], dtype=numpy.float64)
A *= B
I get:
TypeError: Cannot cast ufunc multiply output from dtype('float64') to dtype('int16') with casting rule 'same_kind'
python arrays numpy floating-point int
python arrays numpy floating-point int
edited Nov 23 '18 at 22:37
seralouk
5,71522339
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
edited Nov 23 '18 at 22:37
seralouk
5,71522339
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
edited Nov 23 '18 at 22:37
seralouk
5,71522339
edited Nov 23 '18 at 22:37
seralouk
5,71522339
edited Nov 23 '18 at 22:37
seralouk
5,71522339
5,71522339
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
asked Jul 30 '16 at 11:39
BasjBasj
5,58729104226
5,58729104226
It seems it's possible withnumpy.multiply(A, B, out=A, casting='unsafe')but that's way much longer syntax! Is there a way to set casting='unsafe' in numpy by default?
– Basj
Jul 30 '16 at 12:10
See github.com/numpy/numpy/pull/6499/files
– Basj
Jul 30 '16 at 12:11
add a comment |
It seems it's possible withnumpy.multiply(A, B, out=A, casting='unsafe')but that's way much longer syntax! Is there a way to set casting='unsafe' in numpy by default?
– Basj
Jul 30 '16 at 12:10
See github.com/numpy/numpy/pull/6499/files
– Basj
Jul 30 '16 at 12:11
It seems it's possible with
numpy.multiply(A, B, out=A, casting='unsafe') but that's way much longer syntax! Is there a way to set casting='unsafe' in numpy by default?– Basj
Jul 30 '16 at 12:10
It seems it's possible with
numpy.multiply(A, B, out=A, casting='unsafe') but that's way much longer syntax! Is there a way to set casting='unsafe' in numpy by default?– Basj
Jul 30 '16 at 12:10
See github.com/numpy/numpy/pull/6499/files
– Basj
Jul 30 '16 at 12:11
See github.com/numpy/numpy/pull/6499/files
– Basj
Jul 30 '16 at 12:11
add a comment |
3 Answers
3
active
oldest
votes
6
2 ways to solve this:
You can solve this by replacing
A *= B
with
A = (A * B)
or with
numpy.multiply(A, B, out=A, casting='unsafe')
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
add a comment |
2
You could use broadcasting to multiply the two arrays and take only the integer part as follows:
In [2]: (A*B).astype(int)
Out[2]: array([ 0, 4, 9, 16])
Timing Constraints:
In [8]: %timeit (A*B).astype(int)
1000000 loops, best of 3: 1.65 µs per loop
In [9]: %timeit np.multiply(A, B, out=A, casting='unsafe')
100000 loops, best of 3: 2.01 µs per loop
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
How fast is it compared tonumpy.multiply(A, B, out=A, casting='unsafe')?
– Basj
Jul 30 '16 at 18:45
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
2
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to typeA =, but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.
– SethMMorton
Jul 30 '16 at 18:59
|
show 1 more comment
2
import numpy as np
A = np.float_(A)
A *= B
try this. I think are different array type you get fail.
Cast
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
2 ways to solve this:
You can solve this by replacing
A *= B
with
A = (A * B)
or with
numpy.multiply(A, B, out=A, casting='unsafe')
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
add a comment |
6
2 ways to solve this:
You can solve this by replacing
A *= B
with
A = (A * B)
or with
numpy.multiply(A, B, out=A, casting='unsafe')
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
add a comment |
6
6
6
2 ways to solve this:
You can solve this by replacing
A *= B
with
A = (A * B)
or with
numpy.multiply(A, B, out=A, casting='unsafe')
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
2 ways to solve this:
You can solve this by replacing
A *= B
with
A = (A * B)
or with
numpy.multiply(A, B, out=A, casting='unsafe')
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
edited Jul 22 '18 at 18:14
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
answered Jun 30 '17 at 14:12
seraloukseralouk
5,71522339
5,71522339
add a comment |
add a comment |
2
You could use broadcasting to multiply the two arrays and take only the integer part as follows:
In [2]: (A*B).astype(int)
Out[2]: array([ 0, 4, 9, 16])
Timing Constraints:
In [8]: %timeit (A*B).astype(int)
1000000 loops, best of 3: 1.65 µs per loop
In [9]: %timeit np.multiply(A, B, out=A, casting='unsafe')
100000 loops, best of 3: 2.01 µs per loop
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
How fast is it compared tonumpy.multiply(A, B, out=A, casting='unsafe')?
– Basj
Jul 30 '16 at 18:45
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
2
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to typeA =, but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.
– SethMMorton
Jul 30 '16 at 18:59
|
show 1 more comment
2
You could use broadcasting to multiply the two arrays and take only the integer part as follows:
In [2]: (A*B).astype(int)
Out[2]: array([ 0, 4, 9, 16])
Timing Constraints:
In [8]: %timeit (A*B).astype(int)
1000000 loops, best of 3: 1.65 µs per loop
In [9]: %timeit np.multiply(A, B, out=A, casting='unsafe')
100000 loops, best of 3: 2.01 µs per loop
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
How fast is it compared tonumpy.multiply(A, B, out=A, casting='unsafe')?
– Basj
Jul 30 '16 at 18:45
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
2
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to typeA =, but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.
– SethMMorton
Jul 30 '16 at 18:59
|
show 1 more comment
2
2
2
You could use broadcasting to multiply the two arrays and take only the integer part as follows:
In [2]: (A*B).astype(int)
Out[2]: array([ 0, 4, 9, 16])
Timing Constraints:
In [8]: %timeit (A*B).astype(int)
1000000 loops, best of 3: 1.65 µs per loop
In [9]: %timeit np.multiply(A, B, out=A, casting='unsafe')
100000 loops, best of 3: 2.01 µs per loop
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
You could use broadcasting to multiply the two arrays and take only the integer part as follows:
In [2]: (A*B).astype(int)
Out[2]: array([ 0, 4, 9, 16])
Timing Constraints:
In [8]: %timeit (A*B).astype(int)
1000000 loops, best of 3: 1.65 µs per loop
In [9]: %timeit np.multiply(A, B, out=A, casting='unsafe')
100000 loops, best of 3: 2.01 µs per loop
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
edited Jul 30 '16 at 18:49
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
answered Jul 30 '16 at 18:40
Nickil MaveliNickil Maveli
17.4k43448
17.4k43448
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
How fast is it compared tonumpy.multiply(A, B, out=A, casting='unsafe')?
– Basj
Jul 30 '16 at 18:45
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
2
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to typeA =, but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.
– SethMMorton
Jul 30 '16 at 18:59
|
show 1 more comment
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
How fast is it compared tonumpy.multiply(A, B, out=A, casting='unsafe')?
– Basj
Jul 30 '16 at 18:45
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
2
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to typeA =, but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.
– SethMMorton
Jul 30 '16 at 18:59
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
OP wants to do the multiplication in place
– ali_m
Jul 30 '16 at 18:45
How fast is it compared to
numpy.multiply(A, B, out=A, casting='unsafe') ?– Basj
Jul 30 '16 at 18:45
How fast is it compared to
numpy.multiply(A, B, out=A, casting='unsafe') ?– Basj
Jul 30 '16 at 18:45
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@ali_m Nowhere does the OP mention they want to do the multiplication in place. Their only comment was that the longer way was "way much longer syntax".
– SethMMorton
Jul 30 '16 at 18:51
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
@SethMMorton Well, both the example given in the question and the workaround the OP posted in the comments are explicitly performing in place multiplication.
– ali_m
Jul 30 '16 at 18:56
2
2
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to type
A = , but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.– SethMMorton
Jul 30 '16 at 18:59
@ali_m I don't disagree with that, but based on this question and the other question it seems the OP is more interested in terseness than an in-place operation. Having said that, I think the assumption that they want an in place operation is good because of the examples given and that in-place saves having to type
A = , but I always think it is safer to wait for the OP to explicitly make it clear what they want rather than make assumptions.– SethMMorton
Jul 30 '16 at 18:59
|
show 1 more comment
2
import numpy as np
A = np.float_(A)
A *= B
try this. I think are different array type you get fail.
Cast
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
add a comment |
2
import numpy as np
A = np.float_(A)
A *= B
try this. I think are different array type you get fail.
Cast
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
add a comment |
2
2
2
import numpy as np
A = np.float_(A)
A *= B
try this. I think are different array type you get fail.
Cast
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
import numpy as np
A = np.float_(A)
A *= B
try this. I think are different array type you get fail.
Cast
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
answered Jul 30 '16 at 19:18
Ahmet İlginAhmet İlgin
254
254
add a comment |
add a comment |
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It seems it's possible with
numpy.multiply(A, B, out=A, casting='unsafe')but that's way much longer syntax! Is there a way to set casting='unsafe' in numpy by default?– Basj
Jul 30 '16 at 12:10
See github.com/numpy/numpy/pull/6499/files
– Basj
Jul 30 '16 at 12:11