Multiply numpy int and float arrays: Cannot cast ufunc multiply output from dtype

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'

edited Nov 23 '18 at 22:37

seralouk

5,71522339

asked Jul 30 '16 at 11:39

Basj

5,58729104226

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

add a comment |

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'

edited Nov 23 '18 at 22:37

seralouk

5,71522339

asked Jul 30 '16 at 11:39

Basj

5,58729104226

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

add a comment |

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'

edited Nov 23 '18 at 22:37

seralouk

5,71522339

asked Jul 30 '16 at 11:39

Basj

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

edited Nov 23 '18 at 22:37

seralouk

5,71522339

asked Jul 30 '16 at 11:39

Basj

5,58729104226

edited Nov 23 '18 at 22:37

seralouk

5,71522339

asked Jul 30 '16 at 11:39

Basj

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

asked Jul 30 '16 at 11:39

Basj

5,58729104226

asked Jul 30 '16 at 11:39

Basj

5,58729104226

asked Jul 30 '16 at 11:39

Basj

5,58729104226

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

add a comment |

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

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

add a comment |

3 Answers
3

active

oldest

votes

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')

edited Jul 22 '18 at 18:14

answered Jun 30 '17 at 14:12

seralouk

5,71522339

add a comment |

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

edited Jul 30 '16 at 18:49

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

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

@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 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

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 İlgin

254

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

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')

edited Jul 22 '18 at 18:14

answered Jun 30 '17 at 14:12

seralouk

5,71522339

add a comment |

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')

edited Jul 22 '18 at 18:14

answered Jun 30 '17 at 14:12

seralouk

5,71522339

add a comment |

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')

edited Jul 22 '18 at 18:14

answered Jun 30 '17 at 14:12

seralouk

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')

edited Jul 22 '18 at 18:14

answered Jun 30 '17 at 14:12

seralouk

5,71522339

edited Jul 22 '18 at 18:14

answered Jun 30 '17 at 14:12

seralouk

5,71522339

answered Jun 30 '17 at 14:12

seralouk

5,71522339

answered Jun 30 '17 at 14:12

seralouk

5,71522339

add a comment |

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

edited Jul 30 '16 at 18:49

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

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

@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 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

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

edited Jul 30 '16 at 18:49

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

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

@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 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

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

edited Jul 30 '16 at 18:49

answered Jul 30 '16 at 18:40

Nickil 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

edited Jul 30 '16 at 18:49

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

edited Jul 30 '16 at 18:49

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

answered Jul 30 '16 at 18:40

Nickil Maveli

17.4k43448

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

@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 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

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

@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 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

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

@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

@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

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 İlgin

254

add a comment |

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 İlgin

254

add a comment |

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 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 İlgin

254

answered Jul 30 '16 at 19:18

Ahmet İlgin

254

answered Jul 30 '16 at 19:18

Ahmet İlgin

254

answered Jul 30 '16 at 19:18

Ahmet İlgin

254

add a comment |

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