b10hazard - 1 year ago 94

Python Question

I have to create a very large grid using numpy meshgrid. In order to save memory I use int8 as my dtype for the arrays I'm trying to mesh. However, meshgrid keeps changing the type to int64 which uses a ton of memory. Here is a simple example of the problem...

`import numpy`

grids = [numpy.arange(1, 4, dtype=numpy.int8), numpy.arange(1, 5, dtype=numpy.int8)]

print grids

print grids[0].dtype, grids[0].nbytes

x1, y1 = numpy.meshgrid(*grids)

print x1.dtype, x1.nbytes

This script prints

`[array([1, 2, 3], dtype=int8), array([1, 2, 3, 4], dtype=int8)]`

int8 3

int64 96

Why does meshgrid do this? Is there any way to stop it? I need to create a huge array so I cannot use meshgrid unless I can control the data type of the output. Is this intended behavior or is it a numpy bug? Every other function I've used in numpy preserves datatype of allows you to change it. The meshgrid function does not appear to allow this.

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Answer Source

You can set the optional `copy`

parameter of `numpy.meshgrid()`

to `False`

(note, however, that it has some constraints):

`meshgrid(*xi, **kwargs)`

...

`copy`

:`bool`

, optionalIf

`False`

, a view into the original arrays are returned in order to conserve memory. Default is`True`

. Please note that`sparse=False`

,`copy=False`

will likely return non-contiguous arrays. Furthermore, more than one element of a broadcast array may refer to a single memory location. If you need to write to the arrays, make copies first.

Proof that it works:

```
>>> import numpy
>>>
>>> grids = [numpy.arange(1, 4, dtype=numpy.int8), numpy.arange(1, 5, dtype=numpy.int8)]
>>>
>>> print grids
[array([1, 2, 3], dtype=int8), array([1, 2, 3, 4], dtype=int8)]
>>> print grids[0].dtype, grids[0].nbytes
int8 3
>>>
>>> x1, y1 = numpy.meshgrid(*grids, copy=False)
>>> # ^^^^^^^^^^
>>> print x1.dtype, x1.nbytes
int8 12
```

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