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Module « numpy.matlib »
Signature de la fonction shares_memory
Description
help(numpy.matlib.shares_memory)
shares_memory(a, b, /, max_work=None)
Determine if two arrays share memory.
.. warning::
This function can be exponentially slow for some inputs, unless
`max_work` is set to zero or a positive integer.
If in doubt, use `numpy.may_share_memory` instead.
Parameters
----------
a, b : ndarray
Input arrays
max_work : int, optional
Effort to spend on solving the overlap problem (maximum number
of candidate solutions to consider). The following special
values are recognized:
max_work=-1 (default)
The problem is solved exactly. In this case, the function returns
True only if there is an element shared between the arrays. Finding
the exact solution may take extremely long in some cases.
max_work=0
Only the memory bounds of a and b are checked.
This is equivalent to using ``may_share_memory()``.
Raises
------
numpy.exceptions.TooHardError
Exceeded max_work.
Returns
-------
out : bool
See Also
--------
may_share_memory
Examples
--------
>>> import numpy as np
>>> x = np.array([1, 2, 3, 4])
>>> np.shares_memory(x, np.array([5, 6, 7]))
False
>>> np.shares_memory(x[::2], x)
True
>>> np.shares_memory(x[::2], x[1::2])
False
Checking whether two arrays share memory is NP-complete, and
runtime may increase exponentially in the number of
dimensions. Hence, `max_work` should generally be set to a finite
number, as it is possible to construct examples that take
extremely long to run:
>>> from numpy.lib.stride_tricks import as_strided
>>> x = np.zeros([192163377], dtype=np.int8)
>>> x1 = as_strided(
... x, strides=(36674, 61119, 85569), shape=(1049, 1049, 1049))
>>> x2 = as_strided(
... x[64023025:], strides=(12223, 12224, 1), shape=(1049, 1049, 1))
>>> np.shares_memory(x1, x2, max_work=1000)
Traceback (most recent call last):
...
numpy.exceptions.TooHardError: Exceeded max_work
Running ``np.shares_memory(x1, x2)`` without `max_work` set takes
around 1 minute for this case. It is possible to find problems
that take still significantly longer.
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