numpy数组中的整数溢出 [英] Integer overflow in numpy arrays

问题描述

import numpy as np
a = np.arange(1000000).reshape(1000,1000)
print(a**2)

使用此代码，我得到了这个答案.为什么我会得到负值?

With this code I get this answer. Why do I get negative values?

[[         0          1          4 ...,     994009     996004     998001]
 [   1000000    1002001    1004004 ...,    3988009    3992004    3996001]
 [   4000000    4004001    4008004 ...,    8982009    8988004    8994001]
 ..., 
 [1871554624 1873548625 1875542628 ..., -434400663 -432404668 -430408671]
 [-428412672 -426416671 -424420668 ..., 1562593337 1564591332 1566589329]
 [1568587328 1570585329 1572583332 ..., -733379959 -731379964 -729379967]]

推荐答案

在您的平台上，np.arange返回dtype'int32'的数组:

On your platform, np.arange returns an array of dtype 'int32' :

In [1]: np.arange(1000000).dtype
Out[1]: dtype('int32')

数组的每个元素都是32位整数.平方导致结果不适合32位.结果被裁剪为32位，但仍解释为32位整数，这就是为什么看到负数的原因.

Each element of the array is a 32-bit integer. Squaring leads to a result which does not fit in 32-bits. The result is cropped to 32-bits and still interpreted as a 32-bit integer, however, which is why you see negative numbers.

在这种情况下，可以通过在平方之前构造dtype'int64'数组来避免整数溢出:

In this case, you can avoid the integer overflow by constructing an array of dtype 'int64' before squaring:

a=np.arange(1000000,dtype='int64').reshape(1000,1000)

请注意，使用numpy时，发现的问题是固有的危险.您必须谨慎选择dtype，并事先知道您的代码不会导致算术溢出.出于速度考虑，numpy不能也不会在出现这种情况时警告您.

Note that the problem you've discovered is an inherent danger when working with numpy. You have to choose your dtypes with care and know before-hand that your code will not lead to arithmetic overflows. For the sake of speed, numpy can not and will not warn you when this occurs.

请参见 http://mail.scipy.org/pipermail /numpy-discussion/2009-April/041691.html 在numpy邮件列表上对此进行了讨论.

See http://mail.scipy.org/pipermail/numpy-discussion/2009-April/041691.html for a discussion of this on the numpy mailing list.

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