添加零时奇怪的 numpy.sum 行为

问题描述

我了解数学上等效的算术运算如何由于数值错误(例如，以不同顺序求和浮点数)而导致不同的结果.

I understand how mathematically-equivalent arithmentic operations can result in different results due to numerical errors (e.g. summing floats in different orders).

然而，令我惊讶的是，将零添加到 sum 会改变结果.我认为这始终适用于浮点数，无论如何:x + 0. == x.

However, it surprises me that adding zeros to sum can change the result. I thought that this always holds for floats, no matter what: x + 0. == x.

这是一个例子.我希望所有的行都完全为零.谁能解释一下为什么会这样?

Here's an example. I expected all the lines to be exactly zero. Can anybody please explain why this happens?

M = 4  # number of random values
Z = 4  # number of additional zeros
for i in range(20):
    a = np.random.rand(M)
    b = np.zeros(M+Z)
    b[:M] = a
    print a.sum() - b.sum()

-4.4408920985e-16
0.0
0.0
0.0
4.4408920985e-16
0.0
-4.4408920985e-16
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
2.22044604925e-16
0.0
4.4408920985e-16
4.4408920985e-16
0.0

M 和 Z 的较小值似乎不会发生这种情况.

It seems not to happen for smaller values of M and Z.

我还确定了 a.dtype==b.dtype.

这里还有一个例子，它也展示了 python 的内置 sum 的行为符合预期:

Here is one more example, which also demonstrates python's builtin sum behaves as expected:

a = np.array([0.1,      1.0/3,      1.0/7,      1.0/13, 1.0/23])
b = np.array([0.1, 0.0, 1.0/3, 0.0, 1.0/7, 0.0, 1.0/13, 1.0/23])
print a.sum() - b.sum()
=> -1.11022302463e-16
print sum(a) - sum(b)
=> 0.0

我正在使用 numpy V1.9.2.

I'm using numpy V1.9.2.

推荐答案

简答:你看到了两者的区别

a + b + c + d

和

(a + b) + (c + d)

因为浮点数不准确所以不一样.

which because of floating point inaccuracies is not the same.

长答案: Numpy 将成对求和作为速度(它允许更容易矢量化)和舍入误差的优化.

Long answer: Numpy implements pair-wise summation as an optimization of both speed (it allows for easier vectorization) and rounding error.

numpy sum-implementation 可以在 here(函数pairwise_sum_@TYPE@).它基本上做了以下事情:

The numpy sum-implementation can be found here (function pairwise_sum_@TYPE@). It essentially does the following:

1. 如果数组的长度小于 8，则执行常规的 for 循环求和.这就是为什么如果 W < 没有观察到奇怪的结果.4 在您的情况下 - 在两种情况下都将使用相同的 for 循环求和.
2. 如果长度在 8 到 128 之间，则在 8 个 bin r[0]-r[7] 中累加总和，然后通过 ((r[0] + r[1]) + (r[2] + r[3])) + ((r[4] + r[5]) + (r[6] + r[7])).
3. 否则，它将递归地对数组的两半求和.

1. If the length of the array is less than 8, a regular for-loop summation is performed. This is why the strange result is not observed if W < 4 in your case - the same for-loop summation will be used in both cases.
2. If the length is between 8 and 128, it accumulates the sums in 8 bins r[0]-r[7] then sums them by ((r[0] + r[1]) + (r[2] + r[3])) + ((r[4] + r[5]) + (r[6] + r[7])).
3. Otherwise, it recursively sums two halves of the array.

因此，在第一种情况下，您会得到 a.sum() = a[0] + a[1] + a[2] + a[3] 而在第二种情况下 b.sum() = (a[0] + a[1]) + (a[2] + a[3]) 这导致 a.sum() - b.sum() != 0.

Therefore, in the first case you get a.sum() = a[0] + a[1] + a[2] + a[3] and in the second case b.sum() = (a[0] + a[1]) + (a[2] + a[3]) which leads to a.sum() - b.sum() != 0.

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