快速Numpy滚动_产品 [英] Fast numpy rolling_product

问题描述

我需要rolling_product函数或expanding_product函数.

I need a rolling_product function, or an expanding_product function.

有各种pandas rolling_XXXX和expanding_XXXX函数，但是我惊讶地发现缺少expanding_product()函数.

There are various pandas rolling_XXXX and expanding_XXXX functions, but I was surprised to discover the absence of an expanding_product() function.

为了让事情正常进行，我一直在使用这种相当慢的替代方法

To get things working I've been using this rather slow alternative

pd.expanding_apply(temp_col, lambda x : x.prod())

我的数组通常包含32,000个元素，因此这被证明是一个瓶颈.我很想尝试log()，cumsum()和exp()，但是我认为我应该在这里提出问题，因为可能会有更好的解决方案.

My arrays often have 32,000 elements so this is proving to be a bit of a bottleneck. I was tempted to try log(), cumsum(), and exp(), but I thought I should ask on here since there might be a much better solution.

推荐答案

我有一个更快的机制，尽管您需要运行一些测试以查看准确性是否足够.

I have a faster mechanism, though you'll need to run some tests to see if the accuracy is sufficient.

这是原始的exp/sum/log版本:

Here's the original exp/sum/log version:

def rolling_prod1(xs, n):
    return np.exp(pd.rolling_sum(np.log(xs), n))

这是一个使用累积乘积，将其转移(用nans预先填充)，然后将其重新分配出去的版本.

And here's a version that takes the cumulative product, shifts it over (pre-filling with nans), and then divides it back out.

def rolling_prod2(xs, n):
    cxs = np.cumprod(xs)
    nans = np.empty(n)
    nans[:] = np.nan
    nans[n-1] = 1.
    a = np.concatenate((nans, cxs[:len(cxs)-n]))
    return cxs / a

在此示例中，两个函数均返回相同的结果:

Both functions return the same result for this example:

In [9]: xs
Out[9]: array([ 1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.])

In [10]: rolling_prod1(xs, 3)
Out[10]: array([  nan,   nan,    6.,   24.,   60.,  120.,  210.,  336.,  504.])

In [11]: rolling_prod2(xs, 3)
Out[11]: array([  nan,   nan,    6.,   24.,   60.,  120.,  210.,  336.,  504.])

但是第二个版本要快得多:

But the second version is much faster:

In [12]: temp_col = np.random.rand(30000)

In [13]: %timeit rolling_prod1(temp_col, 3)
1000 loops, best of 3: 694 µs per loop

In [14]: %timeit rolling_prod2(temp_col, 3)
10000 loops, best of 3: 162 µs per loop

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