查找最接近数组 2 元素的数组 1 中的元素

问题描述

这个答案解释了如何找到距离单点，以一种对大型数组有效的方式(稍作修改):

def arg_nearest(array, value):idx = np.searchsorted(array, value, side="left")如果 idx >0 和 (idx == len(array) 或 math.fabs(value - array[idx-1]) 

如果相反，我们想要找到最接近集合点的数组元素(即第二个数组)；除了使用 for 循环之外，是否有有效的(速度，对于大型数组)方法来扩展它?

一些测试用例:

>>>xx = [0.2, 0.8, 1.3, 1.5, 2.0, 3.1, 3.8, 3.9, 4.5, 5.1, 5.5]>>>yy = [1, 2, 3, 4, 5]>>>of_x_nearest_y(xx, yy)[0.5, 2.0, 3.1, 3.9, 5.1]>>>xx = [0.2, 0.8, 1.3, 1.5, 2.0, 3.1, 3.8, 3.9, 4.5, 5.1, 5.5]>>>yy = [-2, -1, 4.6, 5.8]>>>of_x_nearest_y(xx, yy)[0.2, 0.2, 4.5, 5.5]

假设两个数组都已排序，您可以通过排除低于已匹配值的值来比 完全 天真的 for 循环做得更好一点，即

def args_nearest(选项，目标):locs = np.zeros(targets.size, dtype=int)上一个 = 0对于 ii, tt in enumerate(targets):locs[ii] = prev + arg_nearest(options[prev:], tt)上一个 = locs[ii]返回位置

解决方案

你可以做一些改变来扩展 value 中的元素数组，就像这样 -

idx = np.searchsorted(xx, yy, side="left").clip(max=xx.size-1)掩码 = (idx > 0) &\( (idx == len(xx)) | (np.fabs(yy - xx[idx-1]) < np.fabs(yy - xx[idx])) )out = xx[idx-掩码]

说明

命名法:array 是我们希望在其中放置来自 value 的元素以保持 array 的排序性质的数组.>

将单个元素的解决方案扩展到多个元素以进行搜索所需的更改:

1] 最大剪辑从 np.searchsorted 获得的索引数组 idx.array.size-1，因为对于value中大于array最大值的元素，我们需要使idx 可被 array 索引.

2] 引入 numpy 来替换 math 以矢量化方式执行这些操作.

3] 用idx - mask 的技巧替换条件语句.在这种情况下，Python 内部会将 mask 向上转换为 int 数组以匹配 idx 的数据类型.因此，所有 True 元素都变成了 1，因此对于 True 元素，我们将有效地拥有 idx-1，即是原始代码中 IF 条件语句的 True 情况.

This answer explains how to find the nearest (sorted) array element to a single point, in a manner efficient for large arrays (slightly modified):

def arg_nearest(array, value):
    idx = np.searchsorted(array, value, side="left")
    if idx > 0 and (idx == len(array) or math.fabs(value - array[idx-1]) < math.fabs(value - array[idx])):
        return idx-1
    else:
        return idx

If, instead, we want to find the array elements nearest a set of points (i.e. a second array); are there efficient (by speed, for large arrays) ways of extending this besides using a for-loop?

Some test cases:

>>> xx = [0.2, 0.8, 1.3, 1.5, 2.0, 3.1, 3.8, 3.9, 4.5, 5.1, 5.5]
>>> yy = [1, 2, 3, 4, 5]
>>> of_x_nearest_y(xx, yy)
[0.5, 2.0, 3.1, 3.9, 5.1]

>>> xx = [0.2, 0.8, 1.3, 1.5, 2.0, 3.1, 3.8, 3.9, 4.5, 5.1, 5.5]
>>> yy = [-2, -1, 4.6, 5.8]
>>> of_x_nearest_y(xx, yy)
[0.2, 0.2, 4.5, 5.5]

---

Edit: assuming both arrays are sorted, you can do a little better than a completely naive for-loop by excluding values below those already matched, i.e.

def args_nearest(options, targets):
    locs = np.zeros(targets.size, dtype=int)
    prev = 0
    for ii, tt in enumerate(targets):
        locs[ii] = prev + arg_nearest(options[prev:], tt)
        prev = locs[ii]
    return locs

解决方案

You can make few changes to extend it for an array of elements in value, like so -

idx = np.searchsorted(xx, yy, side="left").clip(max=xx.size-1)
mask = (idx > 0) &  \
       ( (idx == len(xx)) | (np.fabs(yy - xx[idx-1]) < np.fabs(yy - xx[idx])) )
out = xx[idx-mask]

Explanation

Nomenclature : array is the array in which we are looking to place elements from value to maintain the sorted nature of array.

Changes needed to extend the solution for a single element to many elements for searching :

1] Clip the indices array idx obtained from np.searchsorted at a max. of array.size-1, because for elements in value that are larger than the maximum of array, we need to make idx indexable by array.

2] Introduce numpy to replace math to do those operations in a vectorized manner.

3] Replace the conditional statement by the trick of idx - mask. In this case, internally Python would up-convert mask to an int array to match up with the datatype of idx. Thus, all the True elements become 1 and thus for True elements we would effectively have idx-1, which is the True case of the IF conditional statement in the original code.

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