numpy数组的bisect_right

Question

我希望在排序的numpy数组(查询值不在数组中)上找到与查询最接近的较高值的索引。

类似于python标准库中的bisect_right，没有将numpy数组转换为python列表，并利用了数组已排序的事实(即运行时应该是O(log )，就像numpy的searchsorted一样)。

熊猫有这个选项，使用get_loc和'bfill‘选项，但是仅仅为了这个而包含它作为一个依赖项似乎有点过分了……我可能不得不将这个数组同时作为python list和numpy数组保存，但我想知道是否有更合理的解决方案。

编辑:看起来searchsorted做的正是我需要的。

Answer 1

我们可以在github上看到bisect_right的代码：

def bisect_right(a, x, lo=0, hi=None):
    """Return the index where to insert item x in list a, assuming a is sorted.
    The return value i is such that all e in a[:i] have e <= x, and all e in
    a[i:] have e > x.  So if x already appears in the list, a.insert(x) will
    insert just after the rightmost x already there.
    Optional args lo (default 0) and hi (default len(a)) bound the
    slice of a to be searched.
    """

    if lo < 0:
        raise ValueError('lo must be non-negative')
    if hi is None:
        hi = len(a)
    while lo < hi:
        mid = (lo+hi)//2
        # Use __lt__ to match the logic in list.sort() and in heapq
        if x < a[mid]: hi = mid
        else: lo = mid+1
    return lo

这是所有numpy兼容的：

import numpy as np

array = np.array([1,2,3,4,5,6])
print(bisect_right(array, 7))
>>> 6
print(bisect_right(array, 0))
>>> 0

要查找与给定数字最接近的较高值的索引，请执行以下操作：

def closest_higher_value(array, value):
    if bisect_right(array, value) < len(array):
        return bisect_right(array, value)
    print("value too large:", value, "is bigger than all elements of:")
    print(array)


print(closest_higher_value(array, 3))
>>> 3
print(closest_higher_value(array, 7))
>>> value too large: 7 is bigger than all elements of:
>>> [1 2 3 4 5 6]
>>> None