下面是一个处理非标量“values”数组的版本:

import numpy as np

def find_nearest(array, values):
    indices = np.abs(np.subtract.outer(array, values)).argmin(0)
    return array[indices]

如果输入是标量，则返回数字类型(例如int, float)的版本:

def find_nearest(array, values):
    values = np.atleast_1d(values)
    indices = np.abs(np.subtract.outer(array, values)).argmin(0)
    out = array[indices]
    return out if len(out) > 1 else out[0]

下面是一个处理非标量“values”数组的版本:

import numpy as np

def find_nearest(array, values):
    indices = np.abs(np.subtract.outer(array, values)).argmin(0)
    return array[indices]

如果输入是标量，则返回数字类型(例如int, float)的版本:

def find_nearest(array, values):
    values = np.atleast_1d(values)
    indices = np.abs(np.subtract.outer(array, values)).argmin(0)
    out = array[indices]
    return out if len(out) > 1 else out[0]

如果你有很多值需要搜索(值可以是多维数组)，下面是@Dimitri的快速向量化解决方案:

# `values` should be sorted
def get_closest(array, values):
    # make sure array is a numpy array
    array = np.array(array)

    # get insert positions
    idxs = np.searchsorted(array, values, side="left")
    
    # find indexes where previous index is closer
    prev_idx_is_less = ((idxs == len(array))|(np.fabs(values - array[np.maximum(idxs-1, 0)]) < np.fabs(values - array[np.minimum(idxs, len(array)-1)])))
    idxs[prev_idx_is_less] -= 1
    
    return array[idxs]

基准

>使用@Demitri的解决方案比使用for循环快100倍”

>>> %timeit ar=get_closest(np.linspace(1, 1000, 100), np.random.randint(0, 1050, (1000, 1000)))
139 ms ± 4.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

>>> %timeit ar=[find_nearest(np.linspace(1, 1000, 100), value) for value in np.random.randint(0, 1050, 1000*1000)]
took 21.4 seconds

稍微修改一下，上面的答案适用于任意维度的数组(1d, 2d, 3d，…):

def find_nearest(a, a0):
    "Element in nd array `a` closest to the scalar value `a0`"
    idx = np.abs(a - a0).argmin()
    return a.flat[idx]

或者，写成一行:

a.flat[np.abs(a - a0).argmin()]

如果你不想使用numpy，可以这样做:

def find_nearest(array, value):
    n = [abs(i-value) for i in array]
    idx = n.index(min(n))
    return array[idx]

对于2d数组，要确定最近元素的i, j位置:

import numpy as np
def find_nearest(a, a0):
    idx = (np.abs(a - a0)).argmin()
    w = a.shape[1]
    i = idx // w
    j = idx - i * w
    return a[i,j], i, j