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()]

对于那些搜索多个最接近的，修改接受的答案:

import numpy as np
def find_nearest(array, value, k):
    array = np.asarray(array)
    idx = np.argsort(abs(array - value))[:k]
    return array[idx]

看到的: https://stackoverflow.com/a/66937734/11671779

这是在向量数组中找到最近向量的扩展。

import numpy as np

def find_nearest_vector(array, value):
  idx = np.array([np.linalg.norm(x+y) for (x,y) in array-value]).argmin()
  return array[idx]

A = np.random.random((10,2))*100
""" A = array([[ 34.19762933,  43.14534123],
   [ 48.79558706,  47.79243283],
   [ 38.42774411,  84.87155478],
   [ 63.64371943,  50.7722317 ],
   [ 73.56362857,  27.87895698],
   [ 96.67790593,  77.76150486],
   [ 68.86202147,  21.38735169],
   [  5.21796467,  59.17051276],
   [ 82.92389467,  99.90387851],
   [  6.76626539,  30.50661753]])"""
pt = [6, 30]  
print find_nearest_vector(A,pt)
# array([  6.76626539,  30.50661753])

所有的答案都有助于收集信息来编写高效的代码。但是，我已经编写了一个小的Python脚本来针对各种情况进行优化。如果提供的数组已排序，则将是最佳情况。如果搜索一个指定值的最近点的索引，那么对半模块是最省时的。当一个索引对应一个数组时，numpy searchsorted是最有效的。

import numpy as np
import bisect
xarr = np.random.rand(int(1e7))

srt_ind = xarr.argsort()
xar = xarr.copy()[srt_ind]
xlist = xar.tolist()
bisect.bisect_left(xlist, 0.3)

In[63]: %时间平分。bisect_left (xlist, 0.3) CPU次数:user 0ns, sys: 0ns, total: 0ns 壁时间:22.2µs

np.searchsorted(xar, 0.3, side="left")

In [64]: %time np。Searchsorted (xar, 0.3, side="left") CPU次数:user 0ns, sys: 0ns, total: 0ns 壁时间:98.9µs

randpts = np.random.rand(1000)
np.searchsorted(xar, randpts, side="left")

%的时间np。Searchsorted (xar, randpts, side="left") CPU次数:用户4ms, sys: 0ns, total: 4ms 壁时间:1.2 ms

如果我们遵循乘法规则，那么numpy应该花费~100 ms，这意味着快了~83倍。

下面是@Ari Onasafari的scipy版本，回答“在向量数组中找到最近的向量”

In [1]: from scipy import spatial

In [2]: import numpy as np

In [3]: A = np.random.random((10,2))*100

In [4]: A
Out[4]:
array([[ 68.83402637,  38.07632221],
       [ 76.84704074,  24.9395109 ],
       [ 16.26715795,  98.52763827],
       [ 70.99411985,  67.31740151],
       [ 71.72452181,  24.13516764],
       [ 17.22707611,  20.65425362],
       [ 43.85122458,  21.50624882],
       [ 76.71987125,  44.95031274],
       [ 63.77341073,  78.87417774],
       [  8.45828909,  30.18426696]])

In [5]: pt = [6, 30]  # <-- the point to find

In [6]: A[spatial.KDTree(A).query(pt)[1]] # <-- the nearest point 
Out[6]: array([  8.45828909,  30.18426696])

#how it works!
In [7]: distance,index = spatial.KDTree(A).query(pt)

In [8]: distance # <-- The distances to the nearest neighbors
Out[8]: 2.4651855048258393

In [9]: index # <-- The locations of the neighbors
Out[9]: 9

#then 
In [10]: A[index]
Out[10]: array([  8.45828909,  30.18426696])

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]