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])

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

使用示例:

array = np.random.random(10)
print(array)
# [ 0.21069679  0.61290182  0.63425412  0.84635244  0.91599191  0.00213826
#   0.17104965  0.56874386  0.57319379  0.28719469]

print(find_nearest(array, value=0.5))
# 0.568743859261

对于大型数组，@Demitri给出的(优秀)答案比目前标记为最佳的答案快得多。我从以下两个方面调整了他的精确算法:

不管输入数组是否排序，下面的函数都有效。 下面的函数返回与最接近的值对应的输入数组的索引，这有点更一般。

请注意，下面的函数还处理了一个特定的边缘情况，这将导致@Demitri编写的原始函数中的错误。否则，我的算法和他的一样。

def find_idx_nearest_val(array, value):
    idx_sorted = np.argsort(array)
    sorted_array = np.array(array[idx_sorted])
    idx = np.searchsorted(sorted_array, value, side="left")
    if idx >= len(array):
        idx_nearest = idx_sorted[len(array)-1]
    elif idx == 0:
        idx_nearest = idx_sorted[0]
    else:
        if abs(value - sorted_array[idx-1]) < abs(value - sorted_array[idx]):
            idx_nearest = idx_sorted[idx-1]
        else:
            idx_nearest = idx_sorted[idx]
    return idx_nearest

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()]

import numpy as np
def find_nearest(array, value):
    array = np.array(array)
    z=np.abs(array-value)
    y= np.where(z == z.min())
    m=np.array(y)
    x=m[0,0]
    y=m[1,0]
    near_value=array[x,y]

    return near_value

array =np.array([[60,200,30],[3,30,50],[20,1,-50],[20,-500,11]])
print(array)
value = 0
print(find_nearest(array, value))

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]