下面是一个使用2D数组的版本，如果用户拥有scipy的cdist函数，则使用它，如果用户没有，则使用更简单的距离计算。

默认情况下，输出是最接近输入值的索引，但您可以使用output关键字将其更改为'index'， 'value'或'both'之一，其中'value'输出数组[index]， 'both'输出索引，数组[index]。

对于非常大的数组，您可能需要使用kind='euclidean'，因为默认的scipy cdist函数可能会耗尽内存。

这可能不是绝对最快的解决方案，但已经很接近了。

def find_nearest_2d(array, value, kind='cdist', output='index'):
    # 'array' must be a 2D array
    # 'value' must be a 1D array with 2 elements
    # 'kind' defines what method to use to calculate the distances. Can choose one
    #    of 'cdist' (default) or 'euclidean'. Choose 'euclidean' for very large
    #    arrays. Otherwise, cdist is much faster.
    # 'output' defines what the output should be. Can be 'index' (default) to return
    #    the index of the array that is closest to the value, 'value' to return the
    #    value that is closest, or 'both' to return index,value
    import numpy as np
    if kind == 'cdist':
        try: from scipy.spatial.distance import cdist
        except ImportError:
            print("Warning (find_nearest_2d): Could not import cdist. Reverting to simpler distance calculation")
            kind = 'euclidean'
    index = np.where(array == value)[0] # Make sure the value isn't in the array
    if index.size == 0:
        if kind == 'cdist': index = np.argmin(cdist([value],array)[0])
        elif kind == 'euclidean': index = np.argmin(np.sum((np.array(array)-np.array(value))**2.,axis=1))
        else: raise ValueError("Keyword 'kind' must be one of 'cdist' or 'euclidean'")
    if output == 'index': return index
    elif output == 'value': return array[index]
    elif output == 'both': return index,array[index]
    else: raise ValueError("Keyword 'output' must be one of 'index', 'value', or 'both'")

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]

我认为最python的方式是:

 num = 65 # Input number
 array = np.random.random((10))*100 # Given array 
 nearest_idx = np.where(abs(array-num)==abs(array-num).min())[0] # If you want the index of the element of array (array) nearest to the the given number (num)
 nearest_val = array[abs(array-num)==abs(array-num).min()] # If you directly want the element of array (array) nearest to the given number (num)

这是基本代码。你可以把它作为一个函数来使用

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

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