如何在numpy数组中找到最近的值?例子:

np.find_nearest(array, value)

当前回答

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'")

其他回答

如果你的数组已经排序并且非常大,这是一个更快的解决方案:

def find_nearest(array,value):
    idx = np.searchsorted(array, value, side="left")
    if idx > 0 and (idx == len(array) or math.fabs(value - array[idx-1]) < math.fabs(value - array[idx])):
        return array[idx-1]
    else:
        return array[idx]

这可以扩展到非常大的阵列。如果不能假定数组已经排序,可以很容易地修改上面的内容以在方法中排序。对于小型数组来说,这是多余的,但一旦它们变大,这就快得多了。

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

这是unutbu答案的矢量化版本:

def find_nearest(array, values):
    array = np.asarray(array)

    # the last dim must be 1 to broadcast in (array - values) below.
    values = np.expand_dims(values, axis=-1) 

    indices = np.abs(array - values).argmin(axis=-1)

    return array[indices]


image = plt.imread('example_3_band_image.jpg')

print(image.shape) # should be (nrows, ncols, 3)

quantiles = np.linspace(0, 255, num=2 ** 2, dtype=np.uint8)

quantiled_image = find_nearest(quantiles, image)

print(quantiled_image.shape) # should be (nrows, ncols, 3)

也许对ndarray有帮助:

def find_nearest(X, value):
    return X[np.unravel_index(np.argmin(np.abs(X - value)), X.shape)]