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]

其他回答

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

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

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])

这个函数使用numpy searchsorted处理任意数量的查询,因此在对输入数组进行排序之后,它的速度也一样快。 它可以在2d, 3d的规则网格上工作…:

#!/usr/bin/env python3
# keywords: nearest-neighbor regular-grid python numpy searchsorted Voronoi

import numpy as np

#...............................................................................
class Near_rgrid( object ):
    """ nearest neighbors on a Manhattan aka regular grid
    1d:
    near = Near_rgrid( x: sorted 1d array )
    nearix = near.query( q: 1d ) -> indices of the points x_i nearest each q_i
        x[nearix[0]] is the nearest to q[0]
        x[nearix[1]] is the nearest to q[1] ...
        nearpoints = x[nearix] is near q
    If A is an array of e.g. colors at x[0] x[1] ...,
    A[nearix] are the values near q[0] q[1] ...
    Query points < x[0] snap to x[0], similarly > x[-1].

    2d: on a Manhattan aka regular grid,
        streets running east-west at y_i, avenues north-south at x_j,
    near = Near_rgrid( y, x: sorted 1d arrays, e.g. latitide longitude )
    I, J = near.query( q: nq × 2 array, columns qy qx )
    -> nq × 2 indices of the gridpoints y_i x_j nearest each query point
        gridpoints = np.column_stack(( y[I], x[J] ))  # e.g. street corners
        diff = gridpoints - querypoints
        distances = norm( diff, axis=1, ord= )
    Values at an array A definded at the gridpoints y_i x_j nearest q: A[I,J]

    3d: Near_rgrid( z, y, x: 1d axis arrays ) .query( q: nq × 3 array )

    See Howitworks below, and the plot Voronoi-random-regular-grid.
    """

    def __init__( self, *axes: "1d arrays" ):
        axarrays = []
        for ax in axes:
            axarray = np.asarray( ax ).squeeze()
            assert axarray.ndim == 1, "each axis should be 1d, not %s " % (
                    str( axarray.shape ))
            axarrays += [axarray]
        self.midpoints = [_midpoints( ax ) for ax in axarrays]
        self.axes = axarrays
        self.ndim = len(axes)

    def query( self, queries: "nq × dim points" ) -> "nq × dim indices":
        """ -> the indices of the nearest points in the grid """
        queries = np.asarray( queries ).squeeze()  # or list x y z ?
        if self.ndim == 1:
            assert queries.ndim <= 1, queries.shape
            return np.searchsorted( self.midpoints[0], queries )  # scalar, 0d ?
        queries = np.atleast_2d( queries )
        assert queries.shape[1] == self.ndim, [
                queries.shape, self.ndim]
        return [np.searchsorted( mid, q )  # parallel: k axes, k processors
                for mid, q in zip( self.midpoints, queries.T )]

    def snaptogrid( self, queries: "nq × dim points" ):
        """ -> the nearest points in the grid, 2d [[y_j x_i] ...] """
        ix = self.query( queries )
        if self.ndim == 1:
            return self.axes[0][ix]
        else:
            axix = [ax[j] for ax, j in zip( self.axes, ix )]
            return np.array( axix )


def _midpoints( points: "array-like 1d, *must be sorted*" ) -> "1d":
    points = np.asarray( points ).squeeze()
    assert points.ndim == 1, points.shape
    diffs = np.diff( points )
    assert np.nanmin( diffs ) > 0, "the input array must be sorted, not %s " % (
            points.round( 2 ))
    return (points[:-1] + points[1:]) / 2  # floats

#...............................................................................
Howitworks = \
"""
How Near_rgrid works in 1d:
Consider the midpoints halfway between fenceposts | | |
The interval [left midpoint .. | .. right midpoint] is what's nearest each post --

    |   |       |                     |   points
    | . |   .   |          .          |   midpoints
      ^^^^^^               .            nearest points[1]
            ^^^^^^^^^^^^^^^             nearest points[2]  etc.

2d:
    I, J = Near_rgrid( y, x ).query( q )
    I = nearest in `x`
    J = nearest in `y` independently / in parallel.
    The points nearest [yi xj] in a regular grid (its Voronoi cell)
    form a rectangle [left mid x .. right mid x] × [left mid y .. right mid y]
    (in any norm ?)
    See the plot Voronoi-random-regular-grid.

Notes
-----
If a query point is exactly halfway between two data points,
e.g. on a grid of ints, the lines (x + 1/2) U (y + 1/2),
which "nearest" you get is implementation-dependent, unpredictable.

"""

Murky = \
""" NaNs in points, in queries ?
"""

__version__ = "2021-10-25 oct  denis-bz-py"
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