如何简化 numpy 条件语句以查找给定值的元素的相邻索引？

Question

如何简化 numpy 数组中的条件语句以在给定搜索列表 bin 的情况下查找所有列表元素 V 的相邻索引。例如，如果 bins 是 100 个元素的列表，范围从 1.1 到 100.1，则 V 是 [23,77.4,42] 的列表。任务是针对 V 中的每个元素，在 bin 中找到其相邻索引。目前下面的代码需要比较两个向量 4 次，然后将结果添加到输出列表中。如何简化这个任务？

def testFindBoundaryVector():

    bins = np.linspace(start=1.1, stop=100.1, num=100)
    V = np.array([23, 77.4, 42])
    V_T = np.reshape(V, (V.shape[0], 1))
    # broadcasting 
    Diff = np.abs(bins - V_T)
    Index = np.argmin(Diff, axis=1)

    print(f"indices: {Index}")
    print(f"value: {bins[Index]}")
    print(f"shape of Diff: {Diff.shape}")

    left, right = [], []

    LeftIndex = Index[V >= bins[Index]]
    RightIndex = LeftIndex + 1

    left.extend(LeftIndex)
    right.extend(RightIndex)

    LeftIndex = Index[V < bins[Index]] - 1
    RightIndex = LeftIndex + 1

    left.extend(LeftIndex)
    right.extend(RightIndex)

    print(f"left Index: {left}")
    print(f"right Index: {right}")
    return

预期结果：左 = [21,76,40] 右 = [22,77,41]

Answer 1

也许是这样？

import numpy as np
L = np.linspace(1.1, 100.1, 100)
V = np.array([23,77.4,42])
Right = np.argmin(L[:,None]<V[None,:],axis=0)
Left = Right - 1

如何简化 numpy 条件语句以查找给定值的元素的相邻索引？

How to simplify numpy conditional statements to find neighbouring indices of elements given a value?

python

numpy

conditional-statements

array-broadcasting