使用 Numpy Python 获取最小值的索引 [英] getting the index of min values with Numpy Python

问题描述

下面的函数将每个值分成由索引 index 和 L_list 中的值分隔的块.所以它输出索引 3-5 之间的最小值，即 -5 和值的索引.numpy_argmin_reduceat(a, b) 和 Drawdown 函数都按计划执行，但是 numpy_argmin_reduceat(a, b) 的索引输出有问题它 Drawdown 的最小值与 numpy_argmin_reduceat(a, b) 的输出索引不匹配.我该如何解决这个问题?数组:

The function below separates each value into chunks separated by indexes index with the values in L_list. So it outputs the minimum value between indexes 3-5 which is -5 and the index of the value. Both the numpy_argmin_reduceat(a, b) and the Drawdown function do as planned however the index output of the numpy_argmin_reduceat(a, b) is faulty it The minimum values of Drawdown do not match with the indexes of the outputs of numpy_argmin_reduceat(a, b).How would I be able to solve this? Arrays:

import numpy as np
# indexes          0, 1, 2,3,4, 5, 6,7, 8, 9,10, 11, 12
L_list = np.array([10,20,30,0,0,-5,11,2,33, 4, 5, 68, 7])
index =  np.array([3,5,7,11])

功能:

#getting the minimum values
Drawdown = np.minimum.reduceat(L_list,index+1)

#Getting the min Index 
def numpy_argmin_reduceat(a, b):
    n = a.max() + 1  # limit-offset
    id_arr = np.zeros(a.size,dtype=int)
    id_arr[b] = 1
    shift = n*id_arr.cumsum()
    sortidx = (a+shift).argsort()
    grp_shifted_argmin = b
    idx =sortidx[grp_shifted_argmin] - b
    min_idx = idx +index
    return min_idx


min_idx =numpy_argmin_reduceat(L_list,index+1)
#printing function
DR_val_index = np.array([np.around(Drawdown,1), min_idx])
DR_result = np.apply_along_axis(lambda x: print(f'Min Values: {x[0]} at index: {x[1]}'), 0, DR_val_index)

输出

Min Values: -5 at index: 4
Min Values: 2 at index: 6
Min Values: 4 at index: 8
Min Values: 7 at index: 11

预期输出:

Min Values: -5 at index: 5
Min Values: 2 at index: 7
Min Values: 4 at index: 9
Min Values: 7 at index: 12

推荐答案

如果换行

id_arr[b[1:]] = 1

到

id_arr[b] = 1

我认为该函数会如您所愿.

I think the function will behave as you hope.

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