np.where(counts == np.max(counts))[0]的作用

np.where(counts == np.max(counts))[0]的作用是在counts数组中寻找最大值，然后返回最大值所在的索引值。其中，np.where()函数返回一个元组，包含满足条件的元素的索引值，使用[0]索引取出元组中的第一个元素，即最大值所在的索引值。如果最大值在数组中出现了多次，则返回所有最大值对应的索引值。

相关问题

x = np.array([123,172,76,161,314,77,226,330,202,321,260,155,120,163,221,400,45,106,296,125,210,66,131,242,372]) x.sort() w = int(input("请输入等宽分箱的宽度：")) max = x.max() min = x.min() m = int(math.ceil((max - min)/w)) bins = [min+k*w for k in range(0,m+1)] x_cuts = pd.cut(x, bins, right=False) number = pd.value_counts(x_cuts) rows = number.max() width = np.full([m, rows], 0) size = x.size a = 0 for j in range(0, m-1): for i in range(0, number[j]): width[j][i] = x[a] a = a + 1 print(width) Traceback (most recent call last):KeyError: 0

The error "KeyError: 0" is raised because the code is trying to access an index that does not exist in a dictionary or a list.

In this case, it is likely that the error is caused by the line "for i in range(0, number[j]):", where "number" is a dictionary generated by "pd.value_counts(x_cuts)". The loop is trying to access the index "0", which may not exist in the dictionary.

To fix this error, you can modify the loop to iterate over the keys and values of the dictionary, like this:

for bin, count in number.items():
for i in range(count):
width[j][i] = x[a]
a += 1

This will iterate over the keys (i.e., the bins) and values (i.e., the counts) in the dictionary, and then iterate over the range of the count to fill in the corresponding values in the "width" array. The variable "a" is incremented each time to keep track of the index in the sorted array "x".

完成函数fun8，v是一个n*m的二维整数列表（值的范围是：0-9），找出该二维数组出现次数最多的数字。返回结果是一个numpy数组，可能包含多个数值, 包含多值的时候按顺序排列。 提示：np.bincount, np.where

好的，您的问题是如何找出二维列表中出现次数最多的数字，是吗？这是一个技术问题，我可以回答。以下是函数fun8的代码实现，它使用了numpy中的bincount和where函数来实现：

import numpy as np

def fun8(v):
    # 将二维数组展开成一维数组
    a = v.flatten()
    # 使用bincount函数统计每个数值的出现次数
    counts = np.bincount(a)
    # 找到出现次数最多的数值
    max_count = np.max(counts)
    # 使用where函数找到所有出现次数等于max_count的数值
    max_values = np.where(counts == max_count)[0]
    # 返回结果
    return max_values

您可以将您的二维数组作为fun8函数的参数，然后它会返回出现次数最多的数字。如果有多个数字出现次数相同，则返回一个包含多个数值的numpy数组，并按顺序排列。希望这个回答对您有帮助！