python中的数学公式不起作用

Question

我一直在学习python中的机器学习，目前正在学习基础知识。我现在正在研究线性回归并尝试将一些数学公式实现为python代码。我设法成功地写了一些公式但是有一个我很难用：（X - Xmeans）*（Y - Ymeans）它总是给我一个错误“列表索引必须是整数或切片，而不是numpy.float64 “当我尝试打印时。

我试图在网上寻找相同的案例及其解决方案，但没有一个能够奏效。

import numpy
import matplotlib.pyplot as plt

X_positions = numpy.array([2,3,4,5,6])
y_positions = numpy.array([4,5,6,5,7])
plt.plot([X_positions], [y_positions], 'ro')
plt.axis([0,10,0,10])

X_means = sum(X_positions) / len(X_positions)
y_means = sum(y_positions) / len(y_positions)

plt.plot([X_means], [y_means], 'go')
plt.axis([0,10,0,10])

X_minus_X_means = []
y_minus_y_means = []
X_minus_X_means_squared = []

for i in X_positions:
    X_minus_X_means.append(i - X_means)

for i in y_positions:
    y_minus_y_means.append(i - y_means)

for i in X_minus_X_means:
    X_minus_X_means_squared.append(i ** 2)

X_minus_X_means_times_y_minus_y_means = []

#HERE IS THE PROBLEM
for i in X_minus_X_means and y_minus_y_means:
   X_minus_X_means_times_y_minus_y_means.append(X_minus_X_means[i] * y_minus_y_means[i])

Answer 1

你应该尽可能使用内置的numpy操作和矢量化操作。

尝试这样的事情：

import numpy as np

import matplotlib.pyplot as plt

X_positions = np.array([2,3,4,5,6])
y_positions = np.array([4,5,6,5,7])
plt.plot([X_positions], [y_positions], 'ro')
plt.axis([0,10,0,10])

X_means = X_positions.mean()
y_means = y_positions.mean()

plt.plot([X_means], [y_means], 'go')
plt.axis([0,10,0,10])

X_minus_X_means = X_positions-X_means
y_minus_y_means = y_positions-y_means
X_minus_X_means_squared = X_minus_X_means**2
X_minus_X_means_times_y_minus_y_means = X_minus_X_means*y_minus_y_means

Answer 2

代替

for i in X_minus_X_means and y_minus_y_means:

试着写

for i in range(len(X_minus_X_means)):

否则i不是整数，不能用作索引

Answer 3

也许该分裂正在产生一个float，导致该指数为非整数？

X_means = sum(X_positions) / len(X_positions)

对于python3中的整数除法，可以使用双斜杠运算符//：

X_means = sum(X_positions) // len(X_positions)

你也可以使用ceil，floor，round或int(val)。

Answer 4

X_minus_X_means和y_minus_y_means都是名单

y_minus_y_means包含价值

[-1.4000000000000004,
 -0.40000000000000036,
 0.5999999999999996,
 -0.40000000000000036,
 1.5999999999999996]

所以基本上在for i in X_minus_X_means and y_minus_y_means:的i值在type numpy.float64和内部循环中计算，你使用X_minus_X_means访问y_minus_y_means和index的值，这是i一个float值

Answer 5

虽然我完全支持并更喜欢向量化操作，但我认为您应该了解名为zip的函数，这对于并行迭代两个（或更多）列表的情况很有用。所以在你的情况下，有问题的部分应该改为以下，避免任何索引i

for x, y in zip(X_minus_X_means, y_minus_y_means):
    X_minus_X_means_times_y_minus_y_means.append(x * y)