# 用 Python 预测今后的北京高考人数

``````# coding: utf-8

# 作者：Wizard <github.com/wizardforcel>
# 预测今后的北京高考人数
# 假设 x 年的出生人数和 (x + 18) 年的高考人数是线性关系

import numpy as np
from matplotlib import pyplot as plt

def unary_linear_fit(x, y):

assert(x.ndim == 1 and y.ndim == 1 and len(x) == len(y))

cov_x_y = np.mean((x - x.mean()) * (y - y.mean()))

k = cov_x_y / x.var()

b = y.mean() - k * x.mean()

return k, b

def r_square(y, y_hat):

assert(y_hat.ndim == 1 and y.ndim == 1 and len(y_hat) == len(y))

return 1 - np.sum((y - y_hat) ** 2) / np.sum((y - y.mean()) ** 2)

# 1988 ~ 2015 年的出生人数（万人）数据
x = np.asarray([15.3, 19, 14, 9.2, 8.3, 7.9, 8.5, 8.5, 7.8, 7.6, 6.7, 6.3, 7.2, \
6, 6, 4.5, 6.6, 7.7, 7.7, 9.9, 10.6, 10.9, 10.2, 12.5, 14.5, 13.6, 17.2, 12.3])

# 考试人数从 2006 年开始统计
y_train = np.asarray([12.6, 12.5, 11.8, 10.1, 8.1, 7.6, 7.3, 7.27, 7.05, 6.8, 6.12, 6])

x_train = x[:len(y_train)]
x_pred = x[len(y_train):]

k, b = unary_linear_fit(x_train, y_train)

y_train_hat = k * x_train + b

rsq = r_square(y_train, y_train_hat)
print rsq
# 0.870773027191
# 拟合度相当高

y_pred = k * x_pred + b

years = np.arange(1988, 2015 + 1) + 18
yr_train = years[:len(y_train)]
yr_pred = years[len(y_train):]

plt.plot(yr_train, y_train, 'b', label='history value')
plt.plot(yr_pred, y_pred, 'r', label='prediction value')

plt.title('Beijing people for CEE')
plt.xlabel('year')
plt.ylabel('population / 10e4')
plt.legend(loc='best')
plt.show()``````

JSmiles

0 文章
0 评论
84935 人气