奥运会分析

奥运数据可视化

应用篇

这是Nature期刊上的一篇文章Nature. 2004 September 30; 431(7008)，

虽然觉得这个结论不太严谨，但我却无力反驳。

于是在文章补充材料里，我找到了文章使用的数据，现在的任务是，重复这张图和文章的分析过程。

导入数据

library(tidyverse)
library(readxl)

d <- read_excel("./demo_data/olympics.xlsx")
d

可视化

我们先画图看看

d %>%
  ggplot() +
  geom_point(aes(x = Olympic_year, y = Men_score), color = "blue") +
  geom_point(aes(x = Olympic_year, y = Women_score), color = "red")

这样写也是可以的，只不过最好先tidy数据

d1 <- d %>%
  pivot_longer(
    cols = -Olympic_year,
    names_to = "sex",
    values_to = "winning_time"
  )

d1

然后在画图

d1 %>%
  ggplot(aes(x = Olympic_year, y = winning_time, color = sex)) +
  geom_point() +
  # geom_smooth(method = "lm") +
  scale_color_manual(
    values = c("Men_score" = "blue", "Women_score" = "red")
  ) +
  scale_x_continuous(
    breaks = seq(1900, 2004, by = 4),
    labels = seq(1900, 2004, by = 4)
  ) +
  theme(axis.text.x = element_text(
    size = 10, angle = 45, colour = "black",
    vjust = 1, hjust = 1
  ))

回归分析

建立年份与成绩的线性关系 $$ \text{score}_i = \alpha + \beta \times \text{year}_i + \epsilon_i; \qquad \epsilon_i\in \text{Normal}(\mu, \sigma) $$

我们需要求出其中系数$\alpha$和$\beta$，写R语言代码如下 (lm(y ~ 1 + x,data = d), 要求得 $\alpha$和$\beta$，就是对应 1 和 x 前的系数)

fit_1 <- lm(Men_score ~ 1 + Olympic_year, data = d)

summary(fit_1)

fit_2 <- lm(Women_score ~ 1 + Olympic_year, data = d)

summary(fit_2)

预测

使用predict()完成预测

df <- data.frame(Olympic_year = 2020)

predict(fit_1, newdata = df)

为了图片中的一致，我们使用1900年到2252年(seq(1900, 2252, by = 4))建立预测项，并整理到数据框里

grid <- tibble(
  Olympic_year = as.numeric(seq(1900, 2252, by = 4))
)
grid

tb <- grid %>%
  mutate(
    Predict_Men = predict(fit_1, newdata = grid),
    Predict_Women = predict(fit_2, newdata = grid)
  )
tb

有时候我喜欢用modelr::add_predictions()函数实现相同的功能

library(modelr)
grid %>%
  add_predictions(fit_1, var = "Predict_Men") %>%
  add_predictions(fit_2, var = "Predict_Women")

再次可视化

tb1 <- tb %>%
  pivot_longer(
    cols = -Olympic_year,
    names_to = "sex",
    values_to = "winning_time"
  )
tb1

tb1 %>%
  ggplot(aes(
    x = Olympic_year,
    y = winning_time,
    color = sex
  )) +
  geom_line(size = 2) +
  geom_point(data = d1) +
  scale_color_manual(
    values = c(
      "Men_score" = "blue",
      "Women_score" = "red",
      "Predict_Men" = "#588B8B",
      "Predict_Women" = "#C8553D"
    ),
    labels = c(
      "Men_score" = "Men score",
      "Women_score" = "Women score",
      "Predict_Men" = "Men Predict score",
      "Predict_Women" = "Women Predict score"
    )
  ) +
  scale_x_continuous(
    breaks = seq(1900, 2252, by = 16),
    labels = as.character(seq(1900, 2252, by = 16))
  ) +
  theme(axis.text.x = element_text(
    size = 10, angle = 45, colour = "black",
    vjust = 1, hjust = 1
  ))

早知道nature文章这么简单，10年前我也可以写啊！

list_column

这里是另外的一种方法

library(modelr)

d1 <- d %>%
  pivot_longer(
    cols = -Olympic_year,
    names_to = "sex",
    values_to = "winning_time"
  )

fit_model <- function(df) lm(winning_time ~ Olympic_year, data = df)

d2 <- d1 %>%
  group_nest(sex) %>%
  mutate(
    mod = map(data, fit_model)
  )
d2



# d2 %>% mutate(p = list(grid, grid))
# d3 <- d2 %>% mutate(p = list(grid, grid))
# d3
# d3 %>%
#   mutate(
#     predictions = map2(p, mod, add_predictions),
#   )

# or
tb4 <- d2 %>%
  mutate(
    predictions = map(mod, ~ add_predictions(grid, .))
  ) %>%
  select(sex, predictions) %>%
  unnest(predictions)

tb4 %>%
  ggplot(aes(
    x = Olympic_year,
    y = pred,
    group = sex,
    color = sex
  )) +
  geom_point() +
  geom_line(size = 2) +
  geom_point(
    data = d1,
    aes(
      x = Olympic_year,
      y = winning_time,
      group = sex,
      color = sex
    )
  ) +
  scale_x_continuous(
    breaks = seq(1900, 2252, by = 16),
    labels = as.character(seq(1900, 2252, by = 16))
  ) +
  theme(axis.text.x = element_text(
    size = 10, angle = 45, colour = "black",
    vjust = 1, hjust = 1
  ))

课后作业

探索数据，建立身高体重的线性模型

# remove the objects
# rm(list=ls())
rm(d, d1, d2, df, fit_1, fit_2, fit_model, grid, tb, tb1, tb4)

pacman::p_unload(pacman::p_loaded(), character.only = TRUE)