使用if語句在R中取平均值

Question

對於示例數據框：

df1 <- structure(list(name = c("a", "b", "c", "d", "e", "f", "g", "h", 
"i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", 
"v", "w", "x", "y", "z", "a", "b", "c", "d", "e", "f", "g", "h", 
"i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", 
"v", "w", "x", "y", "z", "a", "b", "c", "d", "e", "f", "g", "h", 
"i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", 
"v", "w", "x", "y", "z"), amount = c(5.5, 5.4, 5.2, 5.3, 5.1, 
5.1, 5, 5, 4.9, 4.5, 6, 5.9, 5.7, 5.4, 5.3, 5.1, 5.6, 5.4, 5.3, 
5.6, 4.6, 4.2, 4.5, 4.2, 4, 3.8, 6, 5.8, 5.7, 5.6, 5.3, 5.6, 
5.4, 5.5, 5.4, 5.1, 9, 8.8, 8.6, 8.4, 8.2, 8, 7.8, 7.6, 7.4, 
7.2, 6, 5.75, 5.5, 5.25, 5, 4.75, 10, 8.9, 7.8, 6.7, 5.6, 4.5, 
3.4, 2.3, 1.2, 0.1, 6, 5.8, 5.7, 5.6, 5.5, 5.5, 5.4, 5.6, 5.8, 
5.1, 6, 5.5, 5.4, 5.3, 5.2, 5.1), decile = c(1L, 2L, 3L, 4L, 
5L, 6L, 7L, 8L, 9L, 10L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 
10L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 
9L, 10L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 1L, 2L, 3L, 
4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 1L, 2L, 
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 1L, 2L, 3L, 4L, 5L, 6L), time = c(2016L, 
2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 
2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 2016L, 
2016L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 
2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 
2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 
2017L, 2017L, 2017L, 2017L, 2017L, 2017L, 2018L, 2018L, 2018L, 
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
2018L, 2018L, 2018L, 2018L, 2018L)), .Names = c("name", "amount", 
"decile", "time"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-78L), spec = structure(list(cols = structure(list(name = structure(list(), class = c("collector_character", 
"collector")), amount = structure(list(), class = c("collector_double", 
"collector")), decile = structure(list(), class = c("collector_integer", 
"collector")), time = structure(list(), class = c("collector_integer", 
"collector"))), .Names = c("name", "amount", "decile", "time"
)), default = structure(list(), class = c("collector_guess", 
"collector"))), .Names = c("cols", "default"), class = "col_spec"))

我最終希望生成一個ggplot圖表，其中按五分位數詳細說明每年的平均“金額”（即，每年數據的5個小條形圖）。

為此，我需要能夠計算五分位數（將十分位1和2、3和4、5和6、7和8以及9和10的所有值取平均值，並且還包括95％CI。

我過去曾嘗試過濾我的數據，但是我在努力使用if語句將其概念化。

任何幫助，將不勝感激。

Answer 1

您可以使用dplyr函數通過管道執行此操作，方法是將除以2並四舍五入將十分位數轉換為五分位數。 在這里，我只是做了一個非常快而骯臟的置信區間2 x標准偏差，但是您可能需要其他方法。

library(dplyr)
library(ggplot2)

plot_data <- df1 %>% 
  mutate(quintile = ceiling(decile/2)) %>% 
  group_by(time, quintile) %>% 
  summarize(average_amount = mean(amount),
            sd_amount = sd(amount),
            ci_min = average_amount - 2 * sd_amount,
            ci_max = average_amount + 2 * sd_amount)

這是一個（丑陋的）ggplot，帶有按年和五分位數划分的條形圖。

ggplot(plot_data, aes(x = quintile, y = average_amount)) + 
  geom_col() + 
  geom_errorbar(aes(ymin = ci_min, ymax = ci_max)) +
  facet_wrap(~ time)

Answer 2

如果您只是在尋找平均值，請嘗試以下操作：

library(tidyverse)

df1 %>% 
  mutate(quintile = floor((decile - 1) / 2) + 1) %>% 
  group_by(time, quintile) %>% 
  summarise(AvgAmount = mean(amount)) %>% 
  ggplot(aes(quintile, AvgAmount)) + 
  geom_bar(stat = "identity") + 
  facet_grid(time ~ .)

如果您想更好地了解五分位數內的分布，可以使用箱形圖：

df1 %>% 
  mutate(quintile = floor((decile - 1) / 2) + 1) %>% 
  ggplot(aes(quintile, amount, group = quintile)) + 
  geom_boxplot() + 
  facet_grid(time ~ .)