对于示例数据框：
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"))
我希望计算每年十分位数1、5和10 BY的平均结果（2016、17等）。然后，我希望在第一列中创建一个最终年份表，详细说明年份，然后是十分位数1和10的平均结果之间的差距（即十分位数10结果减去十分位数1的结果），然后是十分位数5和10的平均结果之间的梯度。 10（即10个平均结果减去5个平均结果） 这是十分位5和十分位10在均值上的差异。
为了说明这一点，我为2016年的数据创建了一个工作示例。我列出了2016年的十分位1、5和10的值。然后使用这些值计算出间隙和梯度差。
summary2016 <- structure(list(`2016` = c(NA_character_, NA_character_, NA_character_,
NA_character_), `1` = c("5", "10", "Gap", "Gradient"), `5.5` = c(5.1,
4.5, 1.4, 0.3), `6` = c(5.3, 5.6, NA, NA), `11.5` = c(10.4, 10.1,
NA, NA)), .Names = c("2016", "1", "5.5", "6", "11.5"), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -4L), spec = structure(list(
cols = structure(list(`2016` = structure(list(), class = c("collector_character",
"collector")), `1` = structure(list(), class = c("collector_character",
"collector")), `5.5` = structure(list(), class = c("collector_double",
"collector")), `6` = structure(list(), class = c("collector_double",
"collector")), `11.5` = structure(list(), class = c("collector_double",
"collector"))), .Names = c("2016", "1", "5.5", "6", "11.5"
)), default = structure(list(), class = c("collector_guess",
"collector"))), .Names = c("cols", "default"), class = "col_spec"))
这可以一步完成吗，还是我需要分解一下？
答案 0 :(得分：1)
library(tidyverse)
df1 %>% filter(decile %in% c(1, 5, 10)) %>%
group_by(time, decile) %>% summarise(mean = mean(amount)) %>%
mutate(gap1 = mean - mean[1], gap5 = mean - mean[2])
# A tibble: 9 x 5
# Groups: time [3]
# time decile mean gap1 gap5
# <int> <int> <dbl> <dbl> <dbl>
# 1 2016 1 5.75 0 0.55
# 2 2016 5 5.20 -0.55 0
# 3 2016 10 5.05 -0.7 -0.150
# 4 2017 1 6.4 0 0.775
# 5 2017 5 5.62 -0.775 0
# 6 2017 10 6.15 -0.25 0.525
# 7 2018 1 7.33 0 1.90
# 8 2018 5 5.43 -1.90 0
# 9 2018 10 2.60 -4.73 -2.83
数字与您的数字不同，所以也许您正在寻找其他类型的差距。您的示例summary2016的结构也有些不同寻常，而上面的解决方案所产生的效果超出了您的要求，但格式更好。
特别是，gap1是平均值（十分位数i）-均值（十分位数1），其中i = 1、5、10，而gap5是均值（十分位数i）-均值（十分位数5）。