R条件rowSums替换为基于百分比的总和

Question

如果这些行代表

示例数据：

name	Year1	Year2	Year3	Total	Percent
John	1	2	1	4	0.7029877
Paul	230	100	150	480	84.358524
George	41	30	10	81	14.235501
Ringo	2	1	1	4	0.7029877

# Code for example data
name <- c("John", "Paul", "George", "Ringo")
Year1 <- c(1, 230, 41, 2)
Year2 <- c(2, 100, 30, 1)
Year3 <- c(1, 150, 10, 1)
df <- data.frame(name, Year1, Year2, Year3)
df$Total <- rowSums(select(df,Year1:Year3))
df$Percent <- df$Total/sum(df$Total)*100

在解决方案中，John 和 Ringo 将合并为一个“其他”解决方案，因为两者的百分比

# Code for example solution
name <- c("Paul", "George", "Other(n=2)")
Year1 <- c(230, 41, 3)
Year2 <- c(100, 30, 3)
Year3 <- c(150, 10, 2)
df2 <- data.frame(name, Year1, Year2, Year3)
df2$Total <- rowSums(select(df2,Year1:Year3))
df2$Percent <- df2$Total/sum(df2$Total)*100

示例解决方案：

name	Year1	Year2	Year3	Total	Percent
Paul	230	100	150	480	84.358524
George	41	30	10	81	14.235501
Other(n=2)	3	3	2	8	1.405975

Answer 1

library(tidyverse) # or use forcats::fct_lump(...
df %>% 
  mutate(name_lumped = fct_lump(name, w = Percent, prop = 0.01)) %>%
  group_by(name_lumped) %>%
  summarize(across(Year1:Percent, sum))

# A tibble: 3 x 6
  name_lumped Year1 Year2 Year3 Total Percent
  <fct>       <dbl> <dbl> <dbl> <dbl>   <dbl>
1 George         41    30    10    81   14.2 
2 Paul          230   100   150   480   84.4 
3 Other           3     3     2     8    1.41