R 条件 rowSums 替换为基于百分比的总和

如何解决R 条件 rowSums 替换为基于百分比的总和

如果这些行代表数据的

示例数据：

姓名	Year1	Year2	Year3	总计	百分比
约翰	1	2	1	4	0.7029877
保罗	230	100	150	480	84.358524
乔治	41	30	10	81	14.235501
林戈	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，因为两者的百分比都

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

示例解决方案：

姓名	Year1	Year2	Year3	总计	百分比
保罗	230	100	150	480	84.358524
乔治	41	30	10	81	14.235501
其他(n=2)	3	3	2	8	1.405975

解决方法

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