示例数据如下:
x <- read.table(header=T, text="
ID CostType1 Cost1 CostType2 Cost2
1 a 10 c 1
2 b 2 c 20
3 a 1 b 50
4 a 40 c 1
5 c 2 b 30
6 a 60 c 3
7 c 10 d 1
8 a 20 d 2")
我希望第二列和第三列(CostType1和CostType 2)成为新列的名称,并将相应的成本填入特定成本类型。如果没有匹配,请填写NA。理想的格式如下:
a b c d
1 10 NA 1 NA
2 NA 2 20 NA
3 1 50 NA NA
4 40 1 NA NA
5 NA 30 2 NA
6 60 NA 3 NA
7 NA NA 10 1
8 20 NA NA 2
答案 0 :(得分:1)
使用tidyverse的解决方案。我们可以先得到有多少组。在此示例中,有两个组。我们可以转换每个组,组合它们,然后使用列中的第一个非NA值汇总数据框。
library(tidyverse)
# Get the group numbers
g <- (ncol(x) - 1)/2
x2 <- map_dfr(1:g, function(i){
# Transform the data frame one group at a time
x <- x %>%
select(ID, ends_with(as.character(i))) %>%
spread(paste0("CostType", i), paste0("Cost", i))
return(x)
}) %>%
group_by(ID) %>%
# Select the first non-NA value if there are multiple values
summarise_all(funs(first(.[!is.na(.)])))
x2
# # A tibble: 8 x 5
# ID a b c d
# <int> <int> <int> <int> <int>
# 1 1 10 NA 1 NA
# 2 2 NA 2 20 NA
# 3 3 1 50 NA NA
# 4 4 40 NA 1 NA
# 5 5 NA 30 2 NA
# 6 6 60 NA 3 NA
# 7 7 NA NA 10 1
# 8 8 20 NA NA 2
答案 1 :(得分:1)
使用reshape
x1 <- setNames(x[,c("ID", "CostType1", "Cost1")], c("ID", "CostType", "Cost"))
x2 <- setNames(x[,c("ID", "CostType2", "Cost2")], c("ID", "CostType", "Cost"))
reshape(data=rbind(x1, x2), idvar="ID", timevar="CostType", v.names="Cost", direction="wide")