我有一个包含二进制变量的表,我想将其压缩为分类变量。
非常简单,我有一个像这样的数据框:
data <- data.frame(id=c(1,2,3,4,5,6,7,8,9), red=c("1","0","0","0","1","0","0","0","0"),blue=c("0","1","1","1","0","1","1","1","0"),yellow=c("0","0","0","0","0","0","0","0","1"))
data
id red blue yellow
1 1 1 0 0
2 2 0 1 0
3 3 0 1 0
4 4 0 1 0
5 5 1 0 0
6 6 0 1 0
7 7 0 1 0
8 8 0 1 0
9 9 0 0 1
我想回来的是:
id color
1 1 red
2 2 blue
3 3 blue
4 4 blue
5 5 red
6 6 blue
7 7 blue
8 8 blue
9 9 yellow
我希望有一个非常简单的答案。
答案 0 :(得分:4)
您可以使用列names
和as.logical
来获取值。但是,因为你的&#34;二进制&#34;列是因素,你需要更多的箍:
> apply(data[-1], 1, function(x) names(x)[as.logical(as.numeric(as.character(x)))])
[1] "red" "blue" "blue" "blue" "red" "blue" "blue" "blue" "yellow"
将其与第一列(data[1]
)绑定,以获得所需的输出。
cbind(data[1],
color = apply(data[-1], 1,
function(x) names(x)[as.logical(as.numeric(
as.character(x)))]))
# id color
# 1 1 red
# 2 2 blue
# 3 3 blue
# 4 4 blue
# 5 5 red
# 6 6 blue
# 7 7 blue
# 8 8 blue
# 9 9 yellow
或者,您可以尝试以下方法:
data[-1] <- lapply(data[-1], function(x) as.numeric(as.character(x)))
temp <- subset(cbind(data[1], stack(data[-1])), values == 1, c("id", "ind"))
temp[order(temp$id), ]
或者,您可以使用&#34; dplyr&#34;和&#34; tidyr&#34;,像这样:
library(dplyr)
library(tidyr)
data %>%
group_by(id) %>%
mutate_each(funs(an = as.numeric(as.character(.)))) %>%
gather(color, val, -id) %>%
filter(val == 1) %>%
select(-val) %>%
arrange(id)
# Source: local data frame [9 x 2]
#
# id color
# 1 1 red
# 2 2 blue
# 3 3 blue
# 4 4 blue
# 5 5 red
# 6 6 blue
# 7 7 blue
# 8 8 blue
# 9 9 yellow
答案 1 :(得分:4)
这是使用max.col
cbind(data[1L], color = names(data[-1L])[max.col(data[-1L] == 1L)])
# id color
# 1 1 red
# 2 2 blue
# 3 3 blue
# 4 4 blue
# 5 5 red
# 6 6 blue
# 7 7 blue
# 8 8 blue
# 9 9 yellow