我有一个这样的数据框
df= data.frame(a1 = c(1,2,3), a2 = c(4,5,6), b1 = c(1,2,3), b2= c(4,NaN,6), id = c(1,2,3))
我想得到
id a measure1 b measure2
1 1 a1 1 b1 1
2 2 a1 2 b1 2
3 3 a1 3 b1 3
4 1 a2 4 b2 4
5 2 a2 5 b2 NaN
6 3 a2 6 b2 6
我会做
df1 = df[, c(1,2,5)]
df2 = df[, c(3,4,5)]
library(reshape2)
df1_long = melt(df1,id.vars= 'id', measure.vars=c("a1", "a2"),
variable.name="a",
value.name="measure1")
df2_long = melt(df2,id.vars= 'id', measure.vars=c("b1", "b2"),
variable.name="b",
value.name="measure2")
df_new = cbind(df1_long, df2_long[, -1])
但我认为有一个更简单的方法
最佳答案
这是相对的,你认为更容易,但一个选项是使用 dplyr
和 tidyr
:
df %>%
select(id, starts_with("a")) %>%
gather(a, measurement1, -id) %>%
bind_cols(df %>%
select(starts_with("b")) %>%
gather(b, measurement2))
id a measurement1 b measurement2
1 1 a1 1 b1 1
2 2 a1 2 b1 2
3 3 a1 3 b1 3
4 1 a2 4 b2 4
5 2 a2 5 b2 NaN
6 3 a2 6 b2 6
关于r - 在两个变量中将宽转换为长,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/57632058/