在不指定列名的情况下将列转换为行

Question

我有一个具有以下结构的数据框：

bad_df <- data.frame(
id = c("id001", "id002", "id003"),
participant.1 = c("Jana", "Marina", "Vasilei"),
participant.2 = c("Niko", "Micha", "Niko"),
role.1 = c("writer", "writer", "speaker"),
role.2 = c("observer", "observer", "observer"),
stringsAsFactors = F
)
bad_df

我需要把它收集成这样的东西。每行应包含一个 ID、参与者和角色。

good_df <- data.frame(
id = c("id001", "id001", "id002", "id002", "id003", "id003"),
participant = c("Jana", "Niko", "Marina", "Micha", "Vasilei", "Niko"),
role = c("writer", "observer", "writer", "observer", "speaker", "observer"),
stringsAsFactors = F
)
good_df

我看到有无数类似的问题，但我发现很难理解如何将 tidyr 或 reshape2 应用于这种情况。我知道这必须以某种方式通过 gather() 实现。

但是，数据框可以包含更多的参与者和相应的角色，因此理想情况下，该方法不需要指定列名。下面是我想出的一种解决方案，但我认为这不是最优雅的方法。而且我仍然需要处理一些包含 participant.3、role.3 等的数据帧

good_df2 <- rbind(bad_df %>% select(id, participant.1, role.1) %>% 
                    rename(participant = participant.1, role = role.1),
                 bad_df %>% select(id, participant.2, role.2) %>% 
                    rename(participant = participant.2, role = role.2))
good_df2

谢谢！

Answer 1

您可以尝试 data.table 的开发版本，即。 v1.9.5。安装说明是 here

library(data.table)
melt(setDT(bad_df), measure=list(grep('participant', names(bad_df)),
    grep('role', names(bad_df))))[order(id)][, variable:= NULL]
#      id  value1   value2
#1: id001    Jana   writer
#2: id001    Niko observer
#3: id002  Marina   writer
#4: id002   Micha observer
#5: id003 Vasilei  speaker
#6: id003    Niko observer

或者我们可以使用 merged.stack，我们只需要提供唯一列的前缀。根据前缀值，它将具有相同前缀的列分组在一起。

library(splitstackshape)
merged.stack(bad_df, var.stubs=c('participant', 'role'), 
                       sep='var.stubs')[, 2:= NULL]
#      id participant     role
#1: id001        Jana   writer
#2: id001        Niko observer
#3: id002      Marina   writer
#4: id002       Micha observer
#5: id003     Vasilei  speaker
#6: id003        Niko observer

或使用dplyr/tidyr

library(dplyr)
library(tidyr)
gather(bad_df, Var, Val, -id) %>% 
        separate(Var, into=c('Var1', 'Var2')) %>% 
        spread(Var1, Val) %>%
        select(-Var2)
#    id participant     role
#1 id001        Jana   writer
#2 id001        Niko observer
#3 id002      Marina   writer
#4 id002       Micha observer
#5 id003     Vasilei  speaker
#6 id003        Niko observer

Answer 2

我会在 base R:

走这条路

 #find the participant columns
 partCol<-grep("part",colnames(bad_df))
 #... and the role columns
 roleCol<-grep("role",colnames(bad_df))
 data.frame(id=rep(bad_df$id,each=length(partCol)),
            partecipant=as.vector(as.matrix(t(bad_df[,partCol]))),
            role=as.vector(as.matrix(t(bad_df[,roleCol]))))

在不指定列名的情况下将列转换为行

Converting columns into rows without specifying the column names

r

reshape2

tidyr