我创建了一个与您的示例类似的示例数据集

df <- data.frame(ID=c(111,999), Comedy.valence=c(1, 6), Comedy.arousal=c(1,4), Comedy.rating=c(100,82), Drama.valence=c(5,3), Drama.arousal=c(7,8), Drama.rating=c(20,80))

然后，我应用如下所示的整形：

df2 <- reshape(df, varying = 2:7, direction = "long", idvar = "ID", timevar = "Video_Type", v.names = c("valence", "arousal", "rating"), times = c("Comedy", "Drama"))

row.names(df2) <- NULL

df2

输出如下所示：

ID Video_Type valence arousal rating
1 111     Comedy       1     100      1
2 999     Comedy       4      82      6
3 111      Drama       7      20      5
4 999      Drama       8      80      3

我对代码做了一些修改：

df2 <- reshape(df, 
        varying = 2:7, 
        direction = "long", 
        idvar = "ID", 
        timevar = "Video_Type", 
        v.names = unlist(unique(lapply(strsplit(names(df), split="\\.")[2:7], '[[', 2))), 
        times = unlist(unique(lapply(strsplit(names(df), split="\\.")[2:7], '[[', 1))))

row.names(df2) <- NULL

df2

我想这取决于您有多少列，您可以修改varying和lapply中的列索引

只需要切换名称。这不能使用split =参数来反转，因此使用sub函数

reshape(setNames(df, sub("(\\w+).(\\w+)","\\2.\\1", names(df))), -1, dir='long', idv = 'ID')

            ID   time valence arousal rating
111.Comedy 111 Comedy       1       1    100
999.Comedy 999 Comedy       6       4     82
111.Drama  111  Drama       5       7     20
999.Drama  999  Drama       3       8     80