行id到列保持日期完好无损

Question

我有一个名为df_high_A的数据框，如此

            C1  C2
02-01-07    10  23
03-01-07    20  12
04-01-07    14  17

想要像这样做

      C1        C2  C3
    02-01-07    10  23
    03-01-07    20  12
    04-01-07    14  17

当我使用此代码时

library (tibble)
    df_high_A<-tibble::rowid_to_column(data.frame(df_high_A),"date")

看起来像这样

C1  C2  C3
1   10  23
2   20  12
3   14  17

解决方案是什么？

Answer 1

假设您的数据是

dput( df_high_A )
structure(list(C1 = c(10L, 20L, 14L), C2 = c(23L, 12L, 17L)), .Names = c("C1", 
"C2"), class = "data.frame", row.names = c("02-01-07", "03-01-07", 
"04-01-07"))

使用Base R，解决方案将是

df_high_A <- data.frame( "C1" = row.names( df_high_A ), 
                         "C2" = df_high_A$C1, 
                         "C3" = df_high_A$C2 )

产生

df_high_A
        C1 C2 C3
1 02-01-07 10 23
2 03-01-07 20 12
3 04-01-07 14 17

Answer 2

如果你的rownames总是可以尝试将它们绑定在一起的日期，那么重置列名和行名。

df_high_A = cbind(rownames(df_high_A), df_high_A)
colnames(df_high_A) = paste("C",1:ncol(df_high_A),sep="")
rownames(df_high_A) = 1:nrow(df_high_A)