r - 通过 "multiplying"通过其他列的名称转换列的元素

Question

这个问题在这里已经有了答案:

Transpose / reshape dataframe without "timevar" from long to wide format (9 个回答)

关闭 2年前。

下面是一个例子。如何将具有 names 列的数据帧 df 转换为下面的 df.transformed 形式？

> df <- data.frame("names" = c("y1", "y2"), "x1" = 1:2, "x2" = 4:5)
> df
  names x1 x2
1    y1  1  4
2    y2  2  5


> df.transformed <- data.frame("y1x1" = 1, "y1x2" =4, "y2x1" = 2, "y2x2" = 5)
> df.transformed
  y1x1 y1x2 y2x1 y2x2
1    1    4    2    5

Answer 1

代码

require(data.table); setDT(df)

dt = melt(df, id.vars = 'names')[, col := paste0(variable, names)]
out = dt$value; names(out) = dt$col

结果

> data.frame(t(out))

x1y1 x1y2 x2y1 x2y2 
   1    2    4    5