Question

我想通过将所有预先存在的变量彼此除以创建新变量

例如 X1 / X1，X1 / X2，X1 / X3，X1 / X4，X1 / X5，X1 / X6，X1 / X7，X1 / X8，X1 / X9，X1 / X10， X2 / X1，X2 / X2，X2 / X3，X2 / X4，X2 / X5，X2 / X6，X2 / X7，X2 / X8，X2 / X9，X2 / X10， X3 / X1，X3 / X2 ...

我首先尝试分别进行操作，如下所示，但是我需要使用多个变量名来复制它，因此自动化（我假设是函数/ lapply）是理想的选择。

BSTR

我有一个数据框，其中包含40多个变量，具有600万行，下面附加了一个较小的示例数据框。

谢谢！

查理

ds$rom_3_5m <- (ds$roll_open_mean_3m/ds$roll_open_mean_5m)
ds$rom_3_10m <- (ds$roll_open_mean_3m/ds$roll_open_mean_10m)
ds$rom_3_15m <- (ds$roll_open_mean_3m/ds$roll_open_mean_15m)
ds$rom_3_30m <- (ds$roll_open_mean_3m/ds$roll_open_mean_30m)
ds$rom_3_60m <- (ds$roll_open_mean_3m/ds$roll_open_mean_60m)
ds$rom_3_120m <- (ds$roll_open_mean_3m/ds$roll_open_mean_120m)
ds$rom_3_240m <- (ds$roll_open_mean_3m/ds$roll_open_mean_240m)
ds$rom_3_480m <- (ds$roll_open_mean_3m/ds$roll_open_mean_480m)
ds$rom_3_960m <- (ds$roll_open_mean_3m/ds$roll_open_mean_960m)
ds$rom_3_1920m <- (ds$roll_open_mean_3m/ds$roll_open_mean_1920m)
ds$rom_3_3840m <- (ds$roll_open_mean_3m/ds$roll_open_mean_3840m)
ds$rom_3_7680m <- (ds$roll_open_mean_3m/ds$roll_open_mean_7680m)
ds$rom_3_15360m <- (ds$roll_open_mean_3m/ds$roll_open_mean_15360m)
ds$rom_3_30720m <- (ds$roll_open_mean_3m/ds$roll_open_mean_30720m)
ds$rom_3_61440m <- (ds$roll_open_mean_3m/ds$roll_open_mean_61440m)
ds$rom_3_122880m <- (ds$roll_open_mean_3m/ds$roll_open_mean_122880m)
ds$rom_3_245760m <- (ds$roll_open_mean_3m/ds$roll_open_mean_245760m)
ds$rom_3_491520m <- (ds$roll_open_mean_3m/ds$roll_open_mean_491520m)

#5m
ds$rom_5_3m <- (ds$roll_open_mean_5m/ds$roll_open_mean_3m)
ds$rom_5_10m <- (ds$roll_open_mean_5m/ds$roll_open_mean_10m)
ds$rom_5_15m <- (ds$roll_open_mean_5m/ds$roll_open_mean_15m)
ds$rom_5_30m <- (ds$roll_open_mean_5m/ds$roll_open_mean_30m)
ds$rom_5_60m <- (ds$roll_open_mean_5m/ds$roll_open_mean_60m)
ds$rom_5_120m <- (ds$roll_open_mean_5m/ds$roll_open_mean_120m)
ds$rom_5_240m <- (ds$roll_open_mean_5m/ds$roll_open_mean_240m)
ds$rom_5_480m <- (ds$roll_open_mean_5m/ds$roll_open_mean_480m)
ds$rom_5_960m <- (ds$roll_open_mean_5m/ds$roll_open_mean_960m)
ds$rom_5_1920m <- (ds$roll_open_mean_5m/ds$roll_open_mean_1920m)
ds$rom_5_3840m <- (ds$roll_open_mean_5m/ds$roll_open_mean_3840m)
ds$rom_5_7680m <- (ds$roll_open_mean_5m/ds$roll_open_mean_7680m)
ds$rom_5_15360m <- (ds$roll_open_mean_5m/ds$roll_open_mean_15360m)
ds$rom_5_30720m <- (ds$roll_open_mean_5m/ds$roll_open_mean_30720m)
ds$rom_5_61440m <- (ds$roll_open_mean_5m/ds$roll_open_mean_61440m)
ds$rom_5_122880m <- (ds$roll_open_mean_5m/ds$roll_open_mean_122880m)
ds$rom_5_245760m <- (ds$roll_open_mean_5m/ds$roll_open_mean_245760m)
ds$rom_5_491520m <- (ds$roll_open_mean_5m/ds$roll_open_mean_491520m)

#10m
ds$rom_10_3m <- (ds$roll_open_mean_10m/ds$roll_open_mean_3m)
ds$rom_10_5m <- (ds$roll_open_mean_10m/ds$roll_open_mean_5m)
ds$rom_10_15m <- (ds$roll_open_mean_10m/ds$roll_open_mean_15m)

Answer 1

如评论中@ 27 ϕ 9所建议，您应使用该lapply解决方案。

以此，您还可以创建一个具有正确名称的唯一数据框

l <- lapply(df, `/`, df)
l <- unlist(l, recursive = FALSE)
data.frame(l)

通过将所有预先存在的变量除以所有其他变量来创建新变量

1 个答案: