将前25％的行分配为T，将其他行分配为F的列表数据帧(assign first 25 percent of rows as T and others as F from dataframes of l

将前25％的行分配为T，将其他行分配为F的列表数据帧(assign first 25 percent of rows as T and others as F from dataframes of list)

我有一个列表中随机采样的数据帧行列表。 我想将所有数据帧中前25％的行分配为T，将其他行分配为F.例如：

vec.1 <- c(1:574) vec.2 <- c(3001:3574) df.1 <- data.frame(vec.1, vec.2) df.2 <- data.frame(vec.2, vec.1) my_list <- replicate(10, df.1[sample(nrow(df.1)),] , simplify = FALSE)

在这个数据帧列表中，我想将前25％的行分配为F，将所有其他行分配为T.如何执行此操作？

I have a list of randomly sampled rows of dataframes in list. I would like to assign the first 25 percent of rows in all dataframe as T and other rows as F. For example:

vec.1 <- c(1:574) vec.2 <- c(3001:3574) df.1 <- data.frame(vec.1, vec.2) df.2 <- data.frame(vec.2, vec.1) my_list <- replicate(10, df.1[sample(nrow(df.1)),] , simplify = FALSE)

In this list of dataframes I would like to assign the first 25 percent of rows as F and all other rows as T. How to do this?

最满意答案

您可以轻松编写如下函数以在lapply ：

myFun <- function(indf) { indf$vec.3 <- seq_len(nrow(indf)) <= .25*nrow(indf) indf }

然后用法就是lapply(my_list, myFun) 。

You can easily write a function like the following to be used within lapply:

myFun <- function(indf) { indf$vec.3 <- seq_len(nrow(indf)) <= .25*nrow(indf) indf }

Usage would then just be lapply(my_list, myFun).