lapply和mutate_all/for循环 [英] lapply and mutate_all/for loops

问题描述

我在列表中有几个数据框，我必须通过规范化所有列中的所有数据来进行修改(基本上，将每一行/列除以该列数的总和).

I have several data frames within a list which I have to modify by normalizing all the data, in all columns (basically, divide each row/column by the sum of the number of that column).

在用lapply加载所有原始数据帧之后，我想遍历所有列以执行此类操作(即mutate(df，df $ my_column = df $ my_column/sum(df $ my_column))).

After loading all my raw data frames with lapply I want to iterate over all columns to perform such operation (i.e. mutate(df, df$my_column=df$my_column/sum(df$my_column))).

我的代码是:

samplelist <- list(df1 = "path to df1",
                 df2 = "path to df2",
                 df3 = "path to df3")


samples <- lapply(names(samplelist),function(processing){
  aux <- read.csv(samplelist[[processing]], header = T, sep = "") # works
  for (i in colnames(aux)){
    mutate(aux, aux[[i]]=aux[[i]]/sum(aux[[i]]))
  }
})

但是不起作用(意外的"="以及后来出现的意外的"{")，因此我尝试使用dplyr的mutate_all，但我真的不知道该如何使用管道

But doesn't work (unexpected "=" and later on unexpected "{"), so I have tried to use mutate_all from dplyr, but I don't really know how to pipe it

samplelist <- list(df1 = "path to df1",
                     df2 = "path to df2",
                     df3 = "path to df3")


    samples <- lapply(names(samplelist),function(processing){
      aux <- read.csv(samplelist[[processing]], header = T, sep = "") %>% mutate_all(what should I write there?)
    })

我可能只需要添加新行并使用mutate_all，但仍然无法弄清楚要给出什么参数.如果您还知道其他方法也可以.

I could probably just add a new line and use mutate_all, but still I can't figure out what arguments to give. If you know also other ways to do it is fine.

非常感谢您的帮助.

推荐答案

您可以只使用所描述的函数并适应该函数的参数.在dplyr中，在这种情况下，.代表变量.〜定义一个公式.

You can just use your described function and adapt for the function argument. In dplyr, the . stands, in this case, for the variable. The ~ defines a formula.

samples <- lapply(names(samplelist), function(processing){
  aux <- read.csv(samplelist[[processing]], header = T, sep = "") %>% 
    mutate_all(~./sum(.))
})

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