将变量名传递给sapply [英] Passing variable name into sapply

问题描述

这是一个玩具数据集:

df1 <-data.frame(c("267119002","257051033",NA,"267098003","267099020","267047006"))
names(df1)[1]<-"ID"

df2 <-data.frame(c("257051033","267098003","267119002","267047006","267099020"))
names(df2)[1]<-"ID"
df2$vals <-c(11,22,33,44,55)

和玩具代码:

fetcher <-function(x){
  y <- df2$vals[which(match(df2$ID,x)==TRUE)]
  return(y) 
}

sapply(df1$ID,function(x) fetcher(x))

在sapply语句中，而不是使用df1$ID，我需要使用变量名.如:

In the sapply statement, instead of using df1$ID, I need to use a variable name. As in:

col <-"ID"
sapply(df1[col],function(x) fetcher(x))

但是，当我以此方式执行操作时，它不会遍历df1$ID的所有值.这样，它仅对第一个值执行sapply.输出示例:

However when I do it this way it does not iterate through all the values of df1$ID. This way it only does sapply on the first value. Example output:

> sapply(df1[col],function(x) fetcher(x))
ID 
33 
> sapply(df1$ID,function(x) fetcher(x))
[1] 33 11 22 55 44

那为什么会这样呢?我需要使用变量名而不是确切的列名，因为我需要将此变量名应用于每次程序运行时都会有所不同的不同列.但是我需要将它不仅应用于第一行，而且还应用于每一行.

So why is this happening? I need to use the variable name instead of the exact column name as I need to apply this to different columns that will vary each time the program runs. But I need it to apply to each row not just the first row.

推荐答案

区别在于df1[col]返回一列数据帧，而df1$ID返回向量/因数.使用您的代码，您需要一个向量/因数，因此您可以

The difference is that df1[col] returns a one column dataframe and df1$ID returns a vector/factor. Using your code you want a vector/factor, hence you may

使用df1[, col]

sapply(df1[, col],function(x) fetcher(x))

或双括号df1[[col]]

sapply(df1[[col]],function(x) fetcher(x))

.

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