Question

我想根据比率计算为data.table创建新列。我的变量名称略有标准，所以我认为必须有一种方法可以在 data.table 中轻松实现这一点。但是我无法达到如何实现这一目标。以下是我的示例数据和代码 -

set.seed(1200)

ID <- seq(1001,1100)
region <- sample(1:10,100,replace = T)
Q21 <- sample(1:5,100,replace = T)
Q22 <- sample(1:15,100,replace = T)
Q24_LOC_1 <- sample(1:8,100,replace = T)
Q24_LOC_2 <- sample(1:8,100,replace = T)
Q24_LOC_3 <- sample(1:8,100,replace = T)
Q24_LOC_4 <- sample(1:8,100,replace = T)

Q21_PAN <- sample(1:5,100,replace = T)
Q22_PAN <- sample(1:15,100,replace = T)
Q24_LOC_1_PAN <- sample(1:8,100,replace = T)
Q24_LOC_2_PAN <- sample(1:8,100,replace = T)
Q24_LOC_3_PAN <- sample(1:8,100,replace = T)
Q24_LOC_4_PAN <- sample(1:8,100,replace = T)

df1 <- as.data.table(data.frame(ID,region,Q21,Q22,Q24_LOC_1,Q24_LOC_2,Q24_LOC_3,Q24_LOC_4,Q21_PAN,Q22_PAN,Q24_LOC_1_PAN,Q24_LOC_2_PAN,Q24_LOC_3_PAN,Q24_LOC_4_PAN))

col_needed <- c("Q21","Q22","Q24_LOC_1","Q24_LOC_2","Q24_LOC_3","Q24_LOC_4")

check1 <- df1[,Q21_R := mean(Q21,na.rm = T)/mean(Q21_PAN,na.rm = T),by=region]

check1适用于一个变量。我正在寻找一个解决方案，我可以传递所有需要的变量，并在一行中计算新的变量。所以在这种情况下，比如传递 col_needed 。我也试过下面的代码 -

check2 <- df1[,`:=`(paste0(col_needed,"_R"),(mean(col_needed,na.rm = T)/mean(paste0(col_needed,"_PAN"),na.rm = T))),by=region][]

然而，这给了我多次警告，结果是所有的NA。警告是 - 平均值（col_needed，na.rm = T）：参数不是数字或逻辑：返回NA

你能否告诉我哪里出错了。

Answer 1

如果我理解正确，您可以执行以下操作：

df1[, paste(col_needed, "R", sep = "_") := 
      Map(function(x,y) mean(get(x), na.rm = TRUE)/mean(get(y), na.rm=TRUE), 
           col_needed, 
           paste(col_needed, "PAN", sep = "_")),
    by=region]

R data.table使用标准名称创建新列

1 个答案: