Question

我有一个列表，mylist包含5个元素。

我可以使用以下内容合并任何两个元素：

merge(mylist[[1]], mylist[[2]], by = someColumn).

但是，如何在保留名称而不引起警告的情况下一次合并所有这些内容。

谢谢！

修改

经过进一步检查，这里有一些细节可以帮助澄清我的问题：

> mylist
$Alpha
   id count
1 ABC     5
2 DEF    10
3 GHI    15

$Beta
   id count
1 DEF    10
2 ABC    12
3 GHI    14

$Gamma
   id count
1 ABC    13
2 GHI    15
3 DEF    17

$Kappa
   id count
1 GHI    20
2 DEF    21
3 ABC    25
> Reduce(function(x,y){merge(x,y,by="id")}, mylist)
   id count.x count.y count.x count.y
1 ABC       5      12      13      25
2 DEF      10      10      17      21
3 GHI      15      14      15      20
Warning message:
In merge.data.frame(x, y, by = "id") :
  column names ‘count.x’, ‘count.y’ are duplicated in the result

如您所见，事情在结果的顶部重复，R会发出警告。我怎样才能避免这种情况（最好用列表名称调用列名 - 所以第一个count.x将是count.Alpha）。

如果有人想要，请点击这里的mylist副本：

structure(list(Alpha = structure(list(id = structure(1:3, .Label = c("ABC", 
"DEF", "GHI"), class = "factor"), count = c(5, 10, 15)), .Names = c("id", 
"count"), row.names = c(NA, -3L), class = "data.frame"), Beta = structure(list(
    id = structure(c(2L, 1L, 3L), .Label = c("ABC", "DEF", "GHI"
    ), class = "factor"), count = c(10, 12, 14)), .Names = c("id", 
"count"), row.names = c(NA, -3L), class = "data.frame"), Gamma = structure(list(
    id = structure(c(1L, 3L, 2L), .Label = c("ABC", "DEF", "GHI"
    ), class = "factor"), count = c(13, 15, 17)), .Names = c("id", 
"count"), class = "data.frame", row.names = c(NA, -3L)), Kappa = structure(list(
    id = c("GHI", "DEF", "ABC"), count = c(20, 21, 25)), .Names = c("id", 
"count"), row.names = c(NA, -3L), class = "data.frame")), .Names = c("Alpha", 
"Beta", "Gamma", "Kappa"))

Answer 1

您可以快速Map重命名＆＃34;计数＆＃34;柱。例如

Reduce(merge, Map(function(n,x) {names(x)[2]<-n; x}, names(mylist), mylist))

返回

   id Alpha Beta Gamma Kappa
1 ABC     5   12    13    25
2 DEF    10   10    17    21
3 GHI    15   14    15    20

Answer 2

使用match.call()的另一种方式：

 Reduce(merge, lapply(mylist, function(x) {
                   names(x)[2] <- paste(names(x)[2], names(mylist)[match.call()[[2]][[3]]],sep=".")
                 x}))
#   id count.Alpha count.Beta count.Gamma count.Kappa
#1 ABC           5         12          13          25
#2 DEF          10         10          17          21
#3 GHI          15         14          15          20

R中列表的Reduce相当于什么

2 个答案: