3

我有一个这样的清单

list <- structure(list(`1` = structure(c(274L, 173L), .Dim = 2L, .Dimnames = structure(list(
    c("2004", "2005")), .Names = ""), class = "table"), `2` = structure(73L, .Dim = 1L, .Dimnames = structure(list(
    "2005"), .Names = ""), class = "table"), `3` = structure(c(334L, 
365L, 366L, 365L, 365L, 365L, 366L, 365L, 287L), .Dim = 9L, .Dimnames = structure(list(
    c("1990", "1991", "1992", "1993", "1994", "1995", "1996", 
    "1997", "1998")), .Names = ""), class = "table"), `4` = structure(139L, .Dim = 1L, .Dimnames = structure(list(
    "2001"), .Names = ""), class = "table"), `5` = structure(71L, .Dim = 1L, .Dimnames = structure(list(
    "2009"), .Names = ""), class = "table"), `6` = structure(77L, .Dim = 1L, .Dimnames = structure(list(
    "1997"), .Names = ""), class = "table")), .Names = c("1", 
"2", "3", "4", "5", "6"))

该列表的第一级是一个不断上升的数字。在第二级，我们有以年份作为列名的表。

我想构建列名称为 2005 的第二级所有元素的总和。我该怎么做？

4

2 回答 2

3

您可以使用sapply和的组合[。

> sum(sapply(list, `[`, "2005"), na.rm=TRUE)
[1] 246
于 2012-03-02T15:12:21.610 回答
1

这会成功的

get2005 <- function(x){
  x[names(x) %in% 2005]
}

sum(unlist(lapply(list, get2005)))
于 2012-03-02T15:15:38.650 回答