0

当我使用列表功能时：

el_nino_1974_2000_all <- list()
for (k in seq_along(el_nino_start_month)){
     el_nino_1974_2000_all[[k]] = window(Nino3.4_Flow_1974_2000_zoo,
                                         start = (as.Date(el_nino_1974_2000[k,]$el_nino_start_mont)),
                                         end = (as.Date(el_nino_1974_2000[k,]$el_nino_finish_month)))
}

A 给出了一系列从 开始的独立数据子集i = 1。但是，我想以动物园格式或数据框格式将所有子集合并到一帧数据中。

这是 的结构el_nino_1974_2000_all。

    > str(el_nino_1974_2000_all)
List of 7
 $ :‘zoo’ series from 1976-08-15 to 1977-01-15
  Data: num [1:6, 1:2] 0.519 0.874 0.886 0.823 0.734 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:6], format: "1976-08-15" "1976-09-15" ...
 $ :‘zoo’ series from 1982-05-15 to 1983-06-15
  Data: num [1:14, 1:2] 0.961 1.388 0.959 1.171 1.564 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:14], format: "1982-05-15" "1982-06-15" ...
 $ :‘zoo’ series from 1986-09-15 to 1988-01-15
  Data: num [1:17, 1:2] 0.974 1.089 1.322 1.273 1.313 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:17], format: "1986-09-15" "1986-10-15" ...
 $ :‘zoo’ series from 1991-05-15 to 1992-07-15
  Data: num [1:15, 1:2] 0.68 1 0.923 0.773 0.68 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:15], format: "1991-05-15" "1991-06-15" ...
 $ :‘zoo’ series from 1993-02-15 to 1993-07-15
  Data: num [1:6, 1:2] 0.54 0.641 1.01 1.144 0.917 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:6], format: "1993-02-15" "1993-03-15" ...
 $ :‘zoo’ series from 1994-08-15 to 1995-02-15
  Data: num [1:7, 1:2] 0.662 0.746 1.039 1.329 1.301 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:7], format: "1994-08-15" "1994-09-15" ...
 $ :‘zoo’ series from 1997-04-15 to 1998-05-15
  Data: num [1:14, 1:2] 0.601 1.136 1.461 1.668 2.079 ...
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr [1:2] "Nino3.4_degree_1974_2000" "Houlgrave_flow_1974_2000"
  Index:  Date[1:14], format: "1997-04-15" "1997-05-15" ...
>

对不起，我不知道如何进行格式化。

4

2 回答 2

1

如果日期不重叠，您可以使用将它们粘贴在一起rbind（因为每个组件的列数相同）。尝试：

el_nino_1974_2000_all <- c()
for (k in seq_along(el_nino_start_month)){
    el_nino_1974_2000_all <- rbind(el_nino_1974_2000_all,window(...))
}

而不是list你原来的结构。这将返回一个zoo对象。

如果您想返回 a data.frame，请尝试使用rbind, 但 withdata.frame来转换您的对象（即使您的每个数据集之间的日期索引重叠，这也会起作用）：

el_nino_1974_2000_all <- data.frame()
for (k in seq_along(el_nino_start_month)){
    el_nino_1974_2000_all <- rbind(el_nino_1974_2000_all,data.frame(window(...)))
}
于 2012-01-19T03:16:01.760 回答
0

您是否尝试过此功能：

http://rss.acs.unt.edu/Rdoc/library/gtools/html/smartbind.html

out <- smartbind(list_of_dataframes)

注意：list_of_dataframes 应该包含 data.frames 但您可以将您的数据即时转换为 dframes，然后使用此功能。

于 2013-07-05T14:03:51.967 回答