Question

我有一个包含约300,000行和60列的大型数据集。目前，如果我想在我的一个变量中对唯一特征进行子集化，我使用unique()函数创建该变量中所有唯一值的data.frame列表。然后我将它与主数据帧匹配，以从我的主文件中获取相关数据。

然而这个过程有点麻烦，所以我想知道是否有更快的方法来做同样的事情？例如，是否有一个函数可用于选择唯一字段以及连接到这些值的关联数据？

例如：我想创建一个新的数据框，其中只包含唯一的SurveyID_Block ID及其相关的岛代码和丰度。

 structure(list(SurveyID_Block = c("62003713_2", "62003087_2", 
"62003713_2", "62003713_2", "62003713_1", "62003713_2", "62003713_1", 
"62003713_2", "62003713_2", "62003087_1", "62003713_1", "62003713_1", 
"62003713_2", "62003713_2", "62003713_1", "62003087_1", "62003087_2", 
"62003713_2", "62003713_2", "62003713_2", "62003087_2", "62003713_2", 
"62003713_1", "62003713_1", "62003713_1", "62003713_1", "62003713_2", 
"62003713_1", "62003713_2", "62003087_1", "62003713_2", "62003087_1", 
"62003713_1", "62003087_2", "62003087_2", "62003713_2", "62003713_1", 
"62003087_1", "62003713_1", "62003713_1", "62003713_1", "62003087_2", 
"62003087_2", "62003713_2", "62003713_2", "62003713_2", "62003713_1", 
"62003087_1", "62003713_2", "62003087_2", "62003713_1", "62003713_1", 
"62003713_2", "62003713_1", "62003713_2", "62003087_2", "62003087_2", 
"62003087_1", "62003087_1", "62003713_1", "62003087_1", "62003087_1", 
"62003087_2", "62003087_2", "62003713_2", "62003713_1", "62003713_2", 
"62003713_2", "62003713_2", "62003713_1", "62003713_2", "62003087_1", 
"62003713_1", "62003713_1", "62003087_1", "62003087_1", "62003713_1", 
"62003087_2", "62003087_1", "62003087_2", "62003087_2", "62003087_1", 
"62003087_1", "62003087_1", "62003713_2", "62003087_2", "62003713_2", 
"62003087_2", "62003713_1", "62003713_1", "62003087_2", "62003087_1", 
"62003087_1", "62003087_1", "62003713_2", "62003713_2", "62003087_1", 
"62003713_1", "62003087_1", "62003087_2"), IslandCode = c(1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 
1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L, 1391L
), totalAbun = c(667L, 174L, 667L, 667L, 715L, 667L, 715L, 667L, 
667L, 1365L, 715L, 715L, 667L, 667L, 715L, 1365L, 174L, 667L, 
667L, 667L, 174L, 667L, 715L, 715L, 715L, 715L, 667L, 715L, 667L, 
1365L, 667L, 1365L, 715L, 174L, 174L, 667L, 715L, 1365L, 715L, 
715L, 715L, 174L, 174L, 667L, 667L, 667L, 715L, 1365L, 667L, 
174L, 715L, 715L, 667L, 715L, 667L, 174L, 174L, 1365L, 1365L, 
715L, 1365L, 1365L, 174L, 174L, 667L, 715L, 667L, 667L, 667L, 
715L, 667L, 1365L, 715L, 715L, 1365L, 1365L, 715L, 174L, 1365L, 
174L, 174L, 1365L, 1365L, 1365L, 667L, 174L, 667L, 174L, 715L, 
715L, 174L, 1365L, 1365L, 1365L, 667L, 667L, 1365L, 715L, 1365L, 
174L)), .Names = c("SurveyID_Block", "IslandCode", "totalAbun"
), row.names = c(NA, 100L), class = "data.frame")

Answer 1

我们可以通过'SurveyID_Block'split数据集来创建list data.frame个。最好将数据集保存在list中，而不是在全局环境中创建单独的data.frame对象。

lst <- split(df1, df1$SurveyID_Block)

但是，如果我们需要创建单个数据集，可以使用list2env

来完成

list2env(setNames(lst, paste0('dfN', seq_along(lst))),
          envir=.GlobalEnv)

是否有一种简单的方法可以对唯一值及其相关数据进行分组？

1 个答案: