Question

希望在小标题中的单个列上将apply（）系列用于客户功能

我已经阅读了许多与apply（）相关的问题，但到目前为止，它们似乎比我尝试的要复杂。

library(dplyr)
library(stringr)

sample <- tibble(
    id = 1:13, 
    NodeName = c("Tree", "Rock", "Cloud", "Tree.Maple", "Tree.Oak", "Tree.Ash", "Rock.Igneous", "Rock.Sedimentary", "Rock.Metamorphic", "Cloud.Cumulus", "Cloud.Nimbus", "Cloud.Stratus", "Cloud.Cirrus"), 
    NodeType = c("Table", "Table", "Table", "Column", "Column", "Column", "Column", "Column", "Column", "Column", "Column", "Column", "Column")
)

GetTableName2 <- function(tibbleName, rowNum){
    columnName <- unlist(tibbleName[rowNum, 2])
    dotPosition <- str_locate(columnName, "[.]")[1]
    if(!is.na(dotPosition)){
        return(str_sub(columnName, start=1, end=dotPosition - 1))
    }else{
        NA
    }
}

GetParentID <- function(tibbleName, rowNum){
    tibbleName %>% 
      filter(NodeName == GetTableName2(tibbleName, rowNum)) %>% 
      select(id) %>% 
      unlist
}

GetParentID(sample, 11)

apply(sample[,2], MARGIN=2, FUN=GetParentID)

当我自己运行该函数时，我得到了预期的输出：3 当我使用apply（）函数尝试它时，出现此错误：

Error in UseMethod("filter_") : 
  no applicable method for 'filter_' applied to an object of class "character"
trace:
filter_(.data, .dots = compat_as_lazy_dots(...))
filter.default(., NodeName == GetTableName2(tibbleName, rowNum))
filter(., NodeName == GetTableName2(tibbleName, rowNum))
function_list[[i]](value)
freduce(value, `_function_list`)
`_fseq`(`_lhs`)
eval(quote(`_fseq`(`_lhs`)), env, env)
eval(quote(`_fseq`(`_lhs`)), env, env)
withVisible(eval(quote(`_fseq`(`_lhs`)), env, env))
tibbleName %>% filter(NodeName == GetTableName2(tibbleName, rowNum)) %>% 
    select(id) %>% unlist
FUN(newX[, i], ...)
apply(sample[, 2], MARGIN = 2, FUN = GetParentID)

我承认apply（）系列对我来说是一个反复出现的难题。每当我认为得到它，然后在几周或几个月后再回到它时，我都会面临我确实没有“得到它”的证据。

感谢您的关注。

在小标题列上使用apply（）

0 个答案: