Question

我不想重塑它，因为我有很多数据，所以像循环这样的东西会自动翻译它输入-数据框1

Item     LC     ToLC
8T4121  MW92    WK14
8T4121  WK14    RM11
8T4121  WK14    RS11
8T4121  RS11    OY01
AB7651  MW92    RS11
AB7651  RS11    OY01

我想做一个循环，我可以得到这样的输出数据框2

Item     LC1    LC2    LC3    LC4
8T4121  MW92    WK14   RM11  
8T4121  MW92    WK14   RS11   OY01
AB7651  MW92    RS11   OY01

我尝试过这样的事情：

bodlane <- lctolc
colnames(bodlane) <- c("Item","Entry","From")

bodlane$To <- lctolc$To[match(bodlane$From, lctolc$From)]
colnames(bodlane) <- c("Item","Entry","Parent","From")

bodlane$To <- lctolc$To[match(bodlane$From, lctolc$From)]
colnames(bodlane) <- c("Item","Entry","Parent","Parent1","From")

bodlane$To <- lctolc$To[match(bodlane$From, lctolc$From)]
colnames(bodlane) <- c("Item","LC","ToLC","Parent1","From","To")

Answer 1

我相信可以使用igraph来解决此问题，而方法与“recursive” self join in data.table类似，但无需计算。

这里的困难是每个Item都有单独的图形。我的方法是将数据框拆分为图表列表。可能会有更简洁的解决方案使用type顶点属性。

但是，下面的代码创建了预期的结果：

library(igraph)
library(data.table)
library(magrittr)

lapply(
  lapply(split(lctolc, lctolc$Item), function(x) graph.data.frame(x[, 2:3])), 
  function(x) lapply(
    all_simple_paths(x, from = V(x)[degree(x, mode = "in") == 0], 
                     to = V(x)[degree(x, mode = "out") == 0]),
    function(y) as.data.table(t(names(y))) %>% setnames(paste0("LC", seq_along(.)))
  ) %>% rbindlist(fill = TRUE)
) %>% rbindlist(fill = TRUE, idcol = "Item")

     Item  LC1  LC2  LC3  LC4
1: 8T4121 MW92 WK14 RS11 OY01
2: 8T4121 MW92 WK14 RM11 <NA>
3: AB7651 MW92 RS11 OY01 <NA>

说明

待完成

数据

library(data.table)
lctolc <- fread("
Item     LC     ToLC
8T4121  MW92    WK14
8T4121  WK14    RM11
8T4121  WK14    RS11
8T4121  RS11    OY01
AB7651  MW92    RS11
AB7651  RS11    OY01",
data.table = FALSE)

对于Dataframe中的每个项目，要自动循环

1 个答案:

说明

数据