通过 while 循环或 Reduce 的多级员工经理关系,仅基于 R
multilevel employee manager relationship through while loop or Reduce, base R only
我的问题可以被认为是已经解决的 的扩展。
为了重现,我在办公室里有多层次的员工-经理关系,就像这样
manager employee
1 CEO sally
2 sally sue
3 sally paul
4 sue mary
5 mary greg
6 mary don
我必须根据仅在 baseR 中可用的 methods/function 来解决此问题。我尝试 merge
数据本身(3 次)以获得我期望的结果 -
manager employee manager2 manager3 manager4
1 CEO sally <NA> <NA> <NA>
2 sally sue CEO <NA> <NA>
3 sally paul CEO <NA> <NA>
4 sue mary sally CEO <NA>
5 mary greg sue sally CEO
6 mary don sue sally CEO
我试图通过 while loop
来解决它,但由于中间级别的数量未知,我无法告诉循环在哪里停止。由于这个原因,我假设它也无法通过 Reduce
或类似的 purrr
函数解决。我不想要基于 'hR,
data.tree, 'igraph
之类的包的答案,因为这些答案已经在某个地方或另一个地方可用。 BaseR 或 tidyverse 是最受欢迎的。
dput 如下
df <- data.frame(manager = c("CEO","sally","sally","sue","mary", "mary"),
employee = c("sally","sue","paul","mary","greg", "don"),
stringsAsFactors = FALSE)
我试过的是这样的。
df |>
merge(df, by.x = 'manager', by.y = 'employee', suffixes = c('', '2'), all.x = T) |>
merge(df, by.x = 'manager2', by.y = 'employee', suffixes = c('', '3'), all.x = T) |>
merge(df, by.x = 'manager3', by.y = 'employee', suffixes = c('', '4'), all.x = T)
使用相同的合并解决方案,您可以使用终止条件来确定经理的合成稀疏向量是否有任何经理也是员工。这与发布的代码相同,但通过 while 循环执行。
df_new <- df
level <- 2
by_x <- "manager"
next_manager <- df[,"manager"]
while (any(next_manager %in% df$employee)) {
df_new <- merge(df_new, df, by.x = by_x, by.y = "employee", suffixes = c("",level), all.x = T)
next_manager <- df_new[,paste0("manager",level)]
by_x <- paste0("manager",level)
level <- level + 1
}
> df_new
manager3 manager2 manager employee manager4
1 CEO sally sue mary <NA>
2 sally sue mary greg CEO
3 sally sue mary don CEO
4 <NA> CEO sally sue <NA>
5 <NA> CEO sally paul <NA>
6 <NA> <NA> CEO sally <NA>
我的问题可以被认为是已经解决的
为了重现,我在办公室里有多层次的员工-经理关系,就像这样
manager employee
1 CEO sally
2 sally sue
3 sally paul
4 sue mary
5 mary greg
6 mary don
我必须根据仅在 baseR 中可用的 methods/function 来解决此问题。我尝试 merge
数据本身(3 次)以获得我期望的结果 -
manager employee manager2 manager3 manager4
1 CEO sally <NA> <NA> <NA>
2 sally sue CEO <NA> <NA>
3 sally paul CEO <NA> <NA>
4 sue mary sally CEO <NA>
5 mary greg sue sally CEO
6 mary don sue sally CEO
我试图通过 while loop
来解决它,但由于中间级别的数量未知,我无法告诉循环在哪里停止。由于这个原因,我假设它也无法通过 Reduce
或类似的 purrr
函数解决。我不想要基于 'hR,
data.tree, 'igraph
之类的包的答案,因为这些答案已经在某个地方或另一个地方可用。 BaseR 或 tidyverse 是最受欢迎的。
dput 如下
df <- data.frame(manager = c("CEO","sally","sally","sue","mary", "mary"),
employee = c("sally","sue","paul","mary","greg", "don"),
stringsAsFactors = FALSE)
我试过的是这样的。
df |>
merge(df, by.x = 'manager', by.y = 'employee', suffixes = c('', '2'), all.x = T) |>
merge(df, by.x = 'manager2', by.y = 'employee', suffixes = c('', '3'), all.x = T) |>
merge(df, by.x = 'manager3', by.y = 'employee', suffixes = c('', '4'), all.x = T)
使用相同的合并解决方案,您可以使用终止条件来确定经理的合成稀疏向量是否有任何经理也是员工。这与发布的代码相同,但通过 while 循环执行。
df_new <- df
level <- 2
by_x <- "manager"
next_manager <- df[,"manager"]
while (any(next_manager %in% df$employee)) {
df_new <- merge(df_new, df, by.x = by_x, by.y = "employee", suffixes = c("",level), all.x = T)
next_manager <- df_new[,paste0("manager",level)]
by_x <- paste0("manager",level)
level <- level + 1
}
> df_new
manager3 manager2 manager employee manager4
1 CEO sally sue mary <NA>
2 sally sue mary greg CEO
3 sally sue mary don CEO
4 <NA> CEO sally sue <NA>
5 <NA> CEO sally paul <NA>
6 <NA> <NA> CEO sally <NA>