Question

我有一个层次结构，我想创建一个表来捕获最后一个值，之前的值和列号。

我不知道从哪里开始。我可以将列号计算为新列，然后我认为需要融合数据，但是我无法确定要使用什么变量，也不知道要搜索什么。

我的输入看起来像这样：

input = structure(list(V1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L), .Label = "ASIA PACIFIC", class = "factor"), V2 = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "AUSTRALIA", class = "factor"), 
    V3 = structure(c(1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("ACT", 
    "NEW SOUTH WALES"), class = "factor"), V4 = structure(c(1L, 
    3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 4L), .Label = c("CANBERRA", 
    "NEWCASTLE", "SYDNEY", "WOLLONGONG"), class = "factor"), 
    V5 = structure(c(9L, 2L, 6L, 4L, 7L, 3L, 5L, 8L, 10L, 1L), .Label = c("###", 
    "BONDI", "CAMPBELLTOWN", "GEORGE ST", "MAIN ST", "NEWTOWN", 
    "PITT ST", "POKOLBIN", "SMITH ST", "STRANGE PDE"), class = "factor"), 
    V6 = structure(c(1L, 2L, 3L, 1L, 1L, 5L, 1L, 4L, 1L, 1L), .Label = c("###", 
    "CHARLES AVE", "FRANCIS ST", "TOM ST", "TONY LANE"), class = "factor"), 
    V7 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = "###", class = "factor")), class = "data.frame", row.names = c(NA, 
-10L))

我正在寻找创建以下输出：

output = 
structure(list(V1 = structure(c(10L, 3L, 4L, 5L, 9L, 14L, 6L, 
13L, 11L, 15L, 2L, 12L, 8L, 7L, 1L), .Label = c("AUSTRALIA", 
"CANBERRA", "CHARLES AVE", "FRANCIS ST", "GEORGE ST", "MAIN ST", 
"NEW SOUTH WALES", "NEWCASTLE", "PITT ST", "SMITH ST", "STRANGE PDE", 
"SYDNEY", "TOM ST", "TONY LANE", "WOLLONGONG"), class = "factor"), 
    V2 = structure(c(6L, 4L, 9L, 11L, 11L, 5L, 8L, 10L, 8L, 7L, 
    1L, 7L, 7L, 3L, 2L), .Label = c("ACT", "ASIA PACIFIC", "AUSTRALIA", 
    "BONDI", "CAMPBELLTOWN", "CANBERRA", "NEW SOUTH WALES", "NEWCASTLE", 
    "NEWTOWN", "POKOLBIN", "SYDNEY"), class = "factor"), V3 = c(5L, 
    6L, 6L, 5L, 5L, 6L, 5L, 6L, 5L, 4L, 3L, 3L, 3L, 2L, 1L)), class = "data.frame", row.names = c(NA, 
-15L))

我不知道该如何寻找入门。如果有人可以建议我从哪里开始，我将不胜感激。

Answer 1

通常，此类缺失值用NA表示，但这里用###表示，我们可以从每一行中删除这些值，然后使用tail选择最后两个值并返回{{ 1}}。

length(new_x)

一种as.data.frame(t(apply(input, 1, function(x) {new_x = x[x != "###"]; c(rev(tail(new_x, 2)), length(new_x)) }))) # V1 V2 V3 #1 CHARLES AVE BONDI 6 #2 FRANCIS ST NEWTOWN 6 #3 GEORGE ST SYDNEY 5 #4 PITT ST SYDNEY 5 #5 TONY LANE CAMPBELLTOWN 6 #6 MAIN ST NEWCASTLE 5 #7 TOM ST POKOLBIN 6 #8 STRANGE PDE NEWCASTLE 5 #9 WOLLONGONG NEW SOUTH WALES 4的解决方法可能是

tidyverse

从层次结构创建表

1 个答案: