如果fullfiles有两个规则将名称存储在向量中

Question

从数据开始:

> dput(Data1)
structure(list(X1 = structure(c(17L, 14L, 20L, 16L, 1L, 2L, 3L, 
4L, 15L, 8L, 9L, 10L, 11L, 12L, 13L, 21L, 22L, 23L, 18L, 19L, 
5L, 6L, 7L), .Label = c("Astra_1", "Astra_2", "Astra_3", "Astra_4", 
"Audi_1", "Audi_2", "Audi_3", "BMW_1", "BMW_2", "BMW_3", "BMW_4", 
"BMW_5", "Fiat_1", "Mazda_2", "Mercedes_1", "Nexia_1", "Porsche_1", 
"Scania_1", "Scania_2", "Tico_1", "VW_1", "VW_2", "VW_3"), class = "factor"), 
    X2 = structure(c(2L, 3L, 10L, 7L, 8L, 12L, 9L, 14L, 11L, 
    4L, 5L, 6L, 15L, 13L, 4L, 5L, 9L, 14L, 11L, 1L, 3L, 10L, 
    16L), .Label = c("Astra_1", "Astra_3", "Astra_4", "Audi_1", 
    "Audi_2", "Audi_3", "BMW_1", "BMW_2", "Mazda_2", "Mercedes_1", 
    "Nexia_1", "Porsche_1", "Scania_2", "Tico_1", "VW_2", "VW_3"
    ), class = "factor"), AUC_1 = c(5860133.702, 1296009.939, 
    333123.4932, 250348.9407, 1376193.334, 4080502.863, 3777603.233, 
    3503973.487, 99101538.62, 231873.8462, 87258.75465, 147430.9913, 
    1028986.892, 1451482.832, 8136.72382, 25311.41683, 131352.7137, 
    565410.8186, 30196.23792, 70184.82268, 2526321.019, 381643.2138, 
    819687.9824), AUC_2 = c(4849720.322, 928980.4715, 320547.6185, 
    223287.2029, 1340641.323, 4720329.699, 4369150.434, 3371021.243, 
    108591253.3, 266489.7601, 85384.84604, 165726.7626, 1052130.559, 
    1470876.65, 9499.927679, 49309.74984, 138482.765, 444600.7911, 
    25132.73714, 55453.67019, 2038911.81, 422559.3293, 1445477.433
    ), ratio = c(1.20834467, 1.395088463, 1.03923247, 1.121196994, 
    1.02651866, 0.864452935, 0.864608186, 1.039439753, 0.91261069, 
    0.87010415, 1.021946618, 0.889602795, 0.978003046, 0.98681479, 
    0.856503765, 0.513314647, 0.948513078, 1.271726974, 1.201470327, 
    1.265647926, 1.2390536, 0.90317072, 0.567070757), Country = structure(c(1L, 
    1L, 2L, 3L, 5L, 1L, 5L, 1L, 4L, 7L, 4L, 7L, 7L, 7L, 6L, 6L, 
    6L, 6L, 8L, 8L, 6L, 6L, 7L), .Label = c("France", "Germany", 
    "Italy", "Norway", "Poland", "Spain", "Sweden", "Ukraine"
    ), class = "factor"), Comp = structure(c(3L, 5L, 16L, 9L, 
    8L, 9L, 12L, 14L, 4L, 15L, 11L, 14L, 16L, 17L, 10L, 10L, 
    12L, 13L, 1L, 2L, 5L, 6L, 7L), .Label = c("11,12", "12,13", 
    "12,13,14", "14,15", "14,15,16", "15,16,17", "16,17,18", 
    "2,3", "2,3,4", "3,4", "3,4,5", "4,5,6", "5,6", "5,6,7", 
    "5,6,7,8", "6,7,8", "7,8,9"), class = "factor")), .Names = c("X1", 
"X2", "AUC_1", "AUC_2", "ratio", "Country", "Comp"), class = "data.frame", row.names = c(NA, 
-23L))

数据主管看起来像这样:

         X1         X2     AUC_1     AUC_2     ratio Country     Comp
1 Porsche_1    Astra_3 5860133.7 4849720.3 1.2083447  France 12,13,14
2   Mazda_2    Astra_4 1296009.9  928980.5 1.3950885  France 14,15,16
3    Tico_1 Mercedes_1  333123.5  320547.6 1.0392325 Germany    6,7,8
4   Nexia_1      BMW_1  250348.9  223287.2 1.1211970   Italy    2,3,4
5   Astra_1      BMW_2 1376193.3 1340641.3 1.0265187  Poland      2,3
6   Astra_2  Porsche_1 4080502.9 4720329.7 0.8644529  France    2,3,4

现在我们将重点关注最后两列:Country和Comp.我想提取包含相同国家/地区的所有行,然后比较列Comp中的任何数字是否相同,X1和X2中的字符串应该存储在一起 - 可能在单独的向量中或在矩阵中.一行可能属于不同的"簇"/"向量".

期望输出的示例.这只是一个例子,聚类是完全随机的.任何可视化输出的方法都是可以接受的.

    Country         1        2        3         4        5       6
1    France    Astra_3  Scania_2   Tico_1       NA       NA       NA 
2    Poland    Astra_4   Mazda_2   VW_3       Tico_2     NA       NA
3    Sweden Mercedes_1    BMW_1    BMW_2      Audi_1    VW_3      NA  
4    Norway      BMW_1   Astra_1  Scania_2    Audi_3     NA       NA

Answer 1

我想你想要的是：找到给定国家/地区的所有行，比如西班牙。然后在这些行中，获取列 Comp 中出现特定数字的所有行，例如 4。然后在这些行中提取列 X1 和 X2 的内容并将它们放在一起。

也许这段代码就是您想要的：

countries <- levels(data[,"Country"])
results <- list()
cn <- 1
for (i in 1:length(countries))
{
  # find all row numbers with that country:
  idx <- which(data[,"Country"] == countries[i])
  # get all numbers which occur for that country:
  numbers <- unique(as.numeric(unlist(strsplit(as.character(data[idx,"Comp"]), ","))))
  for (j in 1:length(numbers))
  {
    # split all the numbers in the column "Comp" by ",":
    CompList <- strsplit(as.character(data[idx,"Comp"]), ",")
    # get all the row numbers for that country where numbers[j] is contained in the column "Comp":
    rows <- idx[unlist(lapply(CompList, function(x) {any(x == as.character(numbers[j]))}))] 

    # assuming you want a number in the column "Comp" to occur at least in two rows:
    if (length(rows) > 1) 
    {
      results[[cn]] <- list("Country"= countries[i], 
      "Cars"= as.vector(as.matrix(data[rows, c("X1", "X2")])),
      "ValueOfComp"=numbers[j])
      cn <- cn + 1
    }
  }
}

这会给你这样的东西：

> results
[[1]]
[[1]]$Country
[1] "France"

[[1]]$Cars
[1] "Porsche_1" "Mazda_2"   "Astra_3"   "Astra_4"  

[[1]]$ValueOfComp
[1] 14


[[2]]
[[2]]$Country
[1] "Spain"

[[2]]$Cars
[1] "Fiat_1" "VW_1"   "Audi_1" "Audi_2"

[[2]]$ValueOfComp
[1] 3