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))
数据头看起来像这样:
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的所有行，然后比较如果Comp列中的任何数字相同，则应将X1和X2的字符串存储在一起-可能存储在单独的向量中或在矩阵中。一行可能属于不同的“簇”/“向量”。所需输出的示例。这只是一个例子，聚类完全是随机的。可视化输出的任何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
最佳答案
我想您想要的是:找到一个给定国家/地区的所有行，例如西类牙。然后在这些行中，采用在Comp列中出现某个数字的所有行，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
关于r - 如果全文件，则两个规则将名称存储在向量中，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/36620842/