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r - 基于另一个数据框的两行比较

转载 作者:行者123 更新时间:2023-12-04 10:49:37 24 4
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让我们从主要数据集开始:

> dput(tbl_test1)
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, 4L, 6L, 6L, 6L, 6L, 5L, 8L, 8L, 8L, 8L, 8L, 7L, 7L,
7L, 7L, 9L, 9L, 3L, 3L, 3L), .Label = c("France", "Germany",
"Ireland", "Italy", "Norway", "Poland", "Spain", "Sweden",
"Ukraine"), class = "factor")), .Names = c("X1", "X2", "AUC_1",
"AUC_2", "ratio", "Country"), class = "data.frame", row.names = c(NA,
-23L))

我想通过查看另一张表格来比较前两列中的汽车。为了更详细地解释,让我放下一张表:

> dput(tbl_test2)
structure(list(X = structure(c(17L, 14L, 20L, 16L, 1L, 2L, 3L,
8L, 9L, 10L, 11L, 12L, 4L, 15L, 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"),
X10 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), X34 = c(0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 0L, 1L), X59 = c(0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
1L, 1L), X84 = c(0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L,
1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L), X110 = c(0L,
0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 1L, 0L), X134 = c(0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L), X165 = c(1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L),
X199 = c(1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), X234 = c(1L,
0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L,
0L, 1L, 0L, 0L, 0L, 0L, 0L), X257 = c(1L, 0L, 0L, 0L, 1L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 1L, 0L, 0L,
0L, 0L, 0L), X362 = c(0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L, 0L),
X433 = c(0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L, 0L), X506 = c(0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L,
1L, 1L, 1L, 0L, 0L, 0L, 0L), X581 = c(0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L,
0L, 0L, 0L), X652 = c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L,
0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L),
X733 = c(0L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 1L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), X818 = c(0L,
0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 0L, 0L, 0L), X896 = c(0L, 0L, 0L, 0L, 0L,
1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L,
0L, 0L, 0L), X972 = c(0L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 0L,
0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L),
X1039 = c(0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L,
1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L)), .Names = c("X",
"X10", "X34", "X59", "X84", "X110", "X134", "X165", "X199", "X234",
"X257", "X362", "X433", "X506", "X581", "X652", "X733", "X818",
"X896", "X972", "X1039"), class = "data.frame", row.names = c(NA,
-23L))

上表将先前数据集中的汽车存储在第一列中,在接下来的列中,我们有两个数字:01。我想在第一个数据中添加新列,指示第一列中是否有汽车,例如 Porsche_1 (全名很重要。我的意思是 _)和汽车第二列 Astra_3 的数据集 2 的同一列中包含数字 1。这是不正确的,因此我们在此附加列中放入 0 或 NA。如果两者在同一列中都有数字 1,例如 Porsche_1Astra_1 - 数字 1 应放在附加列中。

期望的结果:最后一列的数字是随机生成的!

           X1         X2        AUC_1        AUC_2     ratio Country Comp
1 Porsche_1 Astra_3 5860133.702 4.849720e+06 1.2083447 France 0
2 Mazda_2 Astra_4 1296009.939 9.289805e+05 1.3950885 France 1
3 Tico_1 Mercedes_1 333123.493 3.205476e+05 1.0392325 Germany 1
4 Nexia_1 BMW_1 250348.941 2.232872e+05 1.1211970 Italy 0
5 Astra_1 BMW_2 1376193.334 1.340641e+06 1.0265187 Poland 0
6 Astra_2 Porsche_1 4080502.863 4.720330e+06 0.8644529 Poland 1
7 Astra_3 Mazda_2 3777603.233 4.369150e+06 0.8646082 Poland 0
8 Astra_4 Tico_1 3503973.487 3.371021e+06 1.0394398 Poland 1
9 Mercedes_1 Nexia_1 99101538.620 1.085913e+08 0.9126107 Norway 1
10 BMW_1 Audi_1 231873.846 2.664898e+05 0.8701041 Sweden 0

最佳答案

实现此目的的一种方法是使用 mapply 并检查该行中是否存在 tbl_test2 的任何 1 (第一列除外) X 中第一个汽车名称的行对于 X 中包含第二个汽车名称的行也是 1 (也就是说,检查对于 任何 列,我们是否有 TRUE & TRUE),每对都会得到 TRUEFALSE,然后您可以将其转换为 0/1 as.integer:

tbl_test1$Comp <- mapply(function(x, y) as.integer(any(unlist(tbl_test2[tbl_test2$X==x, -1]) & unlist(tbl_test2[tbl_test2$X==y, -1]))), 
x=as.character(tbl_test1$X1),
y=as.character(tbl_test1$X2))

head(tbl_test1)
# X1 X2 AUC_1 AUC_2 ratio Country Comp
#1 Porsche_1 Astra_3 5860133.702 4.849720e+06 1.2083447 France 0
#2 Mazda_2 Astra_4 1296009.939 9.289805e+05 1.3950885 France 0
#3 Tico_1 Mercedes_1 333123.493 3.205476e+05 1.0392325 Germany 1
#4 Nexia_1 BMW_1 250348.941 2.232872e+05 1.1211970 Italy 0
#5 Astra_1 BMW_2 1376193.334 1.340641e+06 1.0265187 Poland 0
#6 Astra_2 Porsche_1 4080502.863 4.720330e+06 0.8644529 Poland 0

编辑

如果你想知道匹配的列号,可以使用:

mapply(function(x, y) paste(which(unlist(tbl_test2[tbl_test2$X==x, -1]) & unlist(tbl_test2[tbl_test2$X==y, -1]))+1, collapse=","), x=as.character(tbl_test1$X1), y=as.character(tbl_test1$X2))

关于r - 基于另一个数据框的两行比较,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36616981/

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