在这种情况下，我们的大数据集如下：
structure(list(Car = c("Mazda RX4", "Maserati Bora", "Leticia",
"Hornet 4 Drive", "Hornet Sportabout", "Alex", "Duster 360",
"Merc 240D", "Merc 230", "Merc 280", "Merc 280C", "Merc 450SE",
"Merc 450SL", "Merc 450SLC", "Cadillac Fleetwood", "Lincoln Continental",
"Chrysler Imperial", "Fiat 128", "Honda Civic", "Toyota Corolla",
"Toyota Corona", "Datsun 710", "AMC Javelin", "Camaro Z28",
"Datsun 710", "Fiat X1-9", "Mazda RX4", "Lotus Europa",
"Ford Pantera L", "Ferrari Dino", "Mazda RX4 Wag", "Volvo 142E"
), Name = c("Mark", "Random", "Datsun 710", "Trevor", "Joanna",
"Valiant", "Random", "Random", "Random", "Random", "Random",
"Random", "Random", "Random", "Random", "Random", "Random", "Random",
"Random", "Trevor", "Random", "Random", "Random", "Random", "Random",
"Random", "Mazda RX4", "Random", "Alex", "Random", "John", "Random"
), disp = c(160, 160, 108, 258, 360, 225, 360, 146.7, 140.8,
167.6, 167.6, 275.8, 275.8, 275.8, 472, 460, 440, 78.7, 75.7,
71.1, 120.1, 318, 304, 350, 400, 79, 120.3, 95.1, 351, 145, 301,
121), hp = c(110, 110, 93, 110, 175, 105, 245, 62, 95, 123, 123,
180, 180, 180, 205, 215, 230, 66, 52, 65, 97, 150, 150, 245,
175, 66, 91, 113, 264, 175, 335, 109), drat = c(3.9, 3.9, 3.85,
3.08, 3.15, 2.76, 3.21, 3.69, 3.92, 3.92, 3.92, 3.07, 3.07, 3.07,
2.93, 3, 3.23, 4.08, 4.93, 4.22, 3.7, 2.76, 3.15, 3.73, 3.08,
4.08, 4.43, 3.77, 4.22, 3.62, 3.54, 4.11), wt = c(2.62, 2.875,
2.32, 3.215, 3.44, 3.46, 3.57, 3.19, 3.15, 3.44, 3.44, 4.07,
3.73, 3.78, 5.25, 5.424, 5.345, 2.2, 1.615, 1.835, 2.465, 3.52,
3.435, 3.84, 3.845, 1.935, 2.14, 1.513, 3.17, 2.77, 3.57, 2.78
), qsec = c(16.46, 17.02, 18.61, 19.44, 17.02, 20.22, 15.84,
20, 22.9, 18.3, 18.9, 17.4, 17.6, 18, 17.98, 17.82, 17.42, 19.47,
18.52, 19.9, 20.01, 16.87, 17.3, 15.41, 17.05, 18.9, 16.7, 16.9,
14.5, 15.5, 14.6, 18.6), vs = c(0, 0, 1, 1, 0, 1, 0, 1, 1, 1,
1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0,
1), am = c(1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1), gear = c(4, 4,
4, 3, 3, 3, 3, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 3, 3, 3,
3, 3, 4, 5, 5, 5, 5, 5, 4), carb = c(4, 4, 1, 1, 2, 1, 4, 2,
2, 4, 4, 3, 3, 3, 4, 4, 4, 1, 2, 1, 1, 2, 2, 4, 2, 1, 2, 2, 4,
6, 8, 2)), .Names = c("Car", "Name", "disp", "hp", "drat", "wt",
"qsec", "vs", "am", "gear", "carb"), row.names = c(NA, -32L), class = "data.frame")
我想通过从中提取一些行来对此数据集进行子集化。我想要提取的行存储在另一个数据框中：
> dput(list_save)
structure(list(Car = c("Mazda RX4", "Mazda RX4 Wag", "Datsun 710",
"Hornet 4 Drive", "Hornet Sportabout", "Valiant"), Name = c("Mark",
"John", "Leticia", "Trevor", "Joanna", "Alex")), .Names = c("Car",
"Name"), class = "data.frame", row.names = c(NA, -6L))
查看list_save，因为根据df可以在不同的列中找到某些字符串，但也必须提取它。
所需的输出应该是这样的：
Car Name disp hp drat wt qsec vs am gear carb
1 Mazda RX4 Mark 160 110 3.90 2.620 16.46 0 1 4 4
2 Mazda RX4 Wag John 301 335 3.54 3.570 14.60 0 1 5 8
3 Leticia Datsun 710 108 93 3.85 2.320 18.61 1 1 4 1
4 Hornet 4 Drive Trevor 258 110 3.08 3.215 19.44 1 0 3 1
5 Hornet Sportabout Joanna 360 175 3.15 3.440 17.02 0 0 3 2
6 Alex Valiant 225 105 2.76 3.460 20.22 1 0 3 1
我认为功能类似于下面显示的功能：
test <- df[df[,1:2] %in% list_save, ]
答案 0 :(得分：2)
我只会使用data.table使用Car运行两个二进制连接，Name和library(data.table) # v 1.9.6+
res <- setDT(df)[list_save, on = c("Car", "Name")]
res2 <- df[list_save, on = c(Name = "Car", Car = "Name"), nomatch = 0L]
res[is.na(disp), (names(res)) := res2]
# Car Name disp hp drat wt qsec vs am gear carb
# 1: Mazda RX4 Mark 160 110 3.90 2.620 16.46 0 1 4 4
# 2: Mazda RX4 Wag John 301 335 3.54 3.570 14.60 0 1 5 8
# 3: Leticia Datsun 710 108 93 3.85 2.320 18.61 1 1 4 1
# 4: Hornet 4 Drive Trevor 258 110 3.08 3.215 19.44 1 0 3 1
# 5: Hornet Sportabout Joanna 360 175 3.15 3.440 17.02 0 0 3 2
# 6: Alex Valiant 225 105 2.76 3.460 20.22 1 0 3 1
对抗自己，并且一次互相攻击，并将其组合起来。我们将在CRAN上使用最新版本（v 1.9.6 +）
rbind或者，更安全的方法只是res <- setDT(df)[list_save, on = c("Car", "Name"), nomatch = 0L]
res2 <- df[list_save, on = c(Name = "Car", Car = "Name"), nomatch = 0L]
rbind(res, res2)
# Car Name disp hp drat wt qsec vs am gear carb
# 1: Mazda RX4 Mark 160 110 3.90 2.620 16.46 0 1 4 4
# 2: Mazda RX4 Wag John 301 335 3.54 3.570 14.60 0 1 5 8
# 3: Hornet 4 Drive Trevor 258 110 3.08 3.215 19.44 1 0 3 1
# 4: Hornet Sportabout Joanna 360 175 3.15 3.440 17.02 0 0 3 2
# 5: Leticia Datsun 710 108 93 3.85 2.320 18.61 1 1 4 1
# 6: Alex Valiant 225 105 2.76 3.460 20.22 1 0 3 1
只有匹配的结果，但这样你就会丢失原始的行顺序
{{1}}
答案 1 :(得分：1)
sub_df <- df[which(df[,1] %in% list_save[,1] & df[,2] %in% list_save[,2]),]
但是，您的意思是Alex中的Car和Valiant中的Name？我只是问，因为上面假设那些是错误的。如果不是这样的话：
<强> EDITED
sub_df <- df[which(df[,1] %in% list_save[,1] & df[,2] %in% list_save[,2] |
df[,1] %in% list_save[,2] & df[,2] %in% list_save[,1]),]