在一个数据帧中查找字符串并在第二个数据帧中替换它答案

【问题标题】：Find string in one data frame and replace it in second data frame在一个数据帧中查找字符串并在第二个数据帧中替换它
【发布时间】：2015-07-02 21:46:43
【问题描述】：

一般我有两个数据框：

主要的：

structure(list(mpg = c(21, 21, 22.8, 21.4, 18.7, 18.1, 14.3, 
24.4, 22.8, 19.2, 17.8, 16.4, 17.3, 15.2, 10.4, 10.4, 14.7, 32.4, 
30.4, 33.9, 21.5, 15.5, 15.2, 13.3, 19.2, 27.3, 26, 30.4, 15.8, 
19.7, 15, 21.4), cyl = c(6, 6, 4, 6, 8, 6, 8, 4, 4, 6, 6, 8, 
8, 8, 8, 8, 8, 4, 4, 4, 4, 8, 8, 8, 8, 4, 4, 4, 8, 6, 8, 4), 
    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("a", "b", "c", 
    "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", 
    "p", "r", "s", "t", "u", "w", "z", "a1", "b1", "c1", "d1", 
    "e1", "f1", "g1", "h1", "i1"), carb = c("Mazda RX4", "Mazda RX4 Wag", 
    "Datsun 710", "Hornet 4 Drive", "Hornet Sportabout", "Valiant", 
    "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", "Dodge Challenger", "AMC Javelin", 
    "Camaro Z28", "Pontiac Firebird", "Fiat X1-9", "Porsche 914-2", 
    "Lotus Europa", "Ford Pantera L", "Ferrari Dino", "Maserati Bora", 
    "Volvo 142E")), .Names = c("mpg", "cyl", "disp", "hp", "drat", 
"wt", "qsec", "vs", "am", "gear", "carb"), row.names = c(NA, 
-32L), class = "data.frame")

表格名称：

    structure(list(car = structure(c(18L, 18L, 19L, 19L, 5L, 13L, 
14L, 31L, 7L, 21L, 20L, 22L, 23L, 24L, 25L, 26L, 2L, 15L, 4L, 
9L, 12L, 29L, 30L, 6L, 1L, 3L, 27L, 10L, 28L, 16L, 11L, 11L, 
8L, 17L, 32L, 32L), .Label = c("AMC Javelin", "Cadillac Fleetwood", 
"Camaro Z28", "Chrysler Imperial", "Datsun 710", "Dodge Challenger", 
"Duster 360", "Ferrari Dino", "Fiat 128", "Fiat X1-9", "Ford Pantera L", 
"Honda Civic", "Hornet 4 Drive", "Hornet Sportabout", "Lincoln Continental", 
"Lotus Europa", "Maserati Bora", "Mazda RX4", "Mazda RX4 Wag", 
"Merc 230", "Merc 240D", "Merc 280", "Merc 280C", "Merc 450SE", 
"Merc 450SL", "Merc 450SLC", "Pontiac Firebird", "Porsche 914-2", 
"Toyota Corolla", "Toyota Corona", "Valiant", "Volvo 142E"), class = "factor"), 
    owner = structure(c(15L, 25L, 9L, 8L, 4L, 7L, 31L, 22L, 17L, 
    6L, 12L, 32L, 12L, 19L, 1L, 3L, 11L, 20L, 26L, 27L, 24L, 
    29L, 16L, 1L, 2L, 13L, 9L, 26L, 23L, 10L, 18L, 14L, 30L, 
    28L, 5L, 21L), .Label = c("Amid", "Armin", "Crane", "Dietmar", 
    "Gared", "Gratea", "Hank", "Hannea", "Hans", "Heta", "Horse_with_no_name", 
    "Jeff", "Krea", "Marea", "Mark", "Mattheus", "Micha", "Miko", 
    "Myrcella", "Neil", "Nina", "Peter", "Rene", "Robert", "Steffan", 
    "Tim", "Timon", "Timothy", "Uwe", "Vincent", "Wolfram", "Yena"
    ), class = "factor")), .Names = c("car", "owner"), row.names = c(NA, 
-36L), class = "data.frame")

所以我想在我的主数据框中添加一个名为owner 的附加列。您可以在第二个数据中找到每辆车的owners。

但是：

您已经注意到，有些汽车可能有不同的车主。我不想在我的主数据框中创建额外的行，所以我想将两个/三个或更多所有者放在同一行中，但例如用逗号分隔。

【问题讨论】：

你试过merge、match等吗？
您的数据存在缺陷：在主数据框的第 10 列中，缺少一个值。

标签： r

【解决方案1】：

试试

library(data.table)
res <- setDT(df1, key='car')[df2][,c(.SD[1L],
           list(ownerN=toString(owner))), car][,owner:=NULL]
head(res,2)
#             car mpg cyl disp  hp drat    wt  qsec vs am gear carb
#1:     Mazda RX4  21   6  160 110  3.9 2.620 16.46  0  1    4    4
#2: Mazda RX4 Wag  21   6  160 110  3.9 2.875 17.02  0  1    4    4
#          ownerN
#1: Mark, Steffan
#2:  Hans, Hannea

或者paste '所有者' by '汽车' 在'df2' 并加入'df1'

 dt2 <- setDT(df2)[, .(owner=toString(owner)), by= car]
 setkey(setDT(df1), car)[dt2]

数据

df1 <- cbind(mtcars, car=row.names(mtcars))
row.names(df1) <- NULL
df2[] <- lapply(df2, as.character)

【讨论】：

我收到此错误，但我不知道为什么，因为它看起来非常适合您：Error in setDT(df1, key = "car") : unused argument (key = "car")
@ShaxiLiver 您创建的df1 有缺陷（来自ColonelBeauvel 的cmets）。假设您使用与我创建的数据集相同的数据集，key='car' 是开发版本中的一个新功能。没有开发版的可以试试setkey(setDT(df1), car)。

【解决方案2】：

我会让你的第二个数据框成为一个数据表（我称之为 dt2），然后你可以这样做：

dt2 <- dt2[ , list( 'owner' = toString(owner)) , by = car ]

然后将其加入到您的第一个数据表中。

编辑：我确实喜欢@akrun 添加的“toString”

【讨论】：

启动代码后打印出这个错误：Error in [.data.frame(dt2, , list(owner = list(owner)), by = car) : unused argument (by = car)
我认为你需要把它做成一个数据表。所以加载 data.table 库，然后用 data.table() 包装数据来创建数据，例如：dt2

【解决方案3】：

我不知道你为什么要这样做，但如果你想在同一行重新组合所有者并用逗号分隔它们：

library(dplyr)
df2 %>% 
  group_by(car) %>% 
  summarise(owner = paste(owner, collapse = ", ")) %>% 
  inner_join(df1, by = c("car" = "carb"))

# Which gives for example :
        car         owner mpg cyl disp  hp drat   wt  qsec vs am gear
1 Mazda RX4 Mark, Steffan  21   6  160 110  3.9 2.62 16.46  0  1    4

【讨论】：

看来我在使用 RStudio 时遇到了严重问题：Error: cannot join on columns 'car' x 'carb': Can't join on 'car' x 'carb' because of incompatible types (character / numeric) 。与上述类似的问题...
您创建的“主要”数据框有缺陷，您必须加入正确的列。将 car 和 carb 替换为要进行连接的列，如果它们具有相同的名称，则仅使用一个字符串作为示例：by = "car"