具有“相同”名称的行的平均值

时间：2014-05-21 09:23:14

标签： r

我想计算具有相同名称的行的平均值，并在数据框中只留下1（平均值）行。

让我们举个例子：

> mtcars  ## I edited this data by putting a dot and the number
                       mpg cyl  disp  hp drat    wt  qsec vs am gear carb
Mazda RX4.1           21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4.2           21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
Datsun 710.1          22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1
Hornet 4 Drive.2      21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1
Hornet Sportabout.2   18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2
Valiant.3             18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1
Duster 360.1          14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4
Merc 240D.3           24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2
Merc 230.2            22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
Merc 230.4            19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4
Merc 280C.2           17.8   6 167.6 123 3.92 3.440 18.90  1  0    4    4
Merc 450SE.1          16.4   8 275.8 180 3.07 4.070 17.40  0  0    3    3
Merc 450SL.1          17.3   8 275.8 180 3.07 3.730 17.60  0  0    3    3
Merc 450SLC.1         15.2   8 275.8 180 3.07 3.780 18.00  0  0    3    3
Cadillac Fleetwood.1  10.4   8 472.0 205 2.93 5.250 17.98  0  0    3    4
Lincoln Continental.2 10.4   8 460.0 215 3.00 5.424 17.82  0  0    3    4
Chrysler Imperial.1   14.7   8 440.0 230 3.23 5.345 17.42  0  0    3    4
Fiat 128.1            32.4   4  78.7  66 4.08 2.200 19.47  1  1    4    1
Honda Civic.2         30.4   4  75.7  52 4.93 1.615 18.52  1  1    4    2
Toyota Corolla.3      33.9   4  71.1  65 4.22 1.835 19.90  1  1    4    1
Toyota Corona.1       21.5   4 120.1  97 3.70 2.465 20.01  1  0    3    1
Dodge Challenger.3    15.5   8 318.0 150 2.76 3.520 16.87  0  0    3    2
AMC Javelin.1         15.2   8 304.0 150 3.15 3.435 17.30  0  0    3    2
Camaro Z28.1          13.3   8 350.0 245 3.73 3.840 15.41  0  0    3    4
Pontiac Firebird.2    19.2   8 400.0 175 3.08 3.845 17.05  0  0    3    2
Fiat X1-9.1           27.3   4  79.0  66 4.08 1.935 18.90  1  1    4    1
Porsche 914-2.3       26.0   4 120.3  91 4.43 2.140 16.70  0  1    5    2
Lotus Europa.4        30.4   4  95.1 113 3.77 1.513 16.90  1  1    5    2
Ford Pantera L.1      15.8   8 351.0 264 4.22 3.170 14.50  0  1    5    4
Ferrari Dino.1        19.7   6 145.0 175 3.62 2.770 15.50  0  1    5    6
Maserati Bora.3       15.0   8 301.0 335 3.54 3.570 14.60  0  1    5    8
Volvo 142E.1          21.4   4 121.0 109 4.11 2.780 18.60  1  1    4    2

所以对我来说重要的一行是：

                       mpg cyl  disp  hp drat    wt  qsec vs am gear carb
Mazda RX4.1           21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4
Mazda RX4.2           21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4
Merc 230.2            22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2
Merc 230.4            19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4

我想取一列（一个接一个），只留下一个马自达RX4和一个Merc 230的数据。

所以我想通过特定模式Name.Number来计算平均值。对我来说唯一有趣的部分是dot之前的名字。所有相同的名字（在点之前）应该被平均..

编辑：

> dput(mtcars2)
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(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("mpg", "cyl", "disp", "hp", 
"drat", "wt", "qsec", "vs", "am", "gear", "carb"), row.names = c("Mazda RX4.1", 
"Mazda RX4.2", "Datsun 710.3", "Hornet 4 Drive.1", "Hornet Sportabout.2", 
"Valiant.2", "Duster 360.3", "Merc 240D.1", "Merc 230.2", "Merc 230.4", 
"Merc 280C.1", "Merc 450SE.2", "Merc 450SL.3", "Merc 450SLC.2", 
"Cadillac Fleetwood.1", "Lincoln Continental.2", "Chrysler Imperial.3", 
"Fiat 128.1", "Honda Civic.2", "Toyota Corolla.2", "Toyota Corona.3", 
"Dodge Challenger.3", "AMC Javelin.4", "Camaro Z28.1", "Pontiac Firebird.3", 
"Fiat X1-9.1", "Porsche 914-2.2", "Lotus Europa.1", "Ford Pantera L.2", 
"Ferrari Dino.3", "Maserati Bora.2", "Volvo 142E.1"), class = "data.frame")

2 个答案:

答案 0 :(得分：3)

使用data.table 

library(data.table)
mtcars$id <- gsub("\\..*", "", row.names(mtcars))
setDT(mtcars)[, lapply(.SD, mean), by = id]

答案 1 :(得分：1)

对于aggregate()函数来说，这应该是一个简单的任务，你只需要将行名称分解成一个你对使用strsplit感兴趣的部分，所以代码应该是这样的： / p> 

car_types <- sapply(strsplit(rownames(mtcars), "\\."), "[", 1)
aggregate(mtcars,by=list(car_types),FUN=mean)
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