library(plyr)
disp_per_mpg <- splat(function(disp, mpg, ...) disp/mpg)
vehicle <- function(df) rownames(df)
df = quickdf(each(vehicle, disp_per_mpg)(mtcars))
arrange(df, disp_per_mpg)
## vehicle disp_per_mpg
## 1 Toyota Corolla 2.097
## 2 Fiat 128 2.429
## 3 Honda Civic 2.490
## 4 Fiat X1-9 2.894
## 5 Lotus Europa 3.128
## 6 Porsche 914-2 4.627
## 7 Datsun 710 4.737
## 8 Toyota Corona 5.586
## 9 Volvo 142E 5.654
## 10 Merc 240D 6.012
## 11 Merc 230 6.175
## 12 Ferrari Dino 7.360
## 13 Mazda RX4 7.619
## 14 Mazda RX4 Wag 7.619
## 15 Merc 280 8.729
## 16 Merc 280C 9.416
## 17 Hornet 4 Drive 12.056
## 18 Valiant 12.431
## 19 Merc 450SL 15.942
## 20 Merc 450SE 16.817
## 21 Merc 450SLC 18.145
## 22 Hornet Sportabout 19.251
## 23 AMC Javelin 20.000
## 24 Maserati Bora 20.067
## 25 Dodge Challenger 20.516
## 26 Pontiac Firebird 20.833
## 27 Ford Pantera L 22.215
## 28 Duster 360 25.175
## 29 Camaro Z28 26.316
## 30 Chrysler Imperial 29.932
## 31 Lincoln Continental 44.231
## 32 Cadillac Fleetwood 45.385
each(max, min, mean, var)(mtcars$disp)
## max min mean var
## 472.0 71.1 230.7 15360.8
rename(mtcars, replace = c(hp = "Ghp"))
## mpg cyl disp Ghp drat wt qsec vs am gear carb
## Mazda RX4 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
## Mazda RX4 Wag 21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4
## Datsun 710 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1
## Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
## Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
## Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
## Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
## Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
## Merc 230 22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2
## Merc 280 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4
## Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
## Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
## Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
## Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
## Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
## Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
## Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4
## Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
## Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
## Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
## Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
## Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
## AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
## Camaro Z28 13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4
## Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
## Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
## Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
## Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
## Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
## Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
## Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
## Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
# rename(mtcars,replace=c(hp='Ghp')) #旧变量名不需要引号
count(mtcars, "am")
## am freq
## 1 0 19
## 2 1 13
(param = data.frame(n = rep(5, 5), min = c(2, 1, 1, 5, 8), max = c(4, 2, 5,
7, 9)))
## n min max
## 1 5 2 4
## 2 5 1 2
## 3 5 1 5
## 4 5 5 7
## 5 5 8 9
mlply(param, runif)
## $`1`
## [1] 3.699 3.644 2.082 3.129 3.108
##
## $`2`
## [1] 1.147 1.898 1.406 1.414 1.156
##
## $`3`
## [1] 4.705 4.649 3.835 3.338 1.090
##
## $`4`
## [1] 6.494 6.392 6.339 5.582 6.145
##
## $`5`
## [1] 8.652 8.798 8.602 8.297 8.035
##
## attr(,"split_type")
## [1] "array"
## attr(,"split_labels")
## n min max
## 1 5 2 4
## 2 5 1 2
## 3 5 1 5
## 4 5 5 7
## 5 5 8 9
alply(param, 1, splat(runif))
## $`1`
## [1] 2.628 3.472 3.303 3.955 2.990
##
## $`2`
## [1] 1.468 1.598 1.481 1.394 1.438
##
## $`3`
## [1] 3.983 4.304 4.929 3.476 4.134
##
## $`4`
## [1] 5.205 5.786 6.951 5.606 5.352
##
## $`5`
## [1] 8.613 8.752 8.571 8.156 8.796
##
## attr(,"split_type")
## [1] "array"
## attr(,"split_labels")
## n min max
## 1 5 2 4
## 2 5 1 2
## 3 5 1 5
## 4 5 5 7
## 5 5 8 9
(x <- data.frame(k1 = c(NA, NA, 3, 4, 5), k2 = c(1, NA, NA, 4, 5), data = 1:5))
## k1 k2 data
## 1 NA 1 1
## 2 NA NA 2
## 3 3 NA 3
## 4 4 4 4
## 5 5 5 5
(y <- data.frame(k1 = c(NA, 2, NA, 4, 5), k2 = c(NA, NA, 3, 4, 5), data = 1:5))
## k1 k2 data
## 1 NA NA 1
## 2 2 NA 2
## 3 NA 3 3
## 4 4 4 4
## 5 5 5 5
如果参数match=“all"(默认),将保留所有y$k1与x$k1相匹配的组合。
如果参数match="first",将只保留第一个y$k1与x$k1相匹配的组合。
join(x, y, by = "k1")
## k1 k2 data k2 data
## 1 NA 1 1 NA 1
## 2 NA 1 1 3 3
## 3 NA NA 2 NA 1
## 4 NA NA 2 3 3
## 5 3 NA 3 NA NA
## 6 4 4 4 4 4
## 7 5 5 5 5 5
join(x, y, by = "k1", match = "first")
## k1 k2 data k2 data
## 1 NA 1 1 NA 1
## 2 NA NA 2 NA 1
## 3 3 NA 3 NA NA
## 4 4 4 4 4 4
## 5 5 5 5 5 5
match参数的含义同上。
join(x, y, "k1", type = "right")
## k1 k2 data k2 data
## 1 NA 1 1 NA 1
## 2 NA NA 2 NA 1
## 3 2 NA NA NA 2
## 4 NA 1 1 3 3
## 5 NA NA 2 3 3
## 6 4 4 4 4 4
## 7 5 5 5 5 5
这句会报错,原因不详,可参考下面的讨论:
join(x, y, by = "k1", type = "right", match = "first")
## Error: Duplicated key in y
用match="all"就没这个问题,因为所有的行每次都会遍历一边。暂时不能理解的是,用左连接也完全不会出现这个问题,无论match参数是什么。
# 注意这里是大写的X和Y,与题目里面小写的x和y互不影响
X = data.frame(k = c(2, 1, 2), v1 = 1:3)
Y = data.frame(k = rep(2, 3), v2 = 1:3)
join(X, Y, by = "k", type = "right", match = "first") #报错
## Error: Duplicated key in y
join(Y, X, by = "k", type = "right", match = "first") #报错
## Error: Duplicated key in y
# 如果使用key值不重复的Y1,X右连接Y1不报错了,但Y1右连接X仍然报错
Y1 = Y
Y1[, 1] = 1:3
Y1
## k v2
## 1 1 1
## 2 2 2
## 3 3 3
join(X, Y1, by = "k", type = "right", match = "first") #不报错了
## k v1 v2
## 1 1 2 1
## 2 2 1 2
## 3 3 NA 3
join(Y1, X, by = "k", type = "right", match = "first") #仍然报错
## Error: Duplicated key in y
# 用match='all'或type='left'参数时都完全正常
join(X, Y, by = "k", type = "right", match = "all") #正常
## k v1 v2
## 1 2 1 1
## 2 2 3 1
## 3 2 1 2
## 4 2 3 2
## 5 2 1 3
## 6 2 3 3
join(Y, X, by = "k", type = "right", match = "all") #正常
## k v2 v1
## 1 2 1 1
## 2 2 2 1
## 3 2 3 1
## 4 1 NA 2
## 5 2 1 3
## 6 2 2 3
## 7 2 3 3
join(X, Y, by = "k", type = "left", match = "first") #正常
## k v1 v2
## 1 2 1 1
## 2 1 2 NA
## 3 2 3 1
join(Y, X, by = "k", type = "left", match = "first") #正常
## k v2 v1
## 1 2 1 1
## 2 2 2 1
## 3 2 3 1
# 此问题也不是因为X的key乱序排列引起的:
(X1 = arrange(X, k))
## k v1
## 1 1 2
## 2 2 1
## 3 2 3
join(X1, Y, by = "k", type = "right", match = "first") #仍然报错
## Error: Duplicated key in y
join(Y, X1, by = "k", type = "right", match = "first") #仍然报错
## Error: Duplicated key in y
join(X1, Y, by = "k", type = "right", match = "all") #正常
## k v1 v2
## 1 2 1 1
## 2 2 3 1
## 3 2 1 2
## 4 2 3 2
## 5 2 1 3
## 6 2 3 3
join(Y, X1, by = "k", type = "right", match = "all") #正常
## k v2 v1
## 1 1 NA 2
## 2 2 1 1
## 3 2 2 1
## 4 2 3 1
## 5 2 1 3
## 6 2 2 3
## 7 2 3 3
结论是:在右连接中,最好不要使用重复的key值,否则会出现莫名其妙的错误。
join(x, y, "k1", type = "inner")
## k1 k2 data k2 data
## 1 NA 1 1 NA 1
## 2 NA 1 1 3 3
## 3 NA NA 2 NA 1
## 4 NA NA 2 3 3
## 5 4 4 4 4 4
## 6 5 5 5 5 5
join(x, y, by = "k1", type = "inner", match = "first")
## k1 k2 data k2 data
## 1 NA 1 1 NA 1
## 1.1 NA 1 1 NA 1
## 4 4 4 4 4 4
## 5 5 5 5 5 5
match参数的含义同上。
join(x, y, "k1", type = "full")
## k1 k2 data
## 1 NA 1 1
## 2 NA 1 1
## 3 NA NA 2
## 4 NA NA 2
## 5 3 NA 3
## 6 4 4 4
## 7 5 5 5
## 8 2 NA 2
join(x, y, "k1", type = "full", match = "first")
## k1 k2 data
## 1 NA 1 1
## 2 NA NA 2
## 3 3 NA 3
## 4 4 4 4
## 5 5 5 5
## 6 2 NA 2