问题

Question

为了获得两组分数的平均值的绝对偏差，我通常需要在R中编写长代码，如下所示。

问题

我想知道在BASE R中是否有可能Vectorize mad() Vectorized函数，以便与示例I中的每组分数的平均分数的绝对偏差可以使用mad()版set.seed(0) y = as.vector(unlist(mapply(FUN = rnorm, n = c(10, 10)))) # Produces two sets of scores groups = factor( rep(1:2, times = c(10, 10) ) ) # Grouping ID variable G1 = y[groups == 1] # subset y scores for group 1 G2 = y[groups == 2] # subset y scores for group 2 G1.abs.dev = abs(G1 - mean(G1)) # absolute deviation from mean scores for group 1 G2.abs.dev = abs(G2 - mean(G2)) # absolute deviation from mean scores for group 2获得以下显示？其他任何可行的想法都受到高度赞赏吗？

section .data
    qVar1: dq 1


section .bss
    var28: resb  28


section .text
    _main:

        ; Use an MMX instruction

            movq mm0, [qVar1] ; Move quadword from r/m64 to mm.

        ; Read Tag Word

            fstenv [var28]
            mov    ax, [var28 + 8] ; move the Tag Word to ax

Answer 1

怎么样

score <- lapply(split(y, groups), FUN = function (u) abs(u - mean(u)))

或

score <- ave(y, groups, FUN = function (u) abs(u - mean(u)))

结果以不同的方式组织。选择最适合你的那个。

你的措辞有问题。 mad为数据返回单个统计信息/值。例如，

sapply(split(y, groups), mad)

您没有对mad进行矢量化，而只是计算每个数据的偏差，如示例代码所示。

Answer 2

如果你把所有东西都放在data.frame中，那就更清洁了。在基地R，

set.seed(0)

df <- data.frame(y = rnorm(20),
                 group = rep(1:2, each = 10))

df$abs_dev <- with(df, ave(y, group, FUN = function(x){abs(mean(x) - x)}))

df
#>               y  group    abs_dev
#> 1   1.262954285      1 0.90403032
#> 2  -0.326233361      1 0.68515732
#> 3   1.329799263      1 0.97087530
#> 4   1.272429321      1 0.91350536
#> 5   0.414641434      1 0.05571747
#> 6  -1.539950042      1 1.89887401
#> 7  -0.928567035      1 1.28749100
#> 8  -0.294720447      1 0.65364441
#> 9  -0.005767173      1 0.36469114
#> 10  2.404653389      1 2.04572943
#> 11  0.763593461      2 1.12607477
#> 12 -0.799009249      2 0.43652794
#> 13 -1.147657009      2 0.78517570
#> 14 -0.289461574      2 0.07301974
#> 15 -0.299215118      2 0.06326619
#> 16 -0.411510833      2 0.04902952
#> 17  0.252223448      2 0.61470476
#> 18 -0.891921127      2 0.52943981
#> 19  0.435683299      2 0.79816461
#> 20 -1.237538422      2 0.87505711

或dplyr，

library(dplyr)
set.seed(0)

df <- data_frame(y = rnorm(20),
                 group = rep(1:2, each = 10))

df <- df %>% group_by(group) %>% mutate(abs_dev = abs(mean(y) - y))

df
#> # A tibble: 20 x 3
#> # Groups:   group [2]
#>               y  group    abs_dev
#>           <dbl>  <int>      <dbl>
#>  1  1.262954285      1 0.90403032
#>  2 -0.326233361      1 0.68515732
#>  3  1.329799263      1 0.97087530
#>  4  1.272429321      1 0.91350536
#>  5  0.414641434      1 0.05571747
#>  6 -1.539950042      1 1.89887401
#>  7 -0.928567035      1 1.28749100
#>  8 -0.294720447      1 0.65364441
#>  9 -0.005767173      1 0.36469114
#> 10  2.404653389      1 2.04572943
#> 11  0.763593461      2 1.12607477
#> 12 -0.799009249      2 0.43652794
#> 13 -1.147657009      2 0.78517570
#> 14 -0.289461574      2 0.07301974
#> 15 -0.299215118      2 0.06326619
#> 16 -0.411510833      2 0.04902952
#> 17  0.252223448      2 0.61470476
#> 18 -0.891921127      2 0.52943981
#> 19  0.435683299      2 0.79816461
#> 20 -1.237538422      2 0.87505711

或data.table：

library(data.table)
set.seed(0)

dt <- data.table(y = rnorm(20),
                 group = rep(1:2, each = 10))

dt[, abs_dev := abs(mean(y) - y), by = group][]
#>                y group    abs_dev
#>  1:  1.262954285     1 0.90403032
#>  2: -0.326233361     1 0.68515732
#>  3:  1.329799263     1 0.97087530
#>  4:  1.272429321     1 0.91350536
#>  5:  0.414641434     1 0.05571747
#>  6: -1.539950042     1 1.89887401
#>  7: -0.928567035     1 1.28749100
#>  8: -0.294720447     1 0.65364441
#>  9: -0.005767173     1 0.36469114
#> 10:  2.404653389     1 2.04572943
#> 11:  0.763593461     2 1.12607477
#> 12: -0.799009249     2 0.43652794
#> 13: -1.147657009     2 0.78517570
#> 14: -0.289461574     2 0.07301974
#> 15: -0.299215118     2 0.06326619
#> 16: -0.411510833     2 0.04902952
#> 17:  0.252223448     2 0.61470476
#> 18: -0.891921127     2 0.52943981
#> 19:  0.435683299     2 0.79816461
#> 20: -1.237538422     2 0.87505711

获得两组分数的绝对偏差

问题

2 个答案: