Question

我正在尝试计算数据框中某些列的平均值和标准差，并将这些值返回到数据框中的新列。我可以让这个工作的意思是：

library(dplyr)
mtcars = mutate(mtcars, mean=(hp+drat+wt)/3)

然而，当我尝试对标准偏差做同样的事情时，我有一个问题，因为我不能像我的意思那样硬编码等式非常容易。所以，我尝试使用一个函数，如下所示：

mtcars = mutate(mtcars, mean=(hp+drat+wt)/3, stdev = sd(hp,drat,wt))

导致错误“sd（hp，drat，wt）中的错误：未使用的参数（wt）”。我该如何纠正我的语法？谢谢。

Answer 1

你可以尝试

library(dplyr)
library(matrixStats)
nm1 <- c('hp', 'drat', 'wt')
res1 <- mtcars %>% 
           mutate(Mean= rowMeans(.[nm1]), stdev=rowSds(as.matrix(.[nm1])))

head(res1,3)
#   mpg cyl disp  hp drat    wt  qsec vs am gear carb     Mean    stdev
#1 21.0   6  160 110 3.90 2.620 16.46  0  1    4    4 38.84000 61.62969
#2 21.0   6  160 110 3.90 2.875 17.02  0  1    4    4 38.92500 61.55489
#3 22.8   4  108  93 3.85 2.320 18.61  1  1    4    1 33.05667 51.91809

或使用do

res2 <- mtcars %>% 
             rowwise() %>%
             do(data.frame(., Mean=mean(unlist(.[nm1])),
                         stdev=sd(unlist(.[nm1]))))

head(res2,3)
#   mpg cyl disp  hp drat    wt  qsec vs am gear carb     Mean    stdev
#1 21.0   6  160 110 3.90 2.620 16.46  0  1    4    4 38.84000 61.62969
#2 21.0   6  160 110 3.90 2.875 17.02  0  1    4    4 38.92500 61.55489
#3 22.8   4  108  93 3.85 2.320 18.61  1  1    4    1 33.05667 51.91809

Answer 2

您也可以在

中编写自己的矢量化RowSD函数

RowSD <- function(x) {
  sqrt(rowSums((x - rowMeans(x))^2)/(dim(x)[2] - 1))
}

然后

mtcars %>% 
  mutate(mean = (hp + drat + wt)/3, stdev = RowSD(cbind(hp, drat, wt)))
##     mpg cyl  disp  hp drat    wt  qsec vs am gear carb      mean     stdev
## 1  21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4  38.84000  61.62969
## 2  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4  38.92500  61.55489
## 3  22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1  33.05667  51.91809
## 4  21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1  38.76500  61.69136
## 5  18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2  60.53000  99.13403
## 6  18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1  37.07333  58.82726
## ...

Answer 3

不需要进行太多更改，只需添加rowwise()（感谢@akrun的评论）并将您的列名包装在c(...)中（以修复错误）：

library(dplyr)
mtcars %>%
    rowwise() %>%
    mutate(mean=(hp+drat+wt)/3, stdev = sd(c(hp,drat,wt)))
## Source: local data frame [32 x 13]
## Groups: <by row>
##     mpg cyl  disp  hp drat    wt  qsec vs am gear carb     mean     stdev
## 1  21.0   6 160.0 110 3.90 2.620 16.46  0  1    4    4 38.84000  61.62969
## 2  21.0   6 160.0 110 3.90 2.875 17.02  0  1    4    4 38.92500  61.55489
## 3  22.8   4 108.0  93 3.85 2.320 18.61  1  1    4    1 33.05667  51.91809
## 4  21.4   6 258.0 110 3.08 3.215 19.44  1  0    3    1 38.76500  61.69136
## 5  18.7   8 360.0 175 3.15 3.440 17.02  0  0    3    2 60.53000  99.13403
## 6  18.1   6 225.0 105 2.76 3.460 20.22  1  0    3    1 37.07333  58.82726
## 7  14.3   8 360.0 245 3.21 3.570 15.84  0  0    3    4 83.92667 139.49371
## 8  24.4   4 146.7  62 3.69 3.190 20.00  1  0    4    2 22.96000  33.81056
## 9  22.8   4 140.8  95 3.92 3.150 22.90  1  0    4    2 34.02333  52.80875
## 10 19.2   6 167.6 123 3.92 3.440 18.30  1  0    4    4 43.45333  68.88985
## ..  ... ...   ... ...  ...   ...   ... .. ..  ...  ...      ...       ...

Answer 4

@ r2evans使用相同的命令给我相同的sd值。意思是工作正常。见下面的输出

> mtcars %>%
+         rowwise() %>%
+         mutate(mean=(hp+drat+wt)/3, stdev = sd(c(hp,drat,wt)))
Source: local data frame [32 x 13]
Groups: <by row>

# A tibble: 32 x 13
     mpg   cyl  disp    hp  drat    wt  qsec    vs    am  gear  carb     mean    stdev
 * <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>    <dbl>    <dbl>
 1  21.0     6 160.0   110  3.90 2.620 16.46     0     1     4     4 38.84000 78.38681
 2  21.0     6 160.0   110  3.90 2.875 17.02     0     1     4     4 38.92500 78.38681
 3  22.8     4 108.0    93  3.85 2.320 18.61     1     1     4     1 33.05667 78.38681
 4  21.4     6 258.0   110  3.08 3.215 19.44     1     0     3     1 38.76500 78.38681
 5  18.7     8 360.0   175  3.15 3.440 17.02     0     0     3     2 60.53000 78.38681
 6  18.1     6 225.0   105  2.76 3.460 20.22     1     0     3     1 37.07333 78.38681
 7  14.3     8 360.0   245  3.21 3.570 15.84     0     0     3     4 83.92667 78.38681
 8  24.4     4 146.7    62  3.69 3.190 20.00     1     0     4     2 22.96000 78.38681
 9  22.8     4 140.8    95  3.92 3.150 22.90     1     0     4     2 34.02333 78.38681
10  19.2     6 167.6   123  3.92 3.440 18.30     1     0     4     4 43.45333 78.38681
# ... with 22 more rows

R Dplyr变异，计算每行的标准偏差

4 个答案: