我如何找到女性的平均值/中位数（任何其他此类事物）？我尝试了一些代码来访问女性数据，但没有成功。任何帮助都非常感谢。
> jalal <- read.csv("jalal.csv", header=TRUE,sep=",")
> which(jalal$sex==F)
integer(0)
> jalal
age sex weight eye.color hair.color
1 23 F 93.8 blue black
2 21 M 180.8 amber gray
3 22 F 196.5 hazel gray
4 22 M 256.2 amber black
5 21 M 219.6 blue gray
6 16 F 152.1 blue gray
7 21 F 183.3 gray chestnut
8 18 M 179.1 brown blond
9 15 M 206.1 blue white
10 19 M 211.6 brown blond
11 20 F 209.4 blue white
12 21 M 194.0 brown auburn
13 22 F 204.1 green black
14 21 F 157.4 hazel red
15 15 F 238.0 green gray
16 20 F 154.8 gray gray
17 16 F 245.8 gray gray
18 23 M 198.2 gray red
19 19 M 169.1 green brown
20 24 M 198.0 green gray
> subset(jalal, subset=(sex =F)) -> females
> females
[1] age sex weight eye.color hair.color
<0 rows> (or 0-length row.names)
> subset(jalal, subset=(sex ==F)) -> females
> females
[1] age sex weight eye.color hair.color
<0 rows> (or 0-length row.names)
这是jalal.csv中的内容：
"age","sex","weight","eye.color","hair.color"
23,"F",93.8,"blue","black"
21,"M",180.8,"amber","gray"
22,"F",196.5,"hazel","gray"
22,"M",256.2,"amber","black"
21,"M",219.6,"blue","gray"
16,"F",152.1,"blue","gray"
21,"F",183.3,"gray","chestnut"
18,"M",179.1,"brown","blond"
15,"M",206.1,"blue","white"
19,"M",211.6,"brown","blond"
20,"F",209.4,"blue","white"
21,"M",194,"brown","auburn"
22,"F",204.1,"green","black"
21,"F",157.4,"hazel","red"
15,"F",238,"green","gray"
20,"F",154.8,"gray","gray"
16,"F",245.8,"gray","gray"
23,"M",198.2,"gray","red"
19,"M",169.1,"green","brown"
24,"M",198,"green","gray"
答案 0 :(得分：5)
您正在寻找aggregate。这是一个论坛，按性别返回中位年龄和体重：
aggregate(cbind(age, weight) ~ sex, data=jalal, FUN=median)
## sex age weight
## 1 F 20.5 189.9
## 2 M 21.0 198.1
要获取仅包含女性的数据框，以下是[的语法：
jalal[jalal$sex == 'F',]
请注意'F'周围的引号。裸F表示FALSE。这就是你的第二个subset表达式失败的原因。
subset(jalal, subset=(sex =='F'))
## age sex weight eye.color hair.color
## 1 23 F 93.8 blue black
## 3 22 F 196.5 hazel gray
## 6 16 F 152.1 blue gray
...
在评论中，要求提供蓝眼女性的平均值的方法。第一种方法是将数据框过滤到蓝眼睛的人：
aggregate(cbind(age, weight) ~ sex, data=jalal[jalal$eye.color == 'blue',], FUN=mean)
## sex age weight
## 1 F 19.66667 151.7667
## 2 M 18.00000 212.8500
但这似乎是hackish，毕竟，我们并没有过滤女性的数据框架。因此，这是一个公式，通过性别和眼睛颜色给出平均年龄和体重。从这里，你可以找到蓝眼睛的女人，绿眼睛的男人等的平均值：
aggregate(cbind(age, weight) ~ sex + eye.color, data=jalal, FUN=mean)
## sex eye.color age weight
## 1 M amber 21.50000 218.5000
## 2 F blue 19.66667 151.7667
## 3 M blue 18.00000 212.8500
## 4 M brown 19.33333 194.9000
## 5 F gray 19.00000 194.6333
## 6 M gray 23.00000 198.2000
## 7 F green 18.50000 221.0500
## 8 M green 21.50000 183.5500
## 9 F hazel 21.50000 176.9500
注意，这里的第2行和第3行与前一个表达式中的结果相匹配。
答案 1 :(得分：1)
以下是使用data.table包的替代解决方案：
require(data.table)
jalal <- as.data.table(jalal)
关于女性的子集：
jalal[sex == "F"]
计算平均值，中位数等：
> jalal[sex == "F", mean(weight)]
[1] 183.52
> jalal[sex == "F", list(mean(weight), median(age))]
V1 V2
1: 183.52 20.5
答案 2 :(得分：0)
这样您就可以看到所有主要选项，这是dplyr的解决方案：
library(dplyr)
jalal %.%
group_by(sex, eye.color) %.%
summarise(age = mean(age), weight = median(weight))