R 获取有条件的画家的姓名
来自图书馆的弥撒。它有画师的数据集R 获取有条件的画家的姓名,r,dataset,R,Dataset,来自图书馆的弥撒。它有画师的数据集 > painters Composition Drawing Colour Expression School Da Udine 10 8 16 3 A Da Vinci 15 16 4 14 A Del Piombo 8 13
> painters
Composition Drawing Colour Expression School
Da Udine 10 8 16 3 A
Da Vinci 15 16 4 14 A
Del Piombo 8 13 16 7 A
Del Sarto 12 16 9 8 A
Fr. Penni 0 15 8 0 A
Guilio Romano 15 16 4 14 A
Michelangelo 8 17 4 8 A
Perino del Vaga 15 16 7 6 A
Perugino 4 12 10 4 A
如何获得构图=10的画家的名字
我这样做:
painters$Composition[painters$Composition==10]
[1] 10 10 10 10 10 10
或
其中:
> rownames(painters[painters$Composition == 10 ,])
[1] "Da Udine" "F. Zucarro" "Parmigiano" "Josepin"
[5] "J. Jordaens" "Bourdon"
说明:
你就快到了。首先,您需要按照所需的标准标识行的索引:
painters$Composition == 10
> painters$Composition == 10
[1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE
[12] FALSE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
[23] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[34] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[45] TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
接下来,我们使用这些索引对数据帧进行子集划分:
> painters[painters$Composition == 10 ,]
Composition Drawing Colour Expression School
Da Udine 10 8 16 3 A
F. Zucarro 10 13 8 8 B
Parmigiano 10 15 6 6 B
Josepin 10 10 6 2 C
J. Jordaens 10 8 16 6 G
Bourdon 10 8 8 4 H
最后,我们得到名称。它怎么能只返回名称?@RobCal.3 rownamespainters返回名称
painters$Composition == 10
> painters$Composition == 10
[1] TRUE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE
[12] FALSE TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
[23] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[34] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[45] TRUE FALSE FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE
> painters[painters$Composition == 10 ,]
Composition Drawing Colour Expression School
Da Udine 10 8 16 3 A
F. Zucarro 10 13 8 8 B
Parmigiano 10 15 6 6 B
Josepin 10 10 6 2 C
J. Jordaens 10 8 16 6 G
Bourdon 10 8 8 4 H