要为三个分类列创建交叉表，我们可以使用 xtabs 函数。xtabs 函数将为两列中的每个类别创建列联表，并将为第三列中的类别创建列联表。
查看以下示例以了解如何完成。
以下代码段创建了一个示例数据框 -
df1<-data.frame(Status=c("Sold", "Available", "Available", "Sold", "Sold",
"In Queue", "In Queue", "Available", "Sold", "In Queue"), Gender = c("Female",
"Male", "Male", "Female", "Female", "Female", "Male", "Female", "Female",
"Female"), Confirm=c("No", "Yes", "No", "Yes", "No", "Yes", "Yes", "Yes",
"Yes", "No"))
df1创建了以下数据框
Status Gender Confirm 1 Sold Female No 2 Available Male Yes 3 Available Male No 4 Sold Female Yes 5 Sold Female No 6 In Queue Female Yes 7 In Queue Male Yes 8 Available Female Yes 9 Sold Female Yes 10 In Queue Female No
要在上面创建的数据框中为 df1 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df1<-data.frame(Status=c("Sold", "Available", "Available", "Sold", "Sold",
"In Queue", "In Queue", "Available", "Sold", "In Queue"), Gender = c("Female",
"Male", "Male", "Female", "Female", "Female", "Male", "Female", "Female",
"Female"), Confirm=c("No", "Yes", "No", "Yes", "No", "Yes", "Yes", "Yes",
"Yes", "No"))
xtabs(~Confirm+Gender+Status,data=df1)
, , Status = Available输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Gender Confirm Female Male No 0 1 Yes 1 1
要在上面创建的数据框中为 df1 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df1<-data.frame(Status=c("Sold", "Available", "Available", "Sold", "Sold",
"In Queue", "In Queue", "Available", "Sold", "In Queue"), Gender = c("Female",
"Male", "Male", "Female", "Female", "Female", "Male", "Female", "Female",
"Female"), Confirm=c("No", "Yes", "No", "Yes", "No", "Yes", "Yes", "Yes",
"Yes", "No"))
xtabs(~Confirm+Gender+Status,data=df1)
, , Status = In Queue输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Gender Confirm Female Male No 1 0 Yes 1 1
要在上面创建的数据框中为 df1 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df1<-data.frame(Status=c("Sold", "Available", "Available", "Sold", "Sold",
"In Queue", "In Queue", "Available", "Sold", "In Queue"), Gender = c("Female",
"Male", "Male", "Female", "Female", "Female", "Male", "Female", "Female",
"Female"), Confirm=c("No", "Yes", "No", "Yes", "No", "Yes", "Yes", "Yes",
"Yes", "No"))
xtabs(~Confirm+Gender+Status,data=df1)
, , Status = Sold输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Gender Confirm Female Male No 2 0 Yes 2 0
以下代码段创建了一个示例数据框 -
df2<-
data.frame(Class=sample(c("I","II","III","IV"),20,replace=TRUE),Group=sample(c(
"G1","G2","G3"),20,replace=TRUE),Rank=sample(1:5,20,replace=TRUE))
df2创建了以下数据框
Class Group Rank 1 I G2 3 2 III G2 2 3 IV G2 3 4 I G3 4 5 I G1 3 6 IV G1 1 7 IV G2 3 8 II G1 1 9 III G1 1 10 I G2 2 11 IV G1 1 12 I G2 1 13 IV G3 1 14 I G2 1 15 II G2 3 16 III G1 4 17 I G2 2 18 IV G2 4 19 I G2 1 20 I G1 1
要在上面创建的数据框中为 df2 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df2<-
data.frame(Class=sample(c("I","II","III","IV"),20,replace=TRUE),Group=sample(c(
"G1","G2","G3"),20,replace=TRUE),Rank=sample(1:5,20,replace=TRUE))
xtabs(~Rank+Group+Class,data=df2)
, , Class = I输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Group Rank G1 G2 G3 1 1 3 0 2 0 2 0 3 1 1 0 4 0 0 1
要在上面创建的数据框中为 df2 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df2<-
data.frame(Class=sample(c("I","II","III","IV"),20,replace=TRUE),Group=sample(c(
"G1","G2","G3"),20,replace=TRUE),Rank=sample(1:5,20,replace=TRUE))
xtabs(~Rank+Group+Class,data=df2)
, , Class = II输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Group Rank G1 G2 G3 1 1 0 0 2 0 0 0 3 0 1 0 4 0 0 0
要在上面创建的数据框中为 df2 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df2<-
data.frame(Class=sample(c("I","II","III","IV"),20,replace=TRUE),Group=sample(c(
"G1","G2","G3"),20,replace=TRUE),Rank=sample(1:5,20,replace=TRUE))
xtabs(~Rank+Group+Class,data=df2)
, , Class = III输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Group Rank G1 G2 G3 1 1 0 0 2 0 1 0 3 0 0 0 4 1 0 0
要在上面创建的数据框中为 df2 中的数据创建交叉表，请将以下代码添加到上面的代码段中 -
df2<-
data.frame(Class=sample(c("I","II","III","IV"),20,replace=TRUE),Group=sample(c(
"G1","G2","G3"),20,replace=TRUE),Rank=sample(1:5,20,replace=TRUE))
xtabs(~Rank+Group+Class,data=df2)
, , Class = IV输出结果如果您将上述所有给定的片段作为单个程序执行，它会生成以下输出 -
Group Rank G1 G2 G3 1 2 0 1 2 0 0 0 3 0 2 0 4 0 1 0