1

我有一个这样的测试数据集： 

table(spamday)  
FR MO SA SU TH TU WE  
30 96  9  9 45 60 51

结构： 

str(spamday)  
'data.frame':   300 obs. of  1 variable:
$ SPAMreceived: Factor w/ 7 levels "FR","MO","SA",..: 2 5 5 5 6 2 6 2 2 6 ...

我的目标是使用该transform函数使分类变量成为有序因子，并具有levels指令强加的顺序

所以我运行以下代码：

spamday <- transform(spamday, SPAMreceived <- factor(SPAMreceived, levels = c("MO", "TU", "WE", "TH", "FR", "SA", "SU"), ordered = TRUE))

你猜怎么着？什么都没有发生……table使用时还是一样：

table(spamday)  
FR MO SA SU TH TU WE  
30 96  9  9 45 60 51

我很困惑......请问我做错了什么？

4

1 回答 1

3

有时很难跟踪何时使用<-和何时不使用它，但这是您不想使用它的情况。换句话说，只需将代码更改为以下内容即可：

spamday <- transform(spamday, SPAMreceived = factor(SPAMreceived, 
    levels = c("MO", "TU", "WE", "TH", "FR", "SA", "SU"), ordered = TRUE))

这是一个小例子：

set.seed(1)
spamday <- data.frame(SPAMreceived = sample(c("FR", "MO", "SA", "SU", "TH", "TU", "WE"),
                                            50, replace = TRUE))
table(spamday$SPAMreceived)
# 
# FR MO SA SU TH TU WE 
#  5  7  7  6  9 10  6

temp <- transform(spamday, SPAMreceived = factor(SPAMreceived, 
    levels = c("MO", "TU", "WE", "TH", "FR", "SA", "SU"), ordered = TRUE))
table(temp$SPAMreceived)
# 
# MO TU WE TH FR SA SU 
#  7 10  6  9  5  7  6

或者，您甚至不需要创建有序因子列。只需在最后阶段对您的表格进行排序:)

table(spamday$SPAMreceived)[c("MO", "TU", "WE", "TH", "FR", "SA", "SU")]
# 
# MO TU WE TH FR SA SU 
#  7 10  6  9  5  7  6
于 2013-10-05T14:54:35.180 回答