Question

可能重复：
Joining factor levels of two columns in R

我对R来说相当新，我试图使我的重新编码脚本更有效并且＃34;正确＆＃34;。我试过在论坛上搜索，但这让我无处可去 - 也许我使用了错误的术语并错过了它，所以如果问题已经提出，请耐心等待。

我有两个因子变量，我希望将其合并为一个因子变量。它们来自同一调查，都衡量教育水平。我之所以有两个变量的原因是因为一个不幸的调查结构，但这不是重点。要点是它们是互斥的（你只能在一个）。

我的数据如下所示：

education       education2
9th grade       <NA>
9th grade       <NA>
<NA>            9th grade
<NA>            10th grade
10th grade      <NA>
11th grade      <NA>
<NA>            9th grade
<NA>            11th grade
<NA>            <NA>

，我的脚本如下所示：

highest.edu     <- vector(length=length(df$education))
a.grade       <- which(df$education=="9th grade")
a.grade2      <- which(df$education2=="9th grade")
b.grade      <- which(df$education=="10th grade")
b.grade2     <- which(df$education2=="10th grade")
c.grade      <- which(df$education=="11th grade")
c.grade2     <- which(df$education=="11th grade")

highest.edu[a.grade]      <- as.character(df$education)[a.grade]
highest.edu[a.grade2]     <- as.character(df$education2)[a.grade2]
highest.edu[b.grade]     <- as.character(df$education)[b.grade]
highest.edu[b.grade2]    <- as.character(df$education2)[b.grade2]
highest.edu[c.grade]     <- as.character(df$education)[c.grade]
highest.edu[c.grade2]    <- as.character(df$education2)[c.grade2]

highest.edu  <- factor(highest.edu)
highest.edu[highest.edu =="FALSE"] =NA
highest.edu  <- factor(highest.edu)

当然这还不错，但当你有两个因子变量有15个等级或更多时，你就会开始寻找更快的选择。

我尝试过这样的事情，但没有运气：

a.grade   <- which(df$education=="9th grade" | df$education2=="9th grade")
b.grade  <- which(df$education=="10th grade" | df$education=="10th grade")
c.grade  <- which(df$education=="11th grade" | df$education2=="11th grade")

highest.edu[a.grade]      <- as.character(df$education)  
[a.grade]|as.character(df$education2)[a.grade]
highest.edu[b.grade]      <- as.character(df$education)          
[b.grade]|as.character(df$education2)[b.grade]

给我这个： as.character错误（df $ education）[9年级] | as.character（df $ education2）[9年级]：只能对数字，逻辑或复杂类型进行操作

有没有办法克服这个问题？

感谢您提前提出任何建议

编辑：

我瞄准的结果是：

highest.education
9th grade
9th grade
9th grade
10th grade
10th grade
11th grade
9th grade
11th grade
<NA>

帖子：＆＃39; R＆＃39;中两列的连接因子水平似乎是在寻找另一个结果

再次，谢谢

Answer 1

您必须确保结果中包含所有因子水平：

levels(education) <- c(levels(education), levels(education2))
education[is.na(education)] <- education2[is.na(education)]

Answer 2

一旦它们成为字符串就很容易

# make them character types
ed <- levels(df$education)[df$education]
ed2 <- levels(df$education2)[df$education2]
# make one new factor that integrates them
ed[is.na(ed)] <- ed2[is.na(ed)]
# make it a factor again
ed <- factor(ed)

您可以通过首先以字符形式阅读来加速该过程，尤其是如果您已在read.table中设置了列类型。

Answer 3

基本上，您需要确保这些级别既是唯一级别的“联合”或“交集”又是相同的顺序，那么您可以使用c加入它们。搜索：[r]因子联合水平。

将两个因素合二为一？

编辑：

3 个答案: