2

可能重复：
R中两列的连接因子水平

我对 R 相当陌生，我正在努力使我的重新编码脚本更有效和“正确”。我已经尝试搜索论坛，但这让我一无所获 - 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，请多多包涵。

我有两个因子变量，我希望将它们合并为一个因子变量。它们来自同一个调查，都衡量教育水平。首先我有两个变量的原因是不幸的调查结构，但这不是重点。要说明的要点是它们是相互排斥的（你只能在一个中）。

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，但是当您有两个具有 15 个级别的因子变量几次或更多次时，您就会开始寻找更快的替代方案。

我尝试过这样的事情，但没有任何运气：

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(df$education)[9 年级] 中的错误 | as.character(df$education2)[9th Grade]：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：

我的目标是这样的：

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

帖子：“R 中两列的连接因子水平”似乎是为了另一个结果

再次谢谢你

4

3 回答 3

1

您必须确保结果中存在所有因子水平：

levels(education) <- c(levels(education), levels(education2))
education[is.na(education)] <- education2[is.na(education)]
于 2012-11-22T00:54:46.283 回答
1

一旦它们是字符串，就很容易了

# 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)

您可以通过首先将它们作为字符读入来加速该过程，特别是如果您已经在read.table.

于 2012-11-22T01:01:40.147 回答
0

基本上，您需要确保关卡既是独特关卡的“联合”或“交集”，并且顺序相同，然后您可以使用c. 搜索：[r] 因子联合级别。

于 2012-11-21T23:58:12.240 回答