df1 <-
structure(c(3L, NA, 3L, 3L, 3L, 2L, 3L, 2L, 2L, 2L, 3L, 3L, 2L,
2L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 1L, 3L, 1L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 2L, 1L, 2L, 3L, 3L, 2L, 3L, 3L, 3L, 3L, 2L, 1L, 3L,
2L, 2L, 3L, 2L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 3L, 3L, 3L, 2L, 3L,
3L, 1L, 3L, 2L, 2L, 3L, 2L, 3L, 1L, 3L, 3L, 3L, 2L, 3L, 3L, 3L,
3L, 3L, 2L, 1L, 2L), levels = c("aaa", "bbb",
"ccc"), class = c("ordered", "factor"))
df2 <-
structure(c(1L, NA, 3L, 1L, 1L, 3L, 2L, 1L, 3L, 2L, 2L, 3L, 1L,
1L, 2L, 3L, 1L, 3L, 2L, 2L, 1L, 3L, 2L, 3L, 2L, 2L, 2L, 3L, 3L,
1L, 2L, 1L, 3L, 2L, 3L, 1L, 2L, 3L, 2L, 3L, 2L, 1L, 3L, 3L, 3L,
2L, 2L, 1L, 3L, 2L, 1L, 2L, 3L, 2L, 2L, 3L, 2L, 2L, 3L, 2L, 3L,
1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 3L, 3L, 3L, 1L, 2L, 1L, 2L, 2L,
3L, 3L, 1L, 3L, 3L), levels = c("aaa", "bbb",
"ccc"), class = c("ordered", "factor"))
df3 <-
structure(c(3L, 2L, 2L, 3L, 3L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
2L, 3L, 2L, 3L, 3L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 3L, 3L, 2L, 2L,
3L, 1L, 3L, 2L, 2L, 3L, 3L, 3L, 2L, 3L, 2L, 3L, 2L, 3L, 2L, 3L,
1L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 3L, 3L, 3L,
3L, 2L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 2L, 3L, 3L), levels = c("ddd", "eee", "fff"
), class = c("ordered", "factor"))
dftest1 <-
structure(c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, 1L, NA, NA, NA, 1L, NA, NA, NA, 2L, NA, NA, NA,
1L, NA, NA, NA, NA, NA, NA, 1L, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, 1L, NA, 2L), levels = c("AAA", "BBB"
), class = "factor")
dftest2 <-
structure(c(NA, NA, NA, NA, NA, NA, NA, 1L, NA, NA, NA, 1L, NA,
1L, NA, 1L, NA, NA, NA, NA, NA, NA, NA, NA, 1L, NA, NA, NA, NA,
NA, NA, 1L, 1L, NA, 1L, NA, NA, NA, NA, NA, NA, NA, NA, NA, 1L,
NA, NA, NA, NA, 1L, NA, NA, NA, 1L, NA, 1L, NA, 1L, NA, NA, NA,
1L, NA, NA, NA, NA, NA, NA, 1L, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, 1L, 1L, 1L), levels = "CCC", class = "factor")我想把df1、df2和df3（在我的示例中是因子）放在一起作为占位符vars2use。
var2use <- c(var1, var2, var3)然后，应将此占位符与其他因子（dftest1、dftest2）组合成一个数据集。
此实现按预期工作：
df <- data.frame(dftest1, dftest2, df1, df2, df3)我希望它也能以这种形式实现：
df <- data.frame(dftest1, dftest2, var2use)但我得到一个错误：
Error in data.frame(dftest1, dftest2, var2use):
arguments imply differing number of rows: 82, 3背景是我想在不同的地方使用这种类型的占位符，有人知道如何解决这个问题吗？
2条答案
按热度按时间inb24sb21#
更直接的方法是把你的值放到一个单独的data.frame中，然后在你需要的时候把值放进
cbind()中。如果出于某种原因，这些名称确实需要成为字符向量，则可以在使用
cbind之前使用mget()获取值j8yoct9x2#
假设
dftest1和dftest2都是大小与（var 1-var 3）相等的 Dataframe ，那么根据下面的代码，两者都能完全正常工作。我认为您的示例中的问题在于您调用的是df 1-df 3，而您的对象却被称为var 1、var 2和var 3
然而，由于您没有在示例中显示抛出了什么错误，因此我不确定我们应该如何得出正确的结论。