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R $运算符对原子向量无效

[英]R $ operator is invalid for atomic vectors

I have a dataset where one of the columns are only "#" sign. 我有一个数据集,其中一列只有“#”符号。 I used the following code to remove this column. 我使用以下代码删除此列。

ia <- as.data.frame(sapply(ia,gsub,pattern="#",replacement=""))

However, after this operation, one of the integer column I had changed to factor. 但是,在此操作之后,整数列中的一个已更改为factor。

I wonder what happened and how can i avoid that. 我想知道发生了什么,我怎么能避免这种情况。 Appreciate it. 欣赏它。

A more correct version of your code might be something like this: 更正确的代码版本可能是这样的:

d <- data.frame(x = as.character(1:5),y = c("a","b","#","c","d"))
> d[] <- lapply(d,gsub,pattern = "#",replace = "")
> d
  x y
1 1 a
2 2 b
3 3  
4 4 c
5 5 d

But as you'll note, this approach will never actually remove the offending column. 但正如您将注意到的,这种方法实际上永远不会删除违规列。 It's just replacing the # values with empty character strings. 它只是用空字符串替换#值。 To remove a column of all # you might do something like this: 要删除所有#列,您可能会执行以下操作:

d <- data.frame(x = as.character(1:5),
                y = c("a","b","#","c","d"),
                z = rep("#",5))
> d[,!sapply(d,function(x) all(x == "#"))]
  x y
1 1 a
2 2 b
3 3 #
4 4 c
5 5 d

Surely if you want to remove an offending column from a data frame, and you know which column it is, you can just subset. 当然,如果你想从数据框中删除一个有问题的列,并且你知道它是哪一列,你可以只是子集。 So, if it's the first column: 所以,如果它是第一列:

df <- df[,-1]

If it's a later column, increment up. 如果是后面的列,则递增。

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