问题

Sure, here's the translated code part:

我想创建一个循环，如果原始变量的相应值也为NA，则将多个虚拟变量的值替换为NA（以重新编码MCQ调查）。

我有19个问题，标记为Q1到Q19，虚拟变量标记为Q1\_[answer1]，Q1\_[answer2]等等。我创建了值为1和0的虚拟变量，而不是嵌套另一个ifelse函数来查看Q1，Q2等的值，我想创建一个循环，自动获取虚拟变量（通过使用grep("Q", [n], "_")，其中n随着循环的进行而增加）。

这基本上是我的数据框的样子

```R
df1 <- data.frame(Q1 = c("a", "b,c", NA, "a,b", "b", NA, "c"))

为了检查我的Q1的值是否缺失，我想使用以下代码（或等效代码：

is.na(paste0("df$Q",n))

这将允许我循环遍历不同的问题。然而，这会测试"df$Q1"是否等于NA，而不是查看Q1作为变量。我想找到一种方法，使它就像我直接输入"df$Q1"一样，返回变量所有值的is.na()测试结果：

is.na(df$Q1)

是否有类似is.na或paste0的函数可以轻松实现这个？

<details>
<summary>英文:</summary>

I&#39;d like to create a loop which replaces the values of several dummy variables by NA if the corresponding value of the original variable is NA as well (in order to recode a MCQ survey).

I have 19 questions, labeled Q1 through Q19, with dummy variables labeled Q1\_\[answer1\], Q1\_\[answer2\] etc. I made dummy variables with values 1 and 0, and instead of nesting another ifelse function which looks at the value of Q1, Q2 etc, I&#39;d like to create a loop that takes the dummy variables automatically (by using grep(&quot;Q&quot;, \[n\], &quot;\_&quot;) where n increases as the loop progresses).

Here is essentially what my dataframe is like

df1 <- data.frame(Q1 = c("a", "b,c", NA, "a,b", "b", NA, "c"))

#this is done for the purposes of the loop, which I'm not including here
n <- 1

In order to check if the values of my Q1 are missing or not, I'd like to use the following code (or equivalent:

is.na(paste0("df$Q",n))

[1] FALSE

which would allow me to cycle through the different questions. However, this tests if "df$Q1" is equal to NA rather than looking at Q1 as a variable. I would like to find a way for it to be like if I had input "df$Q1" directly, which returns the list of results for the is.na() test for all values of the variable:

is.na(df$Q1)

[1] FALSE FALSE TRUE FALSE FALSE TRUE FALSE

Is there a function like is.na or like paste0 which would allow me to do this easily?

</details>


# 答案1
**得分**: 1

如何处理这个问题可能取决于您打算如何处理结果。如评论中所示，最简单的方法是通过 `is.na()` 函数将向量传递。

如果您有一个大型数据框，这可能会很费力。相反，使用循环或 `sapply` 函数可能会得到您需要的结果。

扩展您的数据框以进行演示：

```R
df1 <- data.frame(Q1 = c("a", "b,c", NA, "a,b", "b", NA, "c"), 
                  Q2 = c("a", "b,c", "a.b", "a,b", "b", "b", "c"),
                  Q3 = c("a", "b,c", NA, "a,b", "b", "c", NA))

for(i in 1:length(names(df1))) {
    print(is.na(df1[[paste0("Q", i)]]))
}

结果如下：

[1] FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[1] FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE

或者使用 sapply(df1, is.na)：

     Q1    Q2    Q3
[1,] FALSE FALSE FALSE
[2,] FALSE FALSE FALSE
[3,]  TRUE FALSE  TRUE
[4,] FALSE FALSE FALSE
[5,] FALSE FALSE FALSE
[6,]  TRUE FALSE FALSE
[7,] FALSE FALSE  TRUE

一个技巧是使用 sapply(df1, is.na) * 1，这样可以轻松进行求和。

df1 <- data.frame(Q1 = c("a", "b,c", NA, "a,b", "b", NA, "c"), 
                  Q2 = c("a", "b,c", "a.b", "a,b", "b", "b", "c"),
                  Q3 = c("a", "b,c", NA, "a,b", "b", "c", NA))


  for(i in 1:length(names(df1))) {
    print(is.na(df1[[paste0("Q", i)]]))
  }

[1] FALSE FALSE  TRUE FALSE FALSE  TRUE FALSE
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[1] FALSE FALSE  TRUE FALSE FALSE FALSE  TRUE

        Q1    Q2    Q3
[1,] FALSE FALSE FALSE
[2,] FALSE FALSE FALSE
[3,]  TRUE FALSE  TRUE
[4,] FALSE FALSE FALSE
[5,] FALSE FALSE FALSE
[6,]  TRUE FALSE FALSE
[7,] FALSE FALSE  TRUE