如何在ifelse中包含NA？ [英] How to include NA in ifelse?

问题描述

我正在尝试根据其他列的值的逻辑语句创建列 ID 。例如，在以下数据框中

I am trying to create a column ID based on logical statements for values of other columns. For example, in the following dataframe

test <- structure(list(time = c(10L, 20L, NA, 30L), type = structure(c(1L, 
2L, 3L, NA), .Label = c("A", "B", "C"), class = "factor"), ID = c(NA, 
"1", NA, NA)), .Names = c("time", "type", "ID"), row.names = c(NA, 
-4L), class = "data.frame")

看起来像

    time    type
1   10      A
2   20      B
3   NA      C
4   30      NA

我想创建一个包含值1的新列 ID 所有时间不是 NA 以及所有类型不是 A 。我使用以下代码：

I want to make a new column ID containing a value of 1 for all time that are not NA and all type that are not A. I am using the following code for this:

test$ID <- ifelse(is.na(test$time) | test$type == "A", NA, "1")

这样的结果为

    time    type    ID
1   10      A       NA
2   20      B       1
3   NA      C       NA
4   30      NA      NA

但是，此代码忽略 NA列中的类型，在 NA 的值> ID 。我需要这个值为1，所以我需要的解决方案应该给：

However, this code ignores the NA in column type, resulting in a value of NA in column ID. I need this to be a value of 1, so my needed solution should give:

    time    type    ID
1   10      A       NA
2   20      B       1
3   NA      C       NA
4   30      NA      1

谁能告诉我怎么做这个？如果我能以某种方式更改 is.na（test $ type）的结果以返回 FALSE，我可以使用现有代码code>而不是 TRUE ，但我不知道该怎么做。或者，我现有代码的结构可能需要完全改变？我感谢任何帮助！

Can anyone tell me how I might do this? I could get this to work with my existing code if I could somehow change the result of is.na(test$type) to return FALSE instead of TRUE, but I'm not sure how to do that. Or, maybe the structure of my existing code needs to be entirely changed? I appreciate any help!

推荐答案

你无法真正比​​较 NA 使用其他值，因此使用 == 将无效。请考虑以下因素：

You can't really compare NA with another value, so using == would not work. Consider the following:

NA == NA
# [1] NA

您只需将您的比较从 == 更改为％ in％：

You can just change your comparison from == to %in%:

ifelse(is.na(test$time) | test$type %in% "A", NA, "1")
# [1] NA  "1" NA  "1"

---

关于你的其他问题，

---

Regarding your other question,

我可以让这个工作使用我现有的代码，如果我能以某种方式更改的结果is.na（test $ type）返回 FALSE TRUE ，但我不知道该怎么做。

I could get this to work with my existing code if I could somehow change the result of is.na(test$type) to return FALSE instead of TRUE, but I'm not sure how to do that.

只需使用！来否定结果：

!is.na(test$time)
# [1]  TRUE  TRUE FALSE  TRUE

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