R中字符值的OR函数

我想检查数据帧的变量是否至少包含一个值。我想要OR函数的R等价物，看起来像这样（这里的OR函数显然是错误的）

损失$League

如果

损失$Drawed，则将具有“Highschool”值。From

包含“Academy”或“H.S.”，否则具有“No Highschool”值

我不熟悉R，英语不是我的第一语言，如果我的问题不太清楚，我很抱歉。

使用

（或）运算符如下：

mydf$lost <- ifelse(lost$Drafted.From == "Academy" | lost$Drafted.From == "H.S.", "Highschool", "Not highschool")

---

mydf$lost这一切都取决于对“contains”一词的解释。如果
Drawed.From
列仅包含单个单词或完整（精确）匹配，则垂直管道
|
或%

运算符中的

%就足够了。否则您将需要
grepl

一些示例数据：

lost <- data.frame(drafted.from.1 = c('Academy','College','H.S.'),
                   drafted.from.2 = c('He studied at the Academy','She went to College','He attended Dartmore H.S.'))

这给出了正确的结果：

[1] "Highschool"     "Not highschool" "Highschool"    

[1] "Highschool"     "Not highschool" "Highschool"    

> ifelse(lost$drafted.from.2 %in% c("Academy","H.S."), "Highschool", "Not highschool")
[1] "Not highschool" "Not highschool" "Not highschool"

但在这种情况下，
grepl
也起作用：

> ifelse(grepl("Academy|H.S.", lost$drafted.from.1), "Highschool", "Not highschool")
[1] "Highschool"     "Not highschool" "Highschool"    

对于
drafted.from.2
列，您需要
grepl
来分配
Highschool
和
Not Highschool
值：

ifelse(grepl("Academy|H.S.", lost$drafted.from.2), "Highschool", "Not highschool")

这给出了正确的结果：

[1] "Highschool"     "Not highschool" "Highschool"    

[1] "Highschool"     "Not highschool" "Highschool"    

> ifelse(lost$drafted.from.2 %in% c("Academy","H.S."), "Highschool", "Not highschool")
[1] "Not highschool" "Not highschool" "Not highschool"

关于从.2

列中起草的

，%

中的

%将不会给出正确的结果（或使用
|
或运算符）：

[1] "Highschool"     "Not highschool" "Highschool"    

[1] "Highschool"     "Not highschool" "Highschool"    

> ifelse(lost$drafted.from.2 %in% c("Academy","H.S."), "Highschool", "Not highschool")
[1] "Not highschool" "Not highschool" "Not highschool"

现在，让我们将这些知识应用于数据帧：

lost$League.1 <- ifelse(lost$drafted.from.1 %in% c("Academy","H.S."), "Highschool", "Not highschool")
lost$League.2 <- ifelse(grepl("Academy|H.S.", lost$drafted.from.2), "Highschool", "Not highschool")

在
grep*
中，您使用垂直管道
|
进行或-
grepl（“Acadamny | H.S.”，x）
（其中x是您正在搜索的数据），您不需要像我所示进行grepl调用。@khrm是的，您需要
grep
/
grepl
，因为
丢失了$Drafted.
可以包含单词“Academy”或“H.S”@这是一个非常简单的问题。不需要grep。我的问题给了我正确的答案。你也可以使用grep。但我通常在使用正则表达式时使用它。@khrm它取决于对“contains”一词的解释。请参阅我的答案以获得解释。