R中的grepl找到与任何字符串列表匹配的匹配项 [英] grepl in R to find matches to any of a list of character strings

问题描述

在引用值列表时可能使用 grepl 参数，可能使用％运算符中的％？我想采取下面的数据，如果动物名称中有狗或猫，我想返回一个确定的值，比如说保持。如果它没有狗或猫，我想返回丢弃。

  data < -  data.frame（animal = sample（c（cat，dog，bird， 'doggy'，'kittycat'），50，replace = T））
  

现在，如果我只需要通过严格匹配值来实现这一点，比如说猫和狗，我可以使用以下方法：

 匹配< -c（cat，dog）
 
 data $ keep<  -  ifelse（数据$ animal％in％匹配，Keep，Discard）
  

但是仅使用grep或grepl指的是列表中的第一个参数：

  data $ keep < -  ifelse（grepl（matches，data $ animal），Keep，Discard）
  
 p> 
 
 
退货
 警告消息：
在grepl（matches，data $ animal）中：
参数'pattern'的长度大于1，只有第一个元素会被使用
  
请注意，我在我的搜索中看到了此线索，但这看起来不起作用：
 使用具有多种模式的字符向量的grep   
 
解决方案
您可以在 grepl 的正则表达式中使用或（ | ）语句。 
 
 
 pre $  ifelse（grepl（dog | cat，data $ animal），keep，discard）
＃[1]keepkeepdiscardkeepkeepkeepkeepdiscard
＃[9]keepkeepkeepkeep保留保留丢弃保留
＃[17]丢弃保留保留丢弃保留保留丢弃保留
＃[25 ]keepkeepkeepkeepkeepkeepkeepkeep
＃[33]keepdiscardkeepdiscardkeep 保持保持
＃[41]保持保持保持保持保持保持保持保持
＃[49]保持 丢弃
  

正则表达式 dog | cat 告诉章程ular表达式引擎查找dog或cat，并返回两个匹配项。 p>

Is it possible to use a grepl argument when referring to a list of values, maybe using the %in% operator? I want to take the data below and if the animal name has "dog" or "cat" in it, I want to return a certain value, say, "keep"; if it doesn't have "dog" or "cat", I want to return "discard".

data <- data.frame(animal = sample(c("cat","dog","bird", 'doggy','kittycat'), 50, replace = T))

Now, if I were just to do this by strictly matching values, say, "cat" and "dog', I could use the following approach:

matches <- c("cat","dog")

data$keep <- ifelse(data$animal %in% matches, "Keep", "Discard")

But using grep or grepl only refers to the first argument in the list:

data$keep <- ifelse(grepl(matches, data$animal), "Keep","Discard")

returns

Warning message:
In grepl(matches, data$animal) :
  argument 'pattern' has length > 1 and only the first element will be used

Note, I saw this thread in my search, but this doesn't appear to work: grep using a character vector with multiple patterns

解决方案

You can use an "or" (|) statement inside the regular expression of grepl.

ifelse(grepl("dog|cat", data$animal), "keep", "discard")
# [1] "keep"    "keep"    "discard" "keep"    "keep"    "keep"    "keep"    "discard"
# [9] "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "discard" "keep"   
#[17] "discard" "keep"    "keep"    "discard" "keep"    "keep"    "discard" "keep"   
#[25] "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"   
#[33] "keep"    "discard" "keep"    "discard" "keep"    "discard" "keep"    "keep"   
#[41] "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"    "keep"   
#[49] "keep"    "discard"

The regular expression dog|cat tells the regular expression engine to look for either "dog" or "cat", and return the matches for both.

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