防止R中的grep处理“。”。作为一封信 [英] Prevent grep in R from treating "." as a letter

问题描述

我有一个包含类似以下内容的字符向量：

  text < -  c .xYz，ge，lmo.qrstu）
  

a 。：

  xyzgeqrstu
  

但是， grep 函数似乎以

$

 $ 
 $ 
 c> pattern<  - （[AZ] | [az]）+ $
 
 grep（pattern，text，value = T）
 
& ABc.def.xYzgelmo.qrstu
  
如 regexpal 。 
 
 
 
如何获得 grep 的行为如预期？
解决方案
  grep 用于查找模式。它返回与模式匹配的向量的索引。如果指定 value = TRUE ，则返回值。从描述中，似乎要删除子字符串，而不是返回初始向量的子集。
 
 
 
如果需要删除子字符串，可以使用 sub  
  sub（'。* \\。'，'' ，text）
＃[1]xYzgeqrstu
  
第一个参数，我们匹配一个模式ie '。* \\。'。它匹配多个字符（。* ）后跟一个点（ \\。）。需要 \\ 来转义。，将其视为符号而不是任何字符。这将匹配，直到字符串中最后的。字符。我们用''替换匹配模式作为替换参数，从而删除子字符串。
 
I have a character vector that contains text similar to the following: 
text <- c("ABc.def.xYz", "ge", "lmo.qrstu")
I would like to remove everything before a .: 
> "xYz" "ge" "qrstu"
However, the grep function seems to be treating . as a letter: 
pattern <- "([A-Z]|[a-z])+$"

grep(pattern, text, value = T)

> "ABc.def.xYz" "ge"          "lmo.qrstu" 
The pattern works elsewhere, such as on regexpal. 

How can I get grep to behave as expected?
解决方案
grep is for finding the pattern.  It returns the index of the vector that matches a pattern.  If, value=TRUE is specified, it returns the value.  From the description, it seems that you want to remove substring instead of returning a subset of the initial vector.

If you need to remove the substring, you can use sub 
 sub('.*\\.', '', text)
 #[1] "xYz"   "ge"    "qrstu"
As the first argument, we match a pattern i.e. '.*\\.'.  It matches one of more characters (.*) followed by a dot (\\.).  The \\ is needed to escape the . to treat it as that symbol instead of any character.  This will match until the last . character in the string.  We replace that matched pattern with a '' as the replacement argument and thereby remove the substring.

                        
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