字符串中去除重复字符

Question

字符串中去除重复字符

4

这个问题可能与这个问题有关。

不幸的是，那里给出的解决方案在我的数据中不起作用。

我有以下矢量示例：

example<-c("ChildrenChildren", "Clothing and shoesClothing and shoes","Education, health and beautyEducation, health and beauty", "Leisure activities, travelingLeisure activities, traveling","LoansLoans","Loans and financial servicesLoans and financial services" ,"Personal transfersPersonal transfers" ,"Savings and investmentsSavings and investments","TransportationTransportation","Utility servicesUtility services")

当然，我希望得到没有重复的相同字符串，即：

  > result
 [1]   "Children" "Clothing and shoes" "Education, health and beauty"

可以的吗？

- Henry Navarro

1

你是如何最初得到这样的向量的？似乎更容易修复导致此类数据的步骤，而不是事后修复它。 - MrFlick

3个回答

5

如果所有字符串都是重复的，那么它们的长度是需要的两倍，因此取每个字符串的前一半：

> substr(example, 1, nchar(example)/2)
 [1] "Children"                      "Clothing and shoes"           
 [3] "Education, health and beauty"  "Leisure activities, traveling"
 [5] "Loans"                         "Loans and financial services" 
 [7] "Personal transfers"            "Savings and investments"      
 [9] "Transportation"                "Utility services"

- Spacedman

2

这确实很大程度上取决于重复。@Cath的解决方案具有这样的属性，即如果一个字符串没有重复，则会将整个字符串返回，而不是像我的代码一样只返回一半。 - Spacedman

所有的答案都很好，但我必须选择一个。每个答案都已经被我点赞了，感谢大家。 - Henry Navarro

3

我们可以尝试以下方法：

stringr::str_remove_all(example,"[a-z].*[A-Z]")

结果：

[1] "Children"                      "Clothing and shoes"            "Education, health and beauty" 
 [4] "Leisure activities, traveling" "Loans"                         "Loans and financial services" 
 [7] "Personal transfers"            "Savings and investments"       "Transportation"               
[10] "Utility services"

- NelsonGon

1

所有的答案都很好，但我必须选择一个。每个答案都已经被我点赞了，感谢大家。 - Henry Navarro

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- Cath · Accepted Answer

你可以使用sub来直接捕获你在pattern中想要的比特位：

sub("(.+)\\1", "\\1", example)
 #[1] "Children"                      "Clothing and shoes"            "Education, health and beauty"  "Leisure activities, traveling" "Loans"                        
 #[6] "Loans and financial services"  "Personal transfers"            "Savings and investments"       "Transportation"                "Utility services"

(.+) 允许捕获某些模式，\\1 显示您刚刚捕获的内容，因此您要查找的是“任何东西重复出现两次”，然后用相同的“任何东西”替换，但只出现一次。