Question

从此L Hernandez

开始

来自包含以下内容的载体：

[1] "HernandezOlaf "    "HernandezLuciano " "HernandezAdrian "

我试过了：

'subset(ABC, str_detect(ABC, "L Hernandez") == TRUE)'

Hernandez这个名字包括首都L anyplace是理想的输出。

所需的输出为HernandezLuciano

Answer 1

可能有帮助：

vec1 <- c("L Hernandez", "HernandezOlaf ","HernandezLuciano ", "HernandezAdrian ")
grep("L ?Hernandez|Hernandez ?L",vec1,value=T)
#[1] "L Hernandez" "HernandezLuciano "

更新

variable <- "L Hernandez"

v1 <- gsub(" ", " ?", variable) #replace space with a space and question mark 
v2 <- gsub("([[:alpha:]]+) ([[:alpha:]]+)", "\\2 ?\\1", variable) #reverse the order of words in the string and add question mark

您还可以使用strsplit将variable拆分为@rawr评论

grep(paste(v1,v2, sep="|"), vec1,value=T)
#[1] "L Hernandez"       "HernandezLuciano "

Answer 2

您可以使用agrep函数进行近似字符串匹配。如果你只是运行这个函数，它匹配每个字符串......

agrep("L Hernandez", c("HernandezOlaf ",    "HernandezLuciano ", "HernandezAdrian "))
[1] 1 2 3

但如果你稍微修改一下＆＃34; L Hernandez＆＃34; - ＆GT; ＆＃34; Hernandez L＆＃34;

agrep("Hernandez L", c("HernandezOlaf ",    "HernandezLuciano ", "HernandezAdrian "))
[1] 1 2 3

并更改最大距离

agrep("Hernandez L", c("HernandezOlaf ",    "HernandezLuciano ", "HernandezAdrian "),0.01)
[1] 2

你得到了正确的答案。这只是一个想法，可能适合您：）

Answer 3

如果您只想在大写字母L之后使用全名，则可以修改以下内容：

vec1[grepl("Hernandez", vec1) & grepl("L\\.*", vec1)]
[1] "L Hernandez"       "HernandezLuciano

或

vec1[grepl("Hernandez", vec1) & grepl("L[[:alpha:]]", vec1)]
[1] "HernandezLuciano "

该表达式在＆＃34; Hernandez＆＃34;然后看看是否有资本＆＃34; L＆＃34;随后是任何角色或空间。第二个版本需要在首都＆＃34; L＆＃34;。

之后写一封信顺便说一下，你似乎无法将这些人聚集在一起。

vec1[grepl("Hernandez", vec1) & grepl("L\\[[:alpha:]]", vec1)]
character(0)

找到与R匹配的最佳字符串

3 个答案:

更新