R grepl returns 不同的结果

Question

我正在尝试检查字符串是否遵循特定模式。字符串必须遵循模式 [check(id("numeric only"),"1 of the specified keywords")]，其中指定的关键字 = D、D5、W1、W2、W3、W4、W5、W6、W0、 M, Q1, Q2, Q3, H1, H2, H3, H4, H5, H6, Y1, Y2, A2, Y4, Y5, A3, Y7, Y8, A4, YA, YB, A1.

freq <- c("D", "D5", "W1", "W2", "W3", "W4", "W5", "W6", "W0", "M", "Q1", 
           "Q2", "Q3", "H1", "H2", "H3", "H4", "H5", "H6", "Y1", "Y2", "A2", 
           "Y4", "Y5", "A3", "Y7", "Y8", "A4", "YA", "YB", "A1")
> grepl("\<check\(id\(+\d+\),\bA1\b|\bA2\b\)\>", "check(id(32131241w),A1)")
[1] FALSE

> grepl("\<check\(id\(+\d+\),\bA1\b|\bA2\b\)\>", "check(id(32131241),A1)")
[1] TRUE

> grepl("\<check\(id\(+\d+\),\b", paste0(freq, collapse = "\b|\b"), "\b\)\>", "check(id(32131241w),A1)")
[1] FALSE

> grepl("\<check\(id\(+\d+\),\b", paste0(freq, collapse = "\b|\b"), "\b\)\>", "check(id(32131241),A1)")
[1] FALSE

第一个、第二个和第三个输出是正确的结果，但我希望第四个结果是 TRUE，但 R return 是 FALSE。

Answer 1

#Second line from you
grepl("\<check\(id\(+\d+\),\bA1\b|\bA2\b\)\>", "check(id(32131241),A1)")
#TRUE

#Change A1 to A2
grepl("\<check\(id\(+\d+\),\bA1\b|\bA2\b\)\>", "check(id(32131241),A2)")
#FALSE

#Adding surrounding <>
grepl("\<check\(id\(+\d+\),\bA1\b|\bA2\b\)\>", "<check(id(32131241),A2)>")
#FALSE

#Adding perl=TRUE
grepl("\<check\(id\(+\d+\),\bA1\b|\bA2\b\)\>", "<check(id(32131241),A2)>", perl=TRUE)
#TRUE

#Add () arround |
grepl("\<check\(id\(+\d+\),(\bA1\b|\bA2\b)\)\>", "<check(id(32131241),A2)>", perl=TRUE)
#TRUE

#Here you have to add one additional paste
grepl(paste0("\<check\(id\(+\d+\),(\b", paste0(freq, collapse = "\b|\b"), "\b)\)\>"), "<check(id(32131241),A1)>", perl=TRUE)
#TRUE

R grepl returns 不同的结果

R grepl returns different result

r

grepl