Question

我正在学习数据表的工作方式，并且我尝试在两列（grep()和id1）上使用id2来删除不在的行＃39 ; t返回TRUE。

我知道我必须使用函数lapply()，但它总是返回跟随的错误：

argument 'pattern' has length > 1 and only the first element will be used

我试过了（我知道这是错的）：

DT[, lapply(.SD, grepl(id1, id2)), by= id]

我正在处理的数据：

structure(list(id = c(52L, 52L, 52L, 52L, 54L, 54L, 84L, 84L, 
87L, 87L, 129L, 129L, 130L, 130L, 130L), id1 = c("8113H187", 
"3505H6", "3505H6", "3505H6", "3505H6", "3505H6", "3505H6", "3505H6", 
"8113H187", "8113H187", "3505H6", "3505H6", "3505H6", "3505H6", 
"3505H6"), id2 = c("3505H6856", "3505H6856", "3505H6856", "3505H6856", 
"3505H67158", "3505H67158", "3505H63188", "3505H63188", "3505H64691", 
"3505H64691", "3505H664133", "3505H664133", "3505H658134", "3505H658134", 
"3505H658134")), .Names = c("id", "id1", "id2"), row.names = c(NA, 
-15L), class = c("data.table", "data.frame"), .internal.selfref = <pointer: 0x00000000064f0788>)

Answer 1

我们可以使用Map将'id1'中的相应元素pattern与'ID2'中的元素进行比较

DT[unlist(Map(grepl, id1, id2))]

Answer 2

DT[mapply( grepl, id1, id2), ]

#     id    id1         id2
# 1:  52 3505H6   3505H6856
# 2:  52 3505H6   3505H6856
# 3:  52 3505H6   3505H6856
# 4:  54 3505H6  3505H67158
# 5:  54 3505H6  3505H67158
# 6:  84 3505H6  3505H63188
# 7:  84 3505H6  3505H63188
# 8: 129 3505H6 3505H664133
# 9: 129 3505H6 3505H664133
# 10: 130 3505H6 3505H658134
# 11: 130 3505H6 3505H658134
# 12: 130 3505H6 3505H658134

使用grep（）删除行并使用data.table进行lapply

2 个答案: