Question

我要删除所有带有*字符的行。

当我尝试删除*时-它只会删除所有内容

（“ *”-如何避免选择所有内容？）

xx <- data.frame(
  c1=c(".", ".", ".", ".", ".", "*", ".", ".", "."),
  c2=c(".", ".", ".", ".", "Q", "Q", "R", ".", "."),
  c3=c(".", ".", ".", ".", "W", "*", ".", ".", "."),
  c4=c("A", "A", ".", ".", "I", ".", "P", ".", "."),
  c5=c(".", ".", ".", ".", "D", "Q", "D", ".", "."),
  c6=c(".", ".", ".", ".", "*", ".", ".", ".", "."),
  c7=c(".", ".", ".", ".", "W", ".", ".", ".", "."),
  row.names = paste0("r", 1:9)
)

xx[!grepl('*', xx),]
dplyr::filter(xx, !grepl('*', c1))
xx %>% 
+     filter(!str_detect(xx, '*'))

结果不应包含带有“ *”的行

（删除第5行和第6行，因为它们带有*）

structure(list(c1 = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("-", 
".", "*"), class = "factor"), c2 = structure(c(1L, 1L, 1L, 1L, 
4L, 1L, 1L), .Label = c(".", "A", "Q", "R", "T"), class = "factor"), 
    c3 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c(".", 
    "*", "S", "W"), class = "factor"), c4 = structure(c(2L, 2L, 
    1L, 1L, 4L, 1L, 1L), .Label = c(".", "A", "I", "P", "V"), class = "factor"), 
    c5 = structure(c(1L, 1L, 1L, 1L, 3L, 1L, 1L), .Label = c(".", 
    "A", "D", "K", "Q"), class = "factor"), c6 = structure(c(2L, 
    2L, 2L, 2L, 2L, 2L, 2L), .Label = c("-", ".", "*"), class = "factor"), 
    c7 = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("-", 
    ".", "W"), class = "factor")), row.names = c("r1", "r2", 
"r3", "r4", "r7", "r8", "r9"), class = "data.frame")

谢谢！

Answer 1

如果我们使用CollectionViewCell，则grep是表示任何零个或多个字符的元字符。我们可以使用*或转义（fixed = TRUE）来获取文字值

\\*

或者另一个选择是xx[!Reduce(`|`, lapply(xx, function(x) grepl("*", x, fixed = TRUE))),]来匹配==，使用*来获取连续的匹配计数以及子集

rowSums

无法删除带有字符*的行

1 个答案: