r语言 - 零位问题:一个选择问题

Question

我想按规则匹配一些结果。在我的例子中:

original.names <- c("TL-13","TL-09","TL-12","TL-19A","TL-11","TL-20","TL-16",
"TL-15","TL-14","TL-10","TL-18","TL-19","TL-08A","TL-07A","TL-17A","TL-17",
"TL-21","TL-09A","TL-22","TL-08","TL-05","TL-03","TL-06","TL-07","TL-02","TL-04","TL-01") 
selection <- c(1)
which(original.names==selection)
#integer(0)

但是在我的例子中，TL-01是1。如果我制造which(gsub("TL-|0","",original.names)==selection)，它也不起作用因为10和01不同，gsub相等。有时我有8A(数字-字母顺序)之类的东西，我也需要考虑这种情况。

请帮忙?

Answer 1

与@benson23类似，但捕获所需值而不是删除。

which(as.integer(sub('TL-(\d+).*', '\1', original.names)) == selection)
#[1] 27

捕获TL-之后出现的数字，将其转换为整数，并将其与selection进行比较。

Answer 2

您需要从original.names中删除所有字符串，然后将其转换为整数类型，然后才能进行比较。

original.names <- c("TL-13","TL-09","TL-12","TL-19A","TL-11","TL-20","TL-16",
"TL-15","TL-14","TL-10","TL-18","TL-19","TL-08A","TL-07A","TL-17A","TL-17",
"TL-21","TL-09A","TL-22","TL-08","TL-05","TL-03","TL-06","TL-07","TL-02","TL-04","TL-01") 
selection <- 1
which(as.integer(gsub("\D", "", original.names)) == selection)
[1] 27

Answer 3

这是使用parse_number的另一个选项，尽管我们必须使用abs包装，因为readr将把破折号视为负值。

library(readr)
which(abs(parse_number(original.names)) == selection)
# [1] 27

Answer 4

我们可以使用str_pad:

library(stringr)
which(str_match(original.names, '\d+')== 
str_pad(selection, 2, pad = "0"))
[1] 27

Answer 5

您可以删除-之前的所有内容，以及0之后的内容。

selection <- 1
which(sub("^.*-0*", "", original.names) %in% sub("^0*", "", selection))
#[1] 27
selection <- "8A"
which(sub("^.*-0*", "", original.names) %in% sub("^0*", "", selection))
#[1] 13
selection <- "01"
which(sub("^.*-0*", "", original.names) %in% sub("^0*", "", selection))
#[1] 27
selection <- c(1, "8A")
which(sub("^.*-0*", "", original.names) %in% sub("^0*", "", selection))
#[1] 13 27