零位置很重要：一个选择问题

Question

我想按规则匹配一些结果。就我而言：

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)

但在我的例子中 TL-01 是 1。如果我做 which(gsub("TL-|0","",original.names)==selection)，它也不起作用因为 10 和 01 不同，而 gsub 相等。有时我有类似8A（数字字母）的东西，我也需要考虑这种情况。

请问有什么帮助吗？

Answer 1

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 2

这是另一个使用 parse_number 的选项，尽管我们必须用 abs 换行，因为 readr 会将破折号视为负数。

library(readr)

which(abs(parse_number(original.names)) == selection)

# [1] 27

Answer 3

与@benson23 类似的方法，但捕获所需的值而不是删除。

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

这会捕获 TL- 之后出现的数字，将其转换为整数并与 selection 进行比较。

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.

之后的所有内容

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

零位置很重要：一个选择问题

Zero position matters: one selection problem

regex

string

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