返回特定字母后直到下一个字母的数字的正则表达式 [英] Regular expression that returns numbers following a specific letter until the next letter

问题描述

我需要一个正则表达式，它返回一个特定的字母和后面的(一个或两个)数字，直到下一个字母.例如，我想使用 R 中的正则表达式提取公式中的碳数 (C)

I need a regular expression that returns a specific letter and the following (one or two) digits until the next letter. For example, I would like to extract how many carbons (C) are in a formula using regular expressions in R

strings <- c("C16H4ClNO2", "CH8O", "F2Ni")

我需要一个表达式来返回 C 的数量，可以是一位或 2 位数字，并且不返回氯 (Cl) 之后的数字.

I need an expression that returns the number of C which can be one or 2 digits and that does not return the number after chlorine (Cl).

substr(strings,regexpr("C[0-9]+",strings) + 1, regexpr("[ABDEFGHIJKLMNOPQRSTUVWXYZ]+",strings) -1)
[1] "16" "C"  ""  

但我想要返回的答案是

"16","1","0"

此外，我希望正则表达式能够自动定位下一个字母并停在它之前，而不是将最终位置指定为不是 C 的字母.

Moreover, I would like the regular expression to automatically locate the next letter and stop before it, instead of having a final position which I specify as a letter not being a C.

推荐答案

makeup 在 CHNOSZ 包中将解析一个化学式.以下是一些使用它的替代方案:

makeup in the CHNOSZ package will parse a chemical formula. Here are some alternatives that use it:

1) 创建一个列表 L 这样完全解析的公式，然后检查每个公式是否有 "C" 组件和如果没有，则返回其值或 0:

1) Create a list L of such fully parsed formulas and then for each one check if it has a "C" component and return its value or 0 if none:

library(CHNOSZ)

L <- Map(makeup, strings)
sapply(L, function(x) if ("C" %in% names(x)) x[["C"]] else 0)
## C16H4ClNO2       CH8O       F2Ni 
##         16          1          0 

请注意，L 是完整解析公式的列表，以防您有其他要求:

Note that L is a list of the fully parsed formulas in case you have other requirements:

> L
$C16H4ClNO2
 C  H Cl  N  O 
16  4  1  1  2 

$CH8O
C H O 
1 8 1 

$F2Ni
 F Ni 
 2  1 

1a) 通过将 c(C = 0) 添加到每个列表组件，我们可以避免测试碳的存在，从而产生以下较短版本的 sapply (1) 中的行:

1a) By adding c(C = 0) to each list component we can avoid having to test for the existence of carbon yielding the following shorter version of the sapply line in (1):

sapply(lapply(L, c, c(C = 0)), "[[", "C")

2) 除了名称之外，(1) 的这一单行变体给出了与 (1) 相同的答案.它将 "C0" 附加到每个公式以避免必须测试碳的存在:

2) This one-line variation of (1) gives the same answer as in (1) except for names. It appends "C0" to each formula to avoid having to test for the existence of carbon:

sapply(lapply(paste0(strings, "C0"), makeup), "[[", "C")
## [1] 16  1  0

2a) 这是 (2) 的变体，它通过使用 makeup 将接受矩阵这一事实消除了 lapply:

2a) Here is a variation of (2) that eliminates the lapply by using the fact that makeup will accept a matrix:

sapply(makeup(as.matrix(paste0(strings, "C0"))), "[[", "C")
## [1] 16  1  0

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