从字符串返回数字 [英] Return number from string

问题描述

我正在尝试提取下面字符串中Humans"的Number"，例如:

I'm trying to extract the "Number" of "Humans" in the string below, for example:

string <- c("ProjectObjectives|Objectives_NA, PublishDate|PublishDate_NA, DeploymentID|DeploymentID_NA, Species|Human|Gender|Female, Species|Cat|Number|1, Species|Human|Number|1, Species|Human|Position|Left")

文本在字符串中的位置会不断变化，所以我需要R来搜索字符串，找到物种|人类|数量|"并返回 1.

The position of the text in the string will constantly change, so I need R to search the string and find "Species|Human|Number|" and return 1.

如果这是另一个线程的重复，我很抱歉，但我看过这里(根据模式在 R 中提取子字符串) 和这里 (R 提取部分字符串).但我没有任何运气.

Apologies if this is a duplicate of another thread, but I've looked here (extract a substring in R according to a pattern) and here (R extract part of string). But I'm not having any luck.

有什么想法吗?

推荐答案

使用捕获方法 - 在已知子字符串之后捕获 1 个或多个数字 (\d+)(只需转义 | 符号):

Use a capturing approach - capture 1 or more digits (\d+) after the known substring (just escape the | symbols):

> string <- c("ProjectObjectives|Objectives_NA, PublishDate|PublishDate_NA, DeploymentID|DeploymentID_NA, Species|Human|Gender|Female, Species|Cat|Number|1, Species|Human|Number|1, Species|Human|Position|Left")
> pattern = "Species\\|Human\\|Number\\|(\\d+)"
> unlist(regmatches(string,regexec(pattern,string)))[2]
[1] "1"

一种变体是使用带有 regmatches/regexpr

> pattern="(?<=Species\\|Human\\|Number\\|)\\d+"
> regmatches(string,regexpr(pattern,string, perl=TRUE))
[1] "1"

这里，左侧上下文被放置在一个非消费模式中，一个积极的回顾，(?<=...).

Here, the left side context is put inside a non-consuming pattern, a positive lookbehind, (?<=...).

使用 \K 运算符可以实现相同的功能:

The same functionality can be achieved with \K operator:

> pattern="Species\\|Human\\|Number\\|\\K\\d+"
> regmatches(string,regexpr(pattern,string, perl=TRUE))
[1] "1"

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