blocks|key|5743770|text|如果您在gsub中设置了perl=TRUE，那么您可以使用正面和负面正面或负面正面或背面查看，这可能会解决您的问题，例如，模式15(?=:)将匹配后面跟着冒号的15+(但不会匹配/替换冒号)，并且不会匹配任何不后面跟着冒号的15。模式`(？|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5743771|有关更多详细信息，请参阅“perl”部分中的?regex。|5743772|entityMap^0|C|9|1R|7|0|M|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

If you set <code>perl=TRUE</code> in gsub then you can use positive and negative look aheads or look behinds which might solve your problem, for example the pattern <code>15(?=:)</code> will match a 15 that is followed by a colon (but will not match/replace the colon) and will not match any 15s that are not followed by a colon. The pattern `(?

See <code>?regex</code> in the 'perl' section for more detail.

blocks|key|5743659|text|我可以向您展示如何只替换“前导15”，但是您还没有提供生成测试用例的代码，在这些测试用例中，您可以替换或不替换这些嵌套模式，所以这只是一种猜测。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5743660|+vec+<-+c('15:+.0234'+,+'12:+.0151')
+gsub("%5E15",+"aa",+vec)
#[1]+"aa:+.0234"+"12:+.0151"|code-block|syntax|javascript|5743661|正则表达式模式中的"%5E“符号表示字符串的开头。|5743662|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I can show you how to replace only "leading 15"s but you have not offered code that produces test cases where you do or don't wnat these nested patterns replaced, so this is a bit of a guess.

<pre><code> vec &lt;- c('15: .0234' , '12: .0151')
 gsub("^15", "aa", vec)
#[1] "aa: .0234" "12: .0151"
</code></pre>

The "^" symbol in a regex pattern indicates the beginning of the string.

blocks|key|3629425|text|我会考虑将字符串解析为(a,+b,+c,+d,+...)向量，而不是使用正则表达式。例如，您可以这样做：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3629426|strsplit("15:+.0234,+12:+.0151",+":+%7C,+")
#+[[1]]
#+[1]+"15"+++++".0234"+"12"++++".0151"|code-block|syntax|javascript|3629427|那么使用==、match、%25in%25、[等函数进行替换应该会容易得多。|3629428|完成后，您可以使用paste将所有内容重新组合在一起。|3629429|entityMap^0|B|H|0|0|4|2|7|5|D|4|I|1|0|9|5|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

Instead of working with regular expressions, I would consider parsing the strings into <code>(a, b, c, d, ...)</code> vectors. For example, you could do:

<pre><code>strsplit("15: .0234, 12: .0151", ": |, ")
# [[1]]
# [1] "15" ".0234" "12" ".0151"
</code></pre>

Then it should be much easier to do substitutions using functions like <code>==</code>, <code>match</code>, <code>%in%</code>, <code>[</code>, etc.

When you are done, you can put things back together using <code>paste</code>.

blocks|key|2430303|text|做一个包含可能比做一个排除更容易。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2430304|首先，设置一些测试数据：|2430305|test+<-+c("15:+.0234+,+12:+.0151")|code-block|syntax|javascript|2430306|替换字符串开头的15的示例(%5E执行此操作)。|2430307|gsub("%5E15","x",test)
[1]+"x:+.0234+,+12:+.0151"|2430308|一个更通用的示例，用于替换字符串开头的任意数字。|2430309|gsub("%5E([0-9]%2B)","x",test)
[1]+"x:+.0234+,+12:+.0151"|2430310|删除后跟冒号(:)的任意数字并替换为x和冒号的示例|2430311|gsub("([0-9]%2B):","x:",test)
[1]+"x:+.0234+,+x:+.0151"|2430312|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

It might be easier to do an inclusion rather than an exclusion.

Firstly, set up some test data:

<pre><code>test &lt;- c("15: .0234 , 12: .0151")
</code></pre>

An example to replace a 15 at the start of the string (the ^ does this).

<pre><code>gsub("^15","x",test)
[1] "x: .0234 , 12: .0151"
</code></pre>

A more general example to replace any number at the start of the string.

<pre><code>gsub("^([0-9]+)","x",test)
[1] "x: .0234 , 12: .0151"
</code></pre>

An example to remove any number followed by a colon (:) and replace with an x and a colon

<pre><code>gsub("([0-9]+):","x:",test)
[1] "x: .0234 , x: .0151"
</code></pre>

blocks|key|526277|text|这对你有用吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|526278|(foo+<-+"15:+.0234+,+12:+.0151")
#+"15:+.0234+,+12:+.0151"
(bar+<-+"a:+.b+,+c:+.d")
#+"a:+.b+,+c:+.d"
gsub("%5E15",bar,+foo)
#+"a:+.b+,+c:+.d:+.0234+,+12:+.0151"|code-block|syntax|javascript|526279|如果没有，请提供一些示例数据和一个示例对象，说明您的最终目标是什么。一串你的数据看起来是什么样子，这是一句话‘简单地告诉它只替换第一次出现的数据是行不通的，因为有很多这些模式嵌套’，这对我来说有点不清楚。你想在一个字符串中找到'15‘，但如果它在以小数点开头的数字的末尾或部分，就不需要了？|526280|你读过帮助文件?gsub吗？|offset|length|style|CODE|526281|entityMap^0|0|0|0|7|5|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

would this work for you?

<pre><code>(foo &lt;- "15: .0234 , 12: .0151")
# "15: .0234 , 12: .0151"
(bar &lt;- "a: .b , c: .d")
# "a: .b , c: .d"
gsub("^15",bar, foo)
# "a: .b , c: .d: .0234 , 12: .0151"
</code></pre>

If not please provide some example data and also an example object showing what your end goal is. A string of how your data could look like, it's the the sentence 'It wouldn't work to simply tell it to replace only the first occurence, because there a lot of these patterns nested' that is a little unclear to me. You want find '15' within a string, but not if it at the end or part of the a number that begins with a decimal point?

Have you read the help file <code>?gsub</code>?

blocks|key|5743687|text|gsub("%5E15","a:+.b+,+c:+.d","15:+.0234+,+12:+.0151")|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5743688|这能解决你的问题吗？|unstyled|5743689|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>gsub("^15","a: .b , c: .d","15: .0234 , 12: .0151")
</code></pre>

Would this solve your problem?

Sorry if this is a dumb question, I'm really new to R. What I'm wondering is, is there any way to specify exclusions within gsub or a similar function? 
My data will look something like this: (15: .0234 , 12: .0151), and I want to replace the leading 15 with another item matching this (a: .b , c: .d) pattern. However, just calling gsub replaces both the leading 15 with the new pattern, and the 15 following the decimal point. It wouldn't work to simply tell it to replace only the first occurence, because there a lot of these patterns nested. What I'm thinking is to exclude numbers following a decimal point and followed by a whitespace from the gsub - is this even possible?

How to exclude something from being replaced by gsub()

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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抱歉，如果这是一个愚蠢的问题，我真的是R的新手，我想知道的是，有没有办法在gsub或类似的函数中指定排除？我的数据将如下所示：( 15：.0234，12：.0151)，我想用另一个与此(a：.b，c：.d)模式匹配的项目替换前15个项目。但是，只需调用gsub就可以用新模式替换前导15和小数点后的15。简单地告诉它只替换第一个出现的模式是行不通的，因为有很多这样的模式是嵌套的。我的想法是从gsub

问如何排除某些内容不被gsub()替换
EN

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问如何排除某些内容不被gsub()替换
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