blocks|key|1120099|text|我真的不确定您希望如何格式化结果，所以我在获取排除两个值出现在一起的组合时停了下来。它依赖于这样一个事实：combn返回一个矩阵，其中每一列都是一个组合。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1120100|mod_headers+<-+c("var1",+"var2",+"var3",+"var4",+"var5",+"var6")


combn_with_exclusion+<-+function(x,+n,+exclude){
++full+<-+combn(x,+n)
++#+remove+any+columns+that+have+all+elements+of+`exclude`
++full[,+!apply(full,+2,+function(y)+all(exclude+%25in%25+y))]
}

combn_with_exclusion(mod_headers,+2,+c("var4",+"var5"))|code-block|syntax|javascript|1120101|entityMap^0|1H|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

I'm not really sure how you want your result to be formatted, so I stopped at getting the combinations that exclude the appearance of two values together. It relies on the fact that <code>combn</code> returns a matrix where each column is a combination.

<pre><code>mod_headers &lt;- c("var1", "var2", "var3", "var4", "var5", "var6")


combn_with_exclusion &lt;- function(x, n, exclude){
 full &lt;- combn(x, n)
 # remove any columns that have all elements of `exclude`
 full[, !apply(full, 2, function(y) all(exclude %in% y))]
}

combn_with_exclusion(mod_headers, 2, c("var4", "var5"))
</code></pre>

blocks|key|4904953|text|这是另一种方法，生成所有组合的列表，然后排除同时包含var4和var5的组合...|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4904954|lapply(
+++lapply(1:length(mod_headers),
++++++++function(i)+combn(mod_headers,+i)),+
+++function(x)+x[,apply(x,+2,+function(y)+!all(c("var4",+"var5")+%25in%25+y))])+

[[1]]
[1]+"var1"+"var2"+"var3"+"var4"+"var5"+"var6"

[[2]]
+++++[,1]+++[,2]+++[,3]+++[,4]+++[,5]+++[,6]+++[,7]+++[,8]+++[,9]+++[,10]++[,11]++[,12]++[,13]++[,14]+
[1,]+"var1"+"var1"+"var1"+"var1"+"var1"+"var2"+"var2"+"var2"+"var2"+"var3"+"var3"+"var3"+"var4"+"var5"
[2,]+"var2"+"var3"+"var4"+"var5"+"var6"+"var3"+"var4"+"var5"+"var6"+"var4"+"var5"+"var6"+"var6"+"var6"

[[3]]
+++++[,1]+++[,2]+++[,3]+++[,4]+++[,5]+++[,6]+++[,7]+++[,8]+++[,9]+++[,10]++[,11]++[,12]++[,13]++[,14]++[,15]++[,16]+
[1,]+"var1"+"var1"+"var1"+"var1"+"var1"+"var1"+"var1"+"var1"+"var1"+"var2"+"var2"+"var2"+"var2"+"var2"+"var3"+"var3"
[2,]+"var2"+"var2"+"var2"+"var2"+"var3"+"var3"+"var3"+"var4"+"var5"+"var3"+"var3"+"var3"+"var4"+"var5"+"var4"+"var5"
[3,]+"var3"+"var4"+"var5"+"var6"+"var4"+"var5"+"var6"+"var6"+"var6"+"var4"+"var5"+"var6"+"var6"+"var6"+"var6"+"var6"

...etc|code-block|syntax|javascript|4904955|entityMap^0|Q|4|V|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Here is another way, generating a list of all combinations, then excluding those containing both <code>var4</code> and <code>var5</code>...

<pre><code>lapply(
 lapply(1:length(mod_headers),
 function(i) combn(mod_headers, i)), 
 function(x) x[,apply(x, 2, function(y) !all(c("var4", "var5") %in% y))]) 

[[1]]
[1] "var1" "var2" "var3" "var4" "var5" "var6"

[[2]]
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] 
[1,] "var1" "var1" "var1" "var1" "var1" "var2" "var2" "var2" "var2" "var3" "var3" "var3" "var4" "var5"
[2,] "var2" "var3" "var4" "var5" "var6" "var3" "var4" "var5" "var6" "var4" "var5" "var6" "var6" "var6"

[[3]]
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] 
[1,] "var1" "var1" "var1" "var1" "var1" "var1" "var1" "var1" "var1" "var2" "var2" "var2" "var2" "var2" "var3" "var3"
[2,] "var2" "var2" "var2" "var2" "var3" "var3" "var3" "var4" "var5" "var3" "var3" "var3" "var4" "var5" "var4" "var5"
[3,] "var3" "var4" "var5" "var6" "var4" "var5" "var6" "var6" "var6" "var4" "var5" "var6" "var6" "var6" "var6" "var6"

...etc
</code></pre>

blocks|key|1120116|text|我只在TIO上尝试过，所以没有基准测试，但我打赌这个版本在大型集合上会更快，如果这应该是重要的话。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1120117|m+<-+c("var2",+"var3",+"var4",+"var5",+"var6")
comb+<-+combn(m,+3)
csums+<-+colSums((comb+==+"var4")+%2B+(comb+==+"var5"))
comb[,+csums+<+2]
#++++++[,1]+++[,2]+++[,3]+++[,4]+++[,5]+++[,6]+++[,7]++
#+[1,]+"var2"+"var2"+"var2"+"var2"+"var2"+"var3"+"var3"
#+[2,]+"var3"+"var3"+"var3"+"var4"+"var5"+"var4"+"var5"
#+[3,]+"var4"+"var5"+"var6"+"var6"+"var6"+"var6"+"var6"|code-block|syntax|javascript|1120118|或者等同于OP的f()|offset|length|style|CODE|1120119|f2+<-+function(m=mod_headers)+{
++++lapply(1:length(m),+function(x)+{
++++++comb+<-+combn(m,+x)
++++++csums+<-+colSums((comb+==+"var4")+%2B+(comb+==+"var5"))
++++++comb[,+csums+<+2]
++++})
}|1120120|entityMap^0|0|0|8|3|0|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

I've only tried this on TIO, so no benchmarking, but I'd wager to bet that this version will be quicker for large sets, if that should be of importance.

<pre><code>m &lt;- c("var2", "var3", "var4", "var5", "var6")
comb &lt;- combn(m, 3)
csums &lt;- colSums((comb == "var4") + (comb == "var5"))
comb[, csums &lt; 2]
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] 
# [1,] "var2" "var2" "var2" "var2" "var2" "var3" "var3"
# [2,] "var3" "var3" "var3" "var4" "var5" "var4" "var5"
# [3,] "var4" "var5" "var6" "var6" "var6" "var6" "var6"
</code></pre>

Or equivalent to OP's <code>f()</code>:

<pre><code>f2 &lt;- function(m=mod_headers) {
 lapply(1:length(m), function(x) {
 comb &lt;- combn(m, x)
 csums &lt;- colSums((comb == "var4") + (comb == "var5"))
 comb[, csums &lt; 2]
 })
}
</code></pre>

blocks|key|3436481|text|这是我的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3436482|f+<-+function(){
++for(i+in+1:length(mod_headers)){
++++++tab+<-+combn(mod_headers,i)
++++++for(j+in+1:ncol(tab)){
++++++++tab_new+<-+c(tab[,j])
++++++++mod_tab_new+<-+c(tab_new,+"newcol")
++++++++if+(all(c("var4","var5")+%25in%25+mod_tab_new))+next
++++++++print(mod_tab_new)
++++}
++}
}

f()|code-block|syntax|javascript|3436483|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is my solution:

<pre><code>f &lt;- function(){
 for(i in 1:length(mod_headers)){
 tab &lt;- combn(mod_headers,i)
 for(j in 1:ncol(tab)){
 tab_new &lt;- c(tab[,j])
 mod_tab_new &lt;- c(tab_new, "newcol")
 if (all(c("var4","var5") %in% mod_tab_new)) next
 print(mod_tab_new)
 }
 }
}

f()
</code></pre>

blocks|key|3436522|text|我使用这个网页来减少N-Way组合的列表，给出了另一组N-Way组合。这是对本杰明代码的一个小小的修改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3436523|mod_headers+<-+c("var1",+"var2",+"var3",+"var4",+"var5",+"var6")

combn_NWayExclusion+<-+function(x,+n,+exclude){
+++full+<-+combn(x,+n);+EXC<-combn(exclude,+n)
+++UU<-lapply(1:ncol(EXC),function(i)+!apply(full,+2,+function(y)+all(EXC[,i]+%25in%25+y)))
+++full[,!apply(do.call(rbind,UU),2,function(u){any(u=="FALSE")})]
+++}

combn_NWayExclusion(mod_headers,+2,+c("var4",+"var5"))|code-block|syntax|javascript|3436524|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I used this webpage to reduce a list of N-Way combination given another set of N-Way combinations. Here's a slight modification of Benjamins Code.
<pre><code>mod_headers &lt;- c(&quot;var1&quot;, &quot;var2&quot;, &quot;var3&quot;, &quot;var4&quot;, &quot;var5&quot;, &quot;var6&quot;)

combn_NWayExclusion &lt;- function(x, n, exclude){
 full &lt;- combn(x, n); EXC&lt;-combn(exclude, n)
 UU&lt;-lapply(1:ncol(EXC),function(i) !apply(full, 2, function(y) all(EXC[,i] %in% y)))
 full[,!apply(do.call(rbind,UU),2,function(u){any(u==&quot;FALSE&quot;)})]
 }

combn_NWayExclusion(mod_headers, 2, c(&quot;var4&quot;, &quot;var5&quot;))
</code></pre>

I am trying to exclude all combinations generated by combn function that consists of "var4" and "var5". Here below is the code that does not work at the moment:

<pre><code>mod_headers &lt;- c("var1", "var2", "var3", "var4", "var5", "var6")

f &lt;- function(){
 for(i in 1:length(mod_headers)){
 tab &lt;- combn(mod_headers,i,function(mod_headers){
 if (combn(mod_headers,i) %in% c("var4","var5")) {return()}
 })
 for(j in 1:ncol(tab)){
 tab_new &lt;- c(tab[,j])
 mod_tab_new &lt;- c(tab_new, "newcol")
 print(mod_tab_new)
 }
 }
}

f()
</code></pre>

Thanks for your help!

Excluding certain combinations from combn function in R

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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我正在尝试排除由包含"var4“和"var5”的combn函数生成的所有组合。下面是目前不起作用的代码：mod_headers <- c("var1", "var2", "var3", "var4", "var5", "var6")f <- function(){ for(i in 1:length(mod_head...

问从R中的combn函数中排除某些组合
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问从R中的combn函数中排除某些组合EN