blocks|key|2806184|text|尝试使用?max，您将看到它实际上有一个na.rm+=参数，缺省情况下设置为FALSE。(这是许多其他R函数的默认设置，包括sum()、mean()等。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2806185|设置na.rm=TRUE就是为了满足您的要求：|2806186|d+<-+c(1,+100,+NA,+10)
max(d,+na.rm=TRUE)|code-block|syntax|javascript|2806187|如果您确实想删除所有的NA，请改用这个习惯用法：|2806188|d+<-+d[!is.na(d)]|2806189|最后注意:其他函数(例如table()、lm()和sort())的NA-related参数使用不同的名称(并提供不同的选项)。因此，如果NA在函数调用中给你带来了问题，那么在函数的参数中寻找一个内置的解决方案是值得的，我发现通常已经有了一个。|2806190|entityMap^0|4|4|K|7|12|5|1Q|5|1W|6|0|2|A|0|0|B|2|0|0|C|7|K|4|P|6|X|2|1W|2|0^^$0|@$1|2|3|4|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]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|19|8|@$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|1C|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|T|$]]

Trying <code>?max</code>, you'll see that it actually has a <code>na.rm =</code> argument, set by default to <code>FALSE</code>. (That's the common default for many other R functions, including <code>sum()</code>, <code>mean()</code>, etc.) 

Setting <code>na.rm=TRUE</code> does just what you're asking for:

<pre><code>d &lt;- c(1, 100, NA, 10)
max(d, na.rm=TRUE)
</code></pre>

If you do want to remove all of the <code>NA</code>s, use this idiom instead:

<pre><code>d &lt;- d[!is.na(d)]
</code></pre>

A final note: Other functions (e.g. <code>table()</code>, <code>lm()</code>, and <code>sort()</code>) have <code>NA</code>-related arguments that use different names (and offer different options). So if <code>NA</code>'s cause you problems in a function call, it's worth checking for a built-in solution among the function's arguments. I've found there's usually one already there.

blocks|key|4703990|text|您可以调用max(vector,+na.rm+=+TRUE)。更广泛地说，您可以使用na.omit()函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4703991|entityMap^0|5|P|16|9|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

You can call <code>max(vector, na.rm = TRUE)</code>. More generally, you can use the <code>na.omit()</code> function.

blocks|key|2806278|text|?max向您展示了一个额外的参数na.rm，您可以将其设置为TRUE。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2806279|除此之外，如果你真的想删除NA，只需使用如下命令：|2806280|myvec[!is.na(myvec)]|code-block|syntax|javascript|2806281|entityMap^0|0|4|G|5|U|4|0|D|2|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

<code>?max</code> shows you that there is an extra parameter <code>na.rm</code> that you can set to <code>TRUE</code>.

Apart from that, if you really want to remove the <code>NA</code>s, just use something like:

<pre><code>myvec[!is.na(myvec)]
</code></pre>

blocks|key|2806334|text|na.omit函数是许多回归例程内部使用的函数：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2806335|vec+<-+1:1000
vec[runif(200,+1,+1000)]+<-+NA
max(vec)
#[1]+NA
max(+na.omit(vec)+)
#[1]+1000|code-block|syntax|javascript|2806336|entityMap^0|0|7|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|$]]

The <code>na.omit</code> function is what a lot of the regression routines use internally:

<pre><code>vec &lt;- 1:1000
vec[runif(200, 1, 1000)] &lt;- NA
max(vec)
#[1] NA
max( na.omit(vec) )
#[1] 1000
</code></pre>

blocks|key|632823|text|以防刚接触R的人想要原始问题的简化答案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|632824|632825|如何从向量中删除NA值？|blockquote|632826|632827|632828|这就是它：|632829|假设您有一个向量foo，如下所示：|offset|length|style|CODE|632830|foo+=+c(1:10,+NA,+20:30)|code-block|syntax|javascript|632831|运行length(foo)给出了22。|632832|nona_foo+=+foo[!is.na(foo)]|632833|length(nona_foo)为21，因为NA值已被删除。|632834|请记住，is.na(foo)返回一个布尔矩阵，因此使用与该值相反的值索引foo将给出所有不是NA的元素。|632835|entityMap^0|0|0|0|0|0|0|8|3|0|0|2|B|0|0|0|G|0|4|A|10|3|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|16|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|G|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|19|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1A|8|@$L|1B|M|1C|N|O]]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1D|8|@]|9|@]|A|$S|T]]|$1|U|3|V|5|6|7|1E|8|@$L|1F|M|1G|N|O]]|9|@]|A|$]]|$1|W|3|X|5|R|7|1H|8|@]|9|@]|A|$S|T]]|$1|Y|3|Z|5|6|7|1I|8|@$L|1J|M|1K|N|O]]|9|@]|A|$]]|$1|10|3|11|5|6|7|1L|8|@$L|1M|M|1N|N|O]|$L|1O|M|1P|N|O]]|9|@]|A|$]]|$1|12|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]]|13|$]]

Just in case someone new to R wants a simplified answer to the original question

<blockquote>
 How can I remove NA values from a vector?
</blockquote>

Here it is:

Assume you have a vector <code>foo</code> as follows:

<pre><code>foo = c(1:10, NA, 20:30)
</code></pre>

running <code>length(foo)</code> gives 22. 

<pre><code>nona_foo = foo[!is.na(foo)]
</code></pre>

<code>length(nona_foo)</code> is 21, because the NA values have been removed. 

Remember <code>is.na(foo)</code> returns a boolean matrix, so indexing <code>foo</code> with the opposite of this value will give you all the elements which are not NA.

blocks|key|632844|text|使用discard+from++(适用于列表和向量)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|632845|discard(v,+is.na)+|code-block|syntax|javascript|632846|好处是它很容易使用管道；或者使用内置子设置函数[|632847|v+%25>%25+discard(is.na)
v+%25>%25+`[`(!is.na(.))|632848|请注意，na.omit不适用于列表：|632849|>+x+<-+list(a=1,+b=2,+c=NA)
>+na.omit(x)
$a
[1]+1

$b
[1]+2

$c
[1]+NA|632850|entityMap|0|LINK|mutability|MUTABLE|url|https://purrr.tidyverse.org/reference/keep.html^0|2|7|2|D|0|0|0|N|1|0|0|4|7|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1E|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

Use <a href="https://purrr.tidyverse.org/reference/keep.html" rel="noreferrer"><code>discard</code> from purrr</a> (works with lists and vectors). 

<pre><code>discard(v, is.na) 
</code></pre>

The benefit is that it is easy to use pipes; alternatively use the built-in subsetting function <code>[</code>:

<pre><code>v %&gt;% discard(is.na)
v %&gt;% `[`(!is.na(.))
</code></pre>

Note that <code>na.omit</code> does not work on lists:

<pre><code>&gt; x &lt;- list(a=1, b=2, c=NA)
&gt; na.omit(x)
$a
[1] 1

$b
[1] 2

$c
[1] NA
</code></pre>

blocks|key|4421356|text|我运行了一个比较两种base方法的快速基准测试，结果发现x[!is.na(x)]比na.omit更快。用户qwr建议我也试试purrr::dicard+--这被证明要慢得多(尽管我会很高兴地对我的实现和测试发表评论！)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4421357|microbenchmark::microbenchmark(
++purrr::map(airquality,function(x)+{x[!is.na(x)]}),+
++purrr::map(airquality,na.omit),
++purrr::map(airquality,+~purrr::discard(.x,+.p+=+is.na)),
++times+=+1e6)

Unit:+microseconds
+++++++++++++++++++++++++++++++++++++++++++++++++++++expr++++min+++++lq++++++mean+median++++++uq+++++++max+neval+cld
+purrr::map(airquality,+function(x)+{+++++x[!is.na(x)]+})+++66.8+++75.9++130.5643+++86.2++131.80++541125.5+1e%2B06+a++
++++++++++++++++++++++++++purrr::map(airquality,+na.omit)+++95.7++107.4++185.5108++129.3++190.50++534795.5+1e%2B06++b+
++purrr::map(airquality,+~purrr::discard(.x,+.p+=+is.na))+3391.7+3648.6+5615.8965+4079.7+6486.45+1121975.4+1e%2B06+++c|code-block|syntax|javascript|4421358|作为参考，这里是x[!is.na(x)]与na.omit的原始测试|4421359|microbenchmark::microbenchmark(
++++purrr::map(airquality,function(x)+{x[!is.na(x)]}),+
++++purrr::map(airquality,na.omit),+
++++times+=+1000000)


Unit:+microseconds
++++++++++++++++++++++++++++++++++++++++++++++expr++min+++lq++++++mean+median++++uq++++++max+neval+cld
+map(airquality,+function(x)+{+++++x[!is.na(x)]+})+53.0+56.6++86.48231+++58.1++64.8+414195.2+1e%2B06++a+
++++++++++++++++++++++++++map(airquality,+na.omit)+85.3+90.4+134.49964+++92.5+104.9+348352.8+1e%2B06+++b|4421360|entityMap^0|A|4|S|C|15|7|1H|3|1Q|D|0|0|8|C|L|7|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|P|$]]

I ran a quick benchmark comparing the two <code>base</code> approaches and it turns out that <code>x[!is.na(x)]</code> is faster than <code>na.omit</code>. User <code>qwr</code> suggested I try <code>purrr::dicard</code> also - this turned out to be massively slower (though I'll happily take comments on my implementation &amp; test!)

<pre><code>microbenchmark::microbenchmark(
 purrr::map(airquality,function(x) {x[!is.na(x)]}), 
 purrr::map(airquality,na.omit),
 purrr::map(airquality, ~purrr::discard(.x, .p = is.na)),
 times = 1e6)

Unit: microseconds
 expr min lq mean median uq max neval cld
 purrr::map(airquality, function(x) { x[!is.na(x)] }) 66.8 75.9 130.5643 86.2 131.80 541125.5 1e+06 a 
 purrr::map(airquality, na.omit) 95.7 107.4 185.5108 129.3 190.50 534795.5 1e+06 b 
 purrr::map(airquality, ~purrr::discard(.x, .p = is.na)) 3391.7 3648.6 5615.8965 4079.7 6486.45 1121975.4 1e+06 c
</code></pre>

For reference, here's the original test of <code>x[!is.na(x)]</code> vs <code>na.omit</code>:

<pre><code>microbenchmark::microbenchmark(
 purrr::map(airquality,function(x) {x[!is.na(x)]}), 
 purrr::map(airquality,na.omit), 
 times = 1000000)


Unit: microseconds
 expr min lq mean median uq max neval cld
 map(airquality, function(x) { x[!is.na(x)] }) 53.0 56.6 86.48231 58.1 64.8 414195.2 1e+06 a 
 map(airquality, na.omit) 85.3 90.4 134.49964 92.5 104.9 348352.8 1e+06 b
</code></pre>

I have a huge vector which has a couple of <code>NA</code> values, and I'm trying to find the max value in that vector (the vector is all numbers), but I can't do this because of the <code>NA</code> values.

How can I remove the <code>NA</code> values so that I can compute the max?

Remove NA values from a vector

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

我有一个巨大的向量，它有几个NA值，我试图找出这个向量中的最大值(向量都是数字)，但由于NA的值，我无法做到这一点。如何删除NA值才能计算最大值？

问从向量中删除NA值
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

问从向量中删除NA值EN