blocks|key|543784|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|543785|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|543828|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|543829|vec+<-+1:1000
vec[runif(200,+1,+1000)]+<-+NA
max(vec)
#[1]+NA
max(+na.omit(vec)+)
#[1]+1000|code-block|syntax|javascript|543830|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|543886|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|543887|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|543888|作为参考，这里是x[!is.na(x)]与na.omit的原始测试|543889|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|543890|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

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
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