blocks|key|1802129|text|1)如果，我们想要if，而不是ifelse。还对代码进行了一些简化。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|1802130|f+<-+function(x)+if+(is.numeric(x))+x*2+else+x
lapply(mylist,+f)|code-block|syntax|javascript|1802131|给予：|1802132|[[1]]
[1]+16

[[2]]
+[1]++2++4++6++8+10+12+14+16+18+20

[[3]]
[1]++8+14+50

[[4]]
+++++[,1]+[,2]+[,3]+[,4]+[,5]
[1,]++++2+++14+++26+++38+++50
[2,]++++4+++16+++28+++40+++52
[3,]++++6+++18+++30+++42+++54
[4,]++++8+++20+++32+++44+++56
[5,]+++10+++22+++34+++46+++58
[6,]+++12+++24+++36+++48+++60

[[5]]
[1]+"my+name+is+not+Donald"|1802133|2)+S3，尽管在这种情况下这可能有点过于复杂，但是如果我们有一组更大的类，这将是有用的，而且它还表明，不需要使用S3的任何if就可以做到这一点。为此，我们使用数值和默认方法定义一个S3泛型，然后S3将自动分派适当的方法。|1802134|dble+<-+function(x,+...)+UseMethod("dble")
dble.numeric+<-+function(x,+...)+2+*+x
dble.default+<-+function(x,+...)+x

lapply(mylist,+dble)|1802135|3)+tryCatch不使用if的另一种方法是将参数加倍，如果它引发错误，则捕获它。同样，在这种情况下，这可能太复杂了，但它说明了如果实际情况更加复杂，可以做些什么。|1802136|f2+<-+function(x)+tryCatch(2+*+x,+error+=+function(e)+x)
lapply(mylist,+f2)|1802137|4)+rrapply+rrapply包中的rrapply函数可以指定这样一个条件，即该函数只应用于满足它的元素。|1802138|library(rrapply)

rrapply(mylist,+is.numeric,+function(x)+2+*+x)|1802139|5)+modify_if+purrr包还有一个函数，它只修改满足条件的元素。|1802140|library(purrr)

modify_if(mylist,+is.numeric,+`*`,+2)|1802141|entityMap^0|0|4|9|2|F|6|0|0|0|0|0|5|1Q|2|0|0|0|B|E|2|0|0|0|A|L|7|0|0|0|C|0|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|D]|$9|1C|A|1D|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|1E|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|1F|8|@]|E|@]|F|$]]|$1|N|3|O|5|I|7|1G|8|@]|E|@]|F|$J|K]]|$1|P|3|Q|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|D]]|E|@]|F|$]]|$1|R|3|S|5|I|7|1M|8|@]|E|@]|F|$J|K]]|$1|T|3|U|5|6|7|1N|8|@$9|1O|A|1P|B|C]|$9|1Q|A|1R|B|D]]|E|@]|F|$]]|$1|V|3|W|5|I|7|1S|8|@]|E|@]|F|$J|K]]|$1|X|3|Y|5|6|7|1T|8|@$9|1U|A|1V|B|C]|$9|1W|A|1X|B|D]]|E|@]|F|$]]|$1|Z|3|10|5|I|7|1Y|8|@]|E|@]|F|$J|K]]|$1|11|3|12|5|6|7|1Z|8|@$9|20|A|21|B|C]]|E|@]|F|$]]|$1|13|3|14|5|I|7|22|8|@]|E|@]|F|$J|K]]|$1|15|3|-4|5|6|7|23|8|@]|E|@]|F|$]]]|16|$]]

1) if We want <code>if</code>, not <code>ifelse</code> here. Have also simplified the code a bit.
<pre><code>f &lt;- function(x) if (is.numeric(x)) x*2 else x
lapply(mylist, f)
</code></pre>
giving:
<pre><code>[[1]]
[1] 16

[[2]]
 [1] 2 4 6 8 10 12 14 16 18 20

[[3]]
[1] 8 14 50

[[4]]
 [,1] [,2] [,3] [,4] [,5]
[1,] 2 14 26 38 50
[2,] 4 16 28 40 52
[3,] 6 18 30 42 54
[4,] 8 20 32 44 56
[5,] 10 22 34 46 58
[6,] 12 24 36 48 60

[[5]]
[1] &quot;my name is not Donald&quot;
</code></pre>
2) S3 Although this is probably a bit too complex to warrant in this case it could be useful if we had a larger set of classes and it is also shows that this could be done without any <code>if</code> using S3. To do that we define an S3 generic with numeric and default methods and then S3 will automatically dispatch the appropriate method.
<pre><code>dble &lt;- function(x, ...) UseMethod(&quot;dble&quot;)
dble.numeric &lt;- function(x, ...) 2 * x
dble.default &lt;- function(x, ...) x

lapply(mylist, dble)
</code></pre>
3) tryCatch Another way which does not use <code>if</code> is to double the argument and if it raises an error catch it. Again this is probably too complex to warrant in this case but it illustrates what could be done if the actual situation were more complex.
<pre><code>f2 &lt;- function(x) tryCatch(2 * x, error = function(e) x)
lapply(mylist, f2)
</code></pre>
4) rrapply The <code>rrapply</code> function in the rrapply package can specify a condition such that the function is only applied to the elements satisfying it.
<pre><code>library(rrapply)

rrapply(mylist, is.numeric, function(x) 2 * x)
</code></pre>
5) modify_if The purrr package also has a function that will modify only elements that satisfy a condition.
<pre><code>library(purrr)

modify_if(mylist, is.numeric, `*`, 2)
</code></pre>

blocks|key|2353072|text|问题是条件只创建一个TRUE，而是尝试如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2353073|#Code
lapply(1:length(mylist),
+++++++function(x)+ifelse(lapply(mylist[[x]],is.numeric),
++++++++++++++++++++++++++mylist[[x]]*2,+mylist[[x]]))|code-block|syntax|javascript|2353074|问题是矩阵结构将丢失，因此@G.Grothendieck的答案将是一个更好的选择。|BOLD|2353075|entityMap^0|A|4|0|0|D|F|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|S|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|T|8|@$9|U|A|V|B|M]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|O|$]]

The issue is that the condition is only creating one <code>TRUE</code>, instead try something like this:
<pre><code>#Code
lapply(1:length(mylist),
 function(x) ifelse(lapply(mylist[[x]],is.numeric),
 mylist[[x]]*2, mylist[[x]]))
</code></pre>
The issue is that the matrix structure will be lost, so the answer from @G.Grothendieck would be a better option.

I am writing a doc about R and I've come up with an example that I cannot understand. Let us say this:
<pre><code>mylist &lt;- list(8, 1:10, c(4,7,25), matrix(1:30, ncol = 5), &quot;my name is not Donald&quot;)
lapply(1:length(mylist), function(x) ifelse(is.numeric(mylist[[x]]), 
 mylist[[x]]*2, mylist[[x]]))
</code></pre>
I would have expected the numeric elements in my list to be 2-folded and the character vector to be left a it is. Instead, the result for numeric elements in the list is just the first component of each element, as if the ifelse's TRUE condition (i.e. 1) had taken over the lapply's x index. Could anybody please tell me the logic behind this, and what should I have typed instead?
Thanks a lot.
Best,
David

lapply and ifelse together not working as expected

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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我正在写一篇关于R的文档，我想出了一个我无法理解的例子。让我们这样说：mylist <- list(8, 1:10, c(4,7,25), matrix(1:30, ncol = 5), "my name is not Donald")lapply(1:length(mylist), function(x) ifels...

问我申请和如果其他在一起没有按预期工作
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问我申请和如果其他在一起没有按预期工作EN