blocks|key|576110|text|旧的离别-替补技巧：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|576111|a<-data.frame(x=1:10,y=1:10)
test<-function(z){
+++mean.x<-mean(z$x)
+++nm+<-deparse(substitute(z))
+++print(nm)
+++return(mean.x)}
+
+test(a)
#[1]+"a"+++...+this+is+the+side-effect+of+the+print()+call
#++++++++++...+you+could+have+done+something+useful+with+that+character+value
#[1]+5.5+++...+this+is+the+result+of+the+function+call|code-block|syntax|javascript|576112|编辑:用新的test-object运行它|576113|注意:当一组列表项从第一个参数传递给lapply时，在本地函数中不会成功(当对象从给定的列表传递给for-loop时，它也会失败)。如果结构结果是一个正在处理的命名向量，那么您将能够从结构结果中提取".Names"-attribute和处理顺序。|offset|length|style|CODE|576114|>+lapply(+list(a=4,b=5),+function(x)+{nm+<-+deparse(substitute(x));+strsplit(nm,+'\\[')}+)
$a++++++#+This+"a"+and+the+next+one+in+the+print+output+are+put+in+after+processing
$a[[1]]
[1]+"X"++++""+++++"1L]]"++#+Notice+that+there+was+no+"a"


$b
$b[[1]]
[1]+"X"++++""+++++"2L]]"

>+lapply(+c(a=4,b=5),+function(x)+{nm+<-+deparse(substitute(x));+strsplit(nm,+'\\[')}+)
$a
$a[[1]]+++#+but+it's+theoretically+possible+to+extract+when+its+an+atomic+vector
[1]+"structure(c(4,+5),+.Names+=+c(\"a\",+\"b\"))"+""++++++++++++++++++++++++++++++++++++++++++++
[3]+"1L]]"++++++++++++++++++++++++++++++++++++++++


$b
$b[[1]]
[1]+"structure(c(4,+5),+.Names+=+c(\"a\",+\"b\"))"+""++++++++++++++++++++++++++++++++++++++++++++
[3]+"2L]]"++|576115|entityMap^0|0|0|0|I|6|1D|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]|$K|Y|L|Z|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

The old deparse-substitute trick:
<pre><code>a&lt;-data.frame(x=1:10,y=1:10)
test&lt;-function(z){
 mean.x&lt;-mean(z$x)
 nm &lt;-deparse(substitute(z))
 print(nm)
 return(mean.x)}
 
 test(a)
#[1] &quot;a&quot; ... this is the side-effect of the print() call
# ... you could have done something useful with that character value
#[1] 5.5 ... this is the result of the function call
</code></pre>
Edit: Ran it with the new test-object
Note: this will not succeed inside a local function when a set of list items are passed from the first argument to <code>lapply</code> (and it also fails when an object is passed from a list given to a <code>for</code>-loop.) You would be able to extract the &quot;.Names&quot;-attribute and the order of processing from the structure result, if it were a named vector that were being processed.
<pre><code>&gt; lapply( list(a=4,b=5), function(x) {nm &lt;- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a # This &quot;a&quot; and the next one in the print output are put in after processing
$a[[1]]
[1] &quot;X&quot; &quot;&quot; &quot;1L]]&quot; # Notice that there was no &quot;a&quot;


$b
$b[[1]]
[1] &quot;X&quot; &quot;&quot; &quot;2L]]&quot;

&gt; lapply( c(a=4,b=5), function(x) {nm &lt;- deparse(substitute(x)); strsplit(nm, '\\[')} )
$a
$a[[1]] # but it's theoretically possible to extract when its an atomic vector
[1] &quot;structure(c(4, 5), .Names = c(\&quot;a\&quot;, \&quot;b\&quot;))&quot; &quot;&quot; 
[3] &quot;1L]]&quot; 


$b
$b[[1]]
[1] &quot;structure(c(4, 5), .Names = c(\&quot;a\&quot;, \&quot;b\&quot;))&quot; &quot;&quot; 
[3] &quot;2L]]&quot; 
</code></pre>

blocks|key|360381|text|请注意，对于print方法，行为可能会有所不同。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|360382|print.foo=function(x){+print(deparse(substitute(x)))+}
test+=+list(a=1,+b=2)
class(test)="foo"
#this+shows+"test"+as+expected
print(test)

#this+(just+typing+'test'+on+the+R+command+line)
test
#shows
#"structure(list(a+=+1,+b+=+2),+.Names+=+c(\"a\",+\"b\"),+class+=+\"foo\")"|code-block|syntax|javascript|360383|我在论坛上看到的其他评论表明，最后一种行为是不可避免的。如果您正在为包编写print方法，这是很不幸的。|360384|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|$]]

Note that for print methods the behavior can be different.
<pre><code>print.foo=function(x){ print(deparse(substitute(x))) }
test = list(a=1, b=2)
class(test)=&quot;foo&quot;
#this shows &quot;test&quot; as expected
print(test)

#this (just typing 'test' on the R command line)
test
#shows
#&quot;structure(list(a = 1, b = 2), .Names = c(\&quot;a\&quot;, \&quot;b\&quot;), class = \&quot;foo\&quot;)&quot;
</code></pre>
Other comments I've seen on forums suggests that the last behavior is unavoidable. This is unfortunate if you are writing print methods for packages.

blocks|key|360432|text|deparse(quote(var))|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|360433|我的直观理解是，引号冻结了计算结果中的var或表达式，离线函数是parse函数的逆函数，它将冻结的符号重新转换为字符串|unstyled|360434|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>deparse(quote(var))
</code></pre>

My intuitive understanding
In which the quote freeze the var or expression from evaluation
and the deparse function which is the inverse of parse function makes that freezed symbol back to String

I am looking for the reverse of <code>get()</code>.

Given an object name, I wish to have the character string representing that object extracted directly from the object.

Trivial example with <code>foo</code> being the placeholder for the function I am looking for.

<pre><code>z &lt;- data.frame(x=1:10, y=1:10)

test &lt;- function(a){
 mean.x &lt;- mean(a$x)
 print(foo(a))
 return(mean.x)}

test(z)
</code></pre>

Would print:

<pre><code> "z"
</code></pre>

My work around, which is harder to implement in my current problem is:

<pre><code>test &lt;- function(a="z"){
 mean.x &lt;- mean(get(a)$x)
 print(a)
 return(mean.x)}

test("z")
</code></pre>

In R, how to get an object's name after it is sent to a function?

我正在寻找get()的反面。给定一个对象名，我希望直接从该对象中提取表示该对象的字符串。一个简单的示例，其中foo是我要查找的函数的占位符。z <- data.frame(x=1:10, y=1:10)test <- function(a){ mean.x <- mean(a$x) print(foo(a)) r...

问在R中，如何在对象被发送到函数后获取对象的名称？
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