我们说我有一个function
需要另外两个functions
,arguments
为arguments
{/ 1}}:
a <- function(x, y = 2){
x + y
}
b <- function(b1, b2 = 7){
b1 + b2
}
x <- function(x1, x2){
# Get arguments of arguments
}
有没有办法从arguments
参数中获取x()
列表?这是在电话之后:
x(a(3,4), b(5))
我想获得如下列表:
$x1
$x1$x
[1] 3
$x1$y
[1] 4
$x2
$x2$b1
[1] 5
$x2$b2
[1] 7
答案 0 :(得分:1)
using (var db = new Entities())
{ var q = from u in db.Sites
where u.SiteURL == result.SiteURL
select u;
if (q.Any()) {
List<webpagespeedtest_flat.DataModel.Sites> l = new List<webpagespeedtest_flat.DataModel.Sites>();
l = q.OrderByDescending(s => s.CheckDate).Take(10).ToList();
webpagespeedtest_flat.DataModel.Sites[] z = l.ToArray();
int x = z.Count();
for( int i=0; i < z.Count(); i++){
z[i].Score; // --> this will be the first attribute(x)
z[i].CheckDate;// -->this will be the second attribute(y)
}
显然,即使使用有效的函数调用,这也很容易被破坏:
x <- function(x1, x2){
theCall <- lapply(as.list(match.call()),as.list)[-1]
args <- lapply(theCall, function(x) as.list(formals(as.character(x))))
Map(function(a, b) {
b <- b[-1]
for (i in seq_along(a)) {
if(i <= length(b)) a[i] <- b[i]
}
a
}, args, theCall)
}
str(x(a(3,4), b(5)))
#List of 2
# $ x1:List of 2
# ..$ x: num 3
# ..$ y: num 4
# $ x2:List of 2
# ..$ b1: num 5
# ..$ b2: num 7
使所有可能输入的此功能正确,留给读者练习。