问题

I want to apply a qualified command utilizing do.call, but get an error message.
('qualified' here means: qualified with a package name.)

fn = function(x) sin(x[1]) + cos(x[2])
x0 = c(pi/2, pi/2)

# qualified command works, but not with `do.call` !
numDeriv::hessian(func = fn, x = x0)
##               [,1]          [,2]
## [1,] -1.000000e+00 -5.910744e-13
## [2,] -5.910744e-13  1.063389e-15

do.call("numDeriv::hessian", list(func = fn, x = x0))
## Error in `numDeriv::hessian`(f = function (x)  : 
##   could not find function "numDeriv::hessian"

# loading the library first works too
library(numDeriv)
do.call("hessian", list(func = fn, x = x0))
##               [,1]          [,2]
## [1,] -1.000000e+00 -5.910744e-13
## [2,] -5.910744e-13  1.063389e-15

How can I make this work without explicitly loading the library?

EDIT: In the comments it is suggested to delete the quotes. This is true.
Unfortunately, I have to assemble the command from package and function name.
(Sorry, I didn't mention this in the original post.)

So the only way that solves my real 'challenge' is the one by Gabor:

fn = function(x) sin(x[1]) + cos(x[2])
x0 = c(pi/2, pi/2)

lib <- "numDeriv"
routine <- "hessian"

do.call(routine, list(func = fn, x = x0),
        envir = asNamespace(lib))
##               [,1]          [,2]
## [1,] -1.000000e+00 -5.910744e-13
## [2,] -5.910744e-13  1.063389e-15

Thanks Gabor, very nice. I'd give you the credits if you formulate it as an official answer.

<details>
<summary>英文:</summary>

I want to apply a qualified command utilizing `do.call`, but get
an error message.\
(&#39;qualified&#39; here means: qualified with a package name.)

    fn = function(x) sin(x[1]) + cos(x[2])
    x0 = c(pi/2, pi/2)
    
    # qualified command works, but not with `do.call` !
    numDeriv::hessian(func = fn, x = x0)
    ##               [,1]          [,2]
    ## [1,] -1.000000e+00 -5.910744e-13
    ## [2,] -5.910744e-13  1.063389e-15

    do.call(&quot;numDeriv::hessian&quot;, list(func = fn, x = x0))
    ## Error in `numDeriv::hessian`(f = function (x)  : 
    ##   could not find function &quot;numDeriv::hessian&quot;

    # loading the library first works too
    library(numDeriv)
    do.call(&quot;hessian&quot;, list(func = fn, x = x0))
    ##               [,1]          [,2]
    ## [1,] -1.000000e+00 -5.910744e-13
    ## [2,] -5.910744e-13  1.063389e-15

How can I make this work without explicitly loading the library?

EDIT: In the comments it is suggested to delete the quotes. This is true.
Unfortunately, I have to assemble the command from package and function name.\
(Sorry, I didn&#39;t mention this in the original post.)

So the only way that solves my real &#39;challenge&#39; is the one by Gabor:

    fn = function(x) sin(x[1]) + cos(x[2])
    x0 = c(pi/2, pi/2)
    
    lib &lt;- &quot;numDeriv&quot;
    routine &lt;- &quot;hessian&quot;

    do.call(routine, list(func = fn, x = x0),
            envir = asNamespace(lib))
    ##               [,1]          [,2]
    ## [1,] -1.000000e+00 -5.910744e-13
    ## [2,] -5.910744e-13  1.063389e-15

Thanks Gabor, very nice. I&#39;d give you the credits if you formulate it as an official answer.


</details>


# 答案1
**得分**: 3

问题在于`numDeriv::hessian`不是一个函数的名称，而是一个对`::`的函数调用，用于从`numDeriv`包中获取名为`hessian`的函数。因此，你可以在调用之前自己执行这个操作，例如：

```R
hessian <- numDeriv::hessian
do.call("hessian", list(func = fn, x = x0))

或者（正如@MikaelJagan在评论中所提到的），传递实际的函数而不是它的名称：

do.call(numDeriv::hessian, list(func = fn, x = x0))

这两种方法在回溯信息中会稍微有些不同；我的第一个建议将显示为hessian(func = ...，而第二个将显示为(function (func, x, method = "Richardson", method.args = list(), ...) UseMethod("hessian"))(func = . 但结果将是相同的。

@G.Grothendieck在评论中提供了另一个解决方案：

do.call("hessian", list(func = fn, x = x0), envir = asNamespace("numDeriv"))

这意味着构建函数调用，然后在numDeriv命名空间中进行评估。一开始我对它有所怀疑，因为

hessian(func = fn, x = x0)

在该命名空间中不一定会正确评估（因为fn和x0在那里没有定义），但它不是尝试评估这个。它首先替换了fn和x0的值，然后评估了：

hessian(func = function(x) sin(x[1]) + cos(x[2]), 
        x = c( 1.570796, 1.570796) )

这也可以正常工作。

hessian <- numDeriv::hessian
do.call("hessian", list(func = fn, x = x0))

do.call(numDeriv::hessian, list(func = fn, x = x0))

do.call("hessian", list(func = fn, x = x0), envir = asNamespace("numDeriv"))

hessian(func = fn, x = x0)

hessian(func = function(x) sin(x[1]) + cos(x[2]), 
x = c( 1.570796, 1.570796) )