在 Rcpp 中，如何从 C 到 R 中获取用户定义的结构 [英] In Rcpp, how to get a user-defined structure from C into R

问题描述

我正在使用 Rcpp 包并且可以让我的 C 函数在 R 中编译和运行，但现在我想将一个大的、用户定义的数据结构返回给 R.结构中的字段是数字或字符串 - 没有新的或结构中的奇数类型.下面的例子是简化的，不能编译，但它传达了我的问题的想法.

Am using Rcpp packages and can get my C function to compile and run in R, but now I want to return a large, user-defined data structure to R. The fields in the structure are either numbers or strings - no new or odd types within the structure. The example below is simplified and doesn't compile, but it conveys the idea of my problem.

    typedef struct {
        char*   firstname[128];
        char*   lastname[128];
        int      nbrOfSamples;
    } HEADER_INFO;

    // [[Rcpp::export]]
    HEADER_INFO* read_header(Rcpp::StringVector strings) {
        FILE *fp;
        MEF_HEADER_INFO *header;

        char * filename = (char*)(strings(0));
        char * password = (char*)(strings(1));

        header = (HEADER_INFO*)malloc(sizeof(HEADER_INFO));
        memset(header, 0, sizeof(HEADER_INFO));

        fp = fopen(filename, "r");
        (void)read_header(header, password);
        return header;
    }

我很确定我可以将标头中的条目打包回 StringVector，但这似乎是一种蛮力方法.我的问题是是否存在更优雅的解决方案.我不清楚这样的结构在 R 中有什么形式:命名列表?

I'm pretty sure that I could package the entries in the header back into a StringVector, but that seems like a brute-force approach. My question is whether a more elegant solution exists. It is not clear to me what form such a structure would even have in R: a named List?

谢谢！

推荐答案

R 中的正确结构取决于您的 struct 究竟是什么样子.命名列表是最通用的列表.这是注释中提到的 wrap 函数的简单示例实现:

The right structure in R depends on what your struct looks like exactly. A named list is the most general one. Here a simple sample implementation for a wrap function as referred to in the comments:

#include <RcppCommon.h>

typedef struct {
  char*   firstname[128];
  char*   lastname[128];
  int      nbrOfSamples;
} HEADER_INFO;

namespace Rcpp {
  template <>
  SEXP wrap(const HEADER_INFO& x);
}

#include <Rcpp.h>

namespace Rcpp {
  template <>
  SEXP wrap(const HEADER_INFO& x) {
    Rcpp::CharacterVector firstname(x.firstname, x.firstname + x.nbrOfSamples);
    Rcpp::CharacterVector lastname(x.lastname, x.lastname + x.nbrOfSamples);
    return Rcpp::wrap(Rcpp::List::create(Rcpp::Named("firstname") = firstname,
                      Rcpp::Named("lastname") = lastname,
                      Rcpp::Named("nbrOfSamples") = Rcpp::wrap(x.nbrOfSamples)));
  };
}

//  [[Rcpp::export]]
HEADER_INFO getHeaderInfo() {
  HEADER_INFO header;
  header.firstname[0] = (char*)"Albert";
  header.lastname[0] = (char*)"Einstein";
  header.firstname[1] = (char*)"Niels";
  header.lastname[1] = (char*)"Bohr";
  header.firstname[2] = (char*)"Werner";
  header.lastname[2] = (char*)"Heisenberg";
  header.nbrOfSamples = 3;
  return header;
}

/*** R
getHeaderInfo()
 */

输出:

> getHeaderInfo()
$firstname
[1] "Albert" "Niels"   "Werner"

$lastname
[1] "Einstein"   "Bohr"       "Heisenberg"

$nbrOfSamples
[1] 3

然而，对于这种特殊情况，使用 data.frame 会更自然，这可以通过将上面的 wrap 替换为:

However, for this particular case a data.frame would be more natural to use, which can be achieved by replacing above wrap with:

  template <>
  SEXP wrap(const HEADER_INFO& x) {
    Rcpp::CharacterVector firstname(x.firstname, x.firstname + x.nbrOfSamples);
    Rcpp::CharacterVector lastname(x.lastname, x.lastname + x.nbrOfSamples);
    return Rcpp::wrap(Rcpp::DataFrame::create(Rcpp::Named("firstname") = firstname,
                                              Rcpp::Named("lastname") = lastname));
  };

输出:

> getHeaderInfo()
  firstname   lastname
1    Albert   Einstein
2     Niels       Bohr
3    Werner Heisenberg

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