使用Boost Python将Python列表传递给接受向量的函数

Question

使用Boost Python将Python列表传递给接受向量的函数

24

我有一个带有以下签名的函数：

function(std::vector<double> vector);

我已经将其公开，但它无法接受Python列表。我查看了其他SO答案，大多涉及更改函数以接受boost :: python :: lists，但我不想更改函数。我想我可以使用vector_indexing_suite来编写一个简单的包装器，但我有许多这种形式的功能，而且不想为每个功能编写包装器。是否有自动创建Python list-> std :: vector映射的方法？

- victor

1个回答

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- Tanner Sansbury · Accepted Answer

有几种解决方案可以实现此功能，而无需修改原始函数。为了以少量样板代码并对Python透明地实现此目的，请考虑注册自定义转换器。Boost.Python在C++和Python类型之间转换时使用已注册的转换器。创建绑定时，例如当class_导出类型时，会隐式创建一些转换器。

以下完整示例使用iterable_converter类型，允许从支持python迭代器协议的Python类型注册转换函数。该示例启用以下转换：

内置类型的集合：std::vector<double>
字符串的二维集合：std::vector<std::vector<std::String> >
用户类型的集合：std::list<foo>

#include <iostream>
#include <list>
#include <vector>
#include <boost/python.hpp>
#include <boost/python/stl_iterator.hpp>

/// @brief Mockup model.
class foo {};

// Test functions demonstrating capabilities.

void test1(std::vector<double> values)
{
  for (auto&& value: values)
    std::cout << value << std::endl;
}

void test2(std::vector<std::vector<std::string> > values)
{
  for (auto&& inner: values)
    for (auto&& value: inner)
      std::cout << value << std::endl;
}


void test3(std::list<foo> values)
{
  std::cout << values.size() << std::endl;
}

/// @brief Type that allows for registration of conversions from
///        python iterable types.
struct iterable_converter
{
  /// @note Registers converter from a python interable type to the
  ///       provided type.
  template <typename Container>
  iterable_converter&
  from_python()
  {
    boost::python::converter::registry::push_back(
      &iterable_converter::convertible,
      &iterable_converter::construct<Container>,
      boost::python::type_id<Container>());

    // Support chaining.
    return *this;
  }

  /// @brief Check if PyObject is iterable.
  static void* convertible(PyObject* object)
  {
    return PyObject_GetIter(object) ? object : NULL;
  }

  /// @brief Convert iterable PyObject to C++ container type.
  ///
  /// Container Concept requirements:
  ///
  ///   * Container::value_type is CopyConstructable.
  ///   * Container can be constructed and populated with two iterators.
  ///     I.e. Container(begin, end)
  template <typename Container>
  static void construct(
    PyObject* object,
    boost::python::converter::rvalue_from_python_stage1_data* data)
  {
    namespace python = boost::python;
    // Object is a borrowed reference, so create a handle indicting it is
    // borrowed for proper reference counting.
    python::handle<> handle(python::borrowed(object));

    // Obtain a handle to the memory block that the converter has allocated
    // for the C++ type.
    typedef python::converter::rvalue_from_python_storage<Container>
                                                                storage_type;
    void* storage = reinterpret_cast<storage_type*>(data)->storage.bytes;

    typedef python::stl_input_iterator<typename Container::value_type>
                                                                    iterator;

    // Allocate the C++ type into the converter's memory block, and assign
    // its handle to the converter's convertible variable.  The C++
    // container is populated by passing the begin and end iterators of
    // the python object to the container's constructor.
    new (storage) Container(
      iterator(python::object(handle)), // begin
      iterator());                      // end
    data->convertible = storage;
  }
};

BOOST_PYTHON_MODULE(example)
{
  namespace python = boost::python;

  // Register interable conversions.
  iterable_converter()
    // Build-in type.
    .from_python<std::vector<double> >()
    // Each dimension needs to be convertable.
    .from_python<std::vector<std::string> >()
    .from_python<std::vector<std::vector<std::string> > >()
    // User type.
    .from_python<std::list<foo> >()
    ;

  python::class_<foo>("Foo");

  python::def("test1", &test1);
  python::def("test2", &test2);
  python::def("test3", &test3);
}

交互式使用：

>>> import example
>>> example.test1([1, 2, 3])
1
2
3
>>> example.test1((4, 5, 6))
4
5
6
>>> example.test2([
...   ['a', 'b', 'c'],
...   ['d', 'e', 'f']
... ])
a
b
c
d
e
f
>>> example.test3([example.Foo(), example.Foo()])
2

这种方法的一些评论：

iterable_converter :: convertible 函数可以更改为仅允许Python列表，而不是允许支持可迭代协议的任何类型。但是，结果可能会使扩展略微不符合Pythonic。
转换是基于C++类型注册的。因此，只需要进行一次注册，因为相同的注册转换将在接受C ++类型作为参数的任意数量的导出函数上选择。
它不会向example扩展命名空间引入不必要的类型。
元编程可以允许多维类型递归地注册每个维度类型。但是，示例代码已经足够复杂，所以我不想添加额外的复杂性。

替代方法包括：

Create a custom function or template function that accepts a boost::python::list for each function accepting a std::vector. This approach causes the bindings to scale based on the amount of functions being exported, rather than the amount of types needing converted.
Using the Boost.Python vector_indexing_suite. The *_indexing_suite classes export a type that is adapted to match some semantics of Python list or dictionaries. Thus, the python code now has to know the exact container type to provide, resulting in a less-pythonic extension. For example, if std::vector<double> is exported as VecDouble, then the resulting Python usage would be:
```
v = example.VecDouble()
v[:] = [1, 2, 3]
example.test1(v)
```
However, the following would not work because the exact types must match, as exporting the class only registers a conversion between VecDouble and std::vector<double>:
```
example.test1([4, 5, 6])
```
While this approach scales to types rather than functions, it results in a less pythonic extension and bloats the example namespace with unnecessary types.