spark dataFrame Row 对象转换json对象

网上查找了一堆没有解决

最终实现:

  from pyspark.sql import Row

查看Row class 源码：

  ...
 def __new__(self, *args, **kwargs):
        if args and kwargs:
            raise ValueError("Can not use both args "
                             "and kwargs to create Row")
        if kwargs:
            # create row objects
            names = sorted(kwargs.keys())
            row = tuple.__new__(self, [kwargs[n] for n in names])
            row.__fields__ = names
            row.__from_dict__ = True
            return row

        else:
            # create row class or objects
            return tuple.__new__(self, args)
def asDict(self, recursive=False):
        """
        Return as an dict

        :param recursive: turns the nested Row as dict (default: False).

        >>> Row(name="Alice", age=11).asDict() == {'name': 'Alice', 'age': 11}
        True
        >>> row = Row(key=1, value=Row(name='a', age=2))
        >>> row.asDict() == {'key': 1, 'value': Row(age=2, name='a')}
        True
        >>> row.asDict(True) == {'key': 1, 'value': {'name': 'a', 'age': 2}}
        True
        """
        if not hasattr(self, "__fields__"):
            raise TypeError("Cannot convert a Row class into dict")

        if recursive:
            def conv(obj):
                if isinstance(obj, Row):
                    return obj.asDict(True)
                elif isinstance(obj, list):
                    return [conv(o) for o in obj]
                elif isinstance(obj, dict):
                    return dict((k, conv(v)) for k, v in obj.items())
                else:
                    return obj
            return dict(zip(self.__fields__, (conv(o) for o in self)))
        else:
            return dict(zip(self.__fields__, self))
...

直接使用 asDict() 这个方法即可