Question

假设我有一个任意嵌套的列表，其中一些嵌套元素可以是生成器。例如：

nested_gens = [
    [1, [2, [3, 4]]],
    [2, (map(int, '123'))],
    [3, (map(str, range(i+1)) for i in range(2))],
    {'a': ({k: (float(i) for i in range(2))} for k in 'xyz')},
    {'b': {'c': dict(zip(range(3), 'abc'))}}
]

如何递归地遍历此结构并使用所有生成器对象？

我想要的输出是：

[
    [1, [2, [3, 4]]],
    [2, [1, 2, 3]],
    [3, [['0'], ['0', '1']]],
    {'a': [{'x': [0.0, 1.0]}, {'y': [0.0, 1.0]}, {'z': [0.0, 1.0]}]},
    {'b': {'c': {0: 'a', 1: 'b', 2: 'c'}}}
]

对于包含发电机的酸洗对象，可以普遍解决该问题。我找到的与TypeError: can't pickle generator objects处理有关的所有答案都与嵌套生成器无关。

更新：该解决方案应该能够处理任何类型的嵌套元素。

Answer 1

一种方法是递归地遍历嵌套对象，然后将生成器转换为list。

from inspect import isgenerator, isgeneratorfunction

def consume_all_generators(row):

    if isinstance(row, str):
        return row
    elif isinstance(row, dict):
        return {k: consume_all_generators(v) for k, v in row.items()}

    output = []
    try:
        for val in row:
            if isgenerator(val) or isgeneratorfunction(val):
                output.append(list(consume_all_generators(val)))
            else:
                output.append(consume_all_generators(val))
        return output
    except TypeError:
        return row

将其应用于问题中的示例：

print(consume_all_generators(nested_gens))
#[[1, [2, [3, 4]]],
# [2, [1, 2, 3]],
# [3, [['0'], ['0', '1']]],
# {'a': [{'x': [0.0, 1.0]}, {'y': [0.0, 1.0]}, {'z': [0.0, 1.0]}]},
# {'b': {'c': {0: 'a', 1: 'b', 2: 'c'}}}]

Answer 2

对于不导入的解决方案，您还可以递归使用列表推导：

def _test(d):
  if isinstance(d, str):
    return d
  try:
    _l = [i for i in d]
    return [_test(i) for i in _l] if not isinstance(d, dict) else {a:_test(b) for a, b in d.items()}
  except:
    return d

nested_gens = [
  [1, [2, [3, 4]]],
  [2, (map(int, '123'))],
  [3, (map(str, range(i+1)) for i in range(2))],
  {'a': ({k: (float(i) for i in range(2))} for k in 'xyz')},
  {'b': {'c': dict(zip(range(3), 'abc'))}}
]
print(_test(nested_gens))

输出：

[[1, [2, [3, 4]]], [2, [1, 2, 3]], [3, [['0'], ['0', '1']]], {'a': [{'x': [0.0, 1.0]}, {'y': [0.0, 1.0]}, {'z': [0.0, 1.0]}]}, {'b': {'c': {0: 'a', 1: 'b', 2: 'c'}}}]

Answer 3

尝试一下

   
def consume(g):
    if hasattr(g, '__iter__') and not isinstance(g, str):
        if isinstance(g, tuple):
            return (consume(e) for e in g)
        elif isinstance(g, dict):
            return { k: consume(v) for k,v in g.items()}
        else:
            return [consume(e) for e in g]
    return g

nested_gens = [ [1, [2, [3, 4]]], 
    [2, (map(int, '123'))], 
    [3, (map(str, range(i+1)) for i in range(2))], 
    {'a': ({k: (float(i) for i in range(2))} for k in 'xyz')},
    {'b': {'c': dict(zip(range(3), 'abc'))}} ]
print(consume(nested_gens))

使用嵌套的生成器

3 个答案: