如何以相同的顺序打乱多个迭代？

Question

我寻求一个简单的 Python 函数，它接受任意数量的迭代（元组、列表、字典），并且 returns 它们以相同的顺序 随机排列 :

a = (1, 2, {3: 4}, 5)
b = [(5,6), [7,8], [9,0], [1,2]]
c = {'arrow': 5, 'knee': 'guard', 0: ('x',2)}

x, y, z = magic(a, b, c)
print(x, y, z, sep='\n')
# ({3: 4}, 1, 2)
# [[9, 0], (5, 6), [7, 8]]
# {0: ('x', 2), 'arrow': 5, 'knee': 'guard'}

函数必须：

Return iterables 以相同的顺序洗牌（见上文）
接受任意数量的迭代器
保留迭代器类型
支持任何 depth 和 type
Not 打乱嵌套元素本身（例如上面的 [7,8] 不会变成 [8,7]）
Return 长度为最短迭代器长度的迭代器 w/o 引发错误（见上文）

如果在洗牌步骤中使用 Numpy、随机等（例如 np.random.shuffle(magic_packing)），则可以，但不能是高级库方法（使用多处理、编码, 等等 - 应该是 'plain')

我见过 related SO's，但无法使它们适应这种普遍情况。如何实现？

Answer 1

这是一个基本方法：

import random
def shuffle_containers(*args):
    min_length = min(map(len, args))
    idx = list(range(min_length))
    random.shuffle(idx)
    results = []
    for arg in args:
        if isinstance(arg, list):
            results.append([arg[i] for i in idx])
        elif isinstance(arg, tuple):
            results.append(tuple(arg[i] for i in idx))
        elif isinstance(arg, dict):
            items = list(arg.items())
            results.append(dict(items[i] for i in idx))
        else:
            raise ValueError(
                "Encountered", type(arg),
                "expecting only list, dict, or tuple"
            )
    return results
a = (1, 2, {3: 4}, 5)
b = [(5,6), [7,8], [9,0], [1,2]]
c = {'arrow': 5, 'knee': 'guard', 0: ('x',2)}
x, y, z = shuffle_containers(a, b, c)
print(x, y, z, sep='\n')

请注意，这将忽略超过最小容器长度的任何项目，如果您不想这样，则需要更复杂的逻辑。

编辑：

这里是两行代码：

def shuffle_containers(*args):
    min_length = min(map(len, args)); idx = list(range(min_length)); random.shuffle(idx)
    return [ [arg[i] for i in idx] if isinstance(arg, list) else tuple(arg[i] for i in idx) if isinstance(arg, tuple) else dict(list(args.items())[i] for i in idx) ]

当然，上面的代码可读性、效率和简单性都比较差。不要那样做。

Answer 2

import random

a = (1, 2, {3: 4}, 5)
b = [(5,6), [7,8], [9,0], [1,2]]
c = {'arrow': 5, 'knee': 'guard', 0: ('x',2)}

def magic(*x):
    out = []
    # 6. length of shortest iterable
    min_len = min(len(a_org) for a_org in x)
    for a_org in x:
        if isinstance(a_org, list) or isinstance(a_org, tuple):
            indices = list(range(len(a_org)))
            random.shuffle(indices)
            a_copy = type(a_org)(a_org[i] for i in indices[:min_len])
        elif isinstance(a_org, dict):
            indices = list(a_org.keys())
            random.shuffle(indices)
            a_copy = {i:a_org[i] for i in indices[:min_len]}
        else:
            raise "not supported type"

        out.append(a_copy)
    return tuple(out)

print(magic(a, b, c))

Answer 3

def ordered_shuffle(*args):
    args_types = [type(arg) for arg in args]                               # [1]
    _args      = [arg if type(arg)!=dict else arg.items() for arg in args] # [2]
    args_split = [arg for arg in zip(*_args)]                              # [3]
    args_shuffled = random.sample(args_split, len(args_split))             # [4]
    args_shuffled = map(tuple, zip(*args_shuffled))                        # [5]
    return [args_types[i](arg) for i, arg in enumerate(args_shuffled)]     # [6]

解释：举个更简单的例子，一步一步来：

a = [1, 2, 3]
b = ([1,2], [3,4], [5,6])
c = {'a': 1, 'b': 2, 'c': 3}

# [1]: list, tuple, dict
# [2]: [[1, 2, 3],
#       ([1, 2], [3, 4], [5, 6]),
#       dict_items([('a', 1), ('b', 2), ('c', 3)])]
# [3]: [(1, [1, 2], ('a', 1)), 
#       (2, [3, 4], ('b', 2)), 
#       (3, [5, 6], ('c', 3))]
# [4]: [(1, [1, 2], ('a', 1)), 
#       (3, [5, 6], ('c', 3)), 
#       (2, [3, 4], ('b', 2))]
# [5]: (1, 2, 3)
#      ([1, 2], [3, 4], [5, 6])
#      (('a', 1), ('b', 2), ('c', 3))
# [6]: [(1, 2, {3: 4}),
#       [(5, 6), [7, 8], [9, 0]],
#       {'arrow': 5, 'knee': 'guard', 0: ('x', 2)}]

存储原始类型以便稍后恢复
默认情况下，Python 仅遍历字典 KEYS - 我们还需要值
Python 字典不能直接改组 - 而是通过首先将键值对转换为元组来间接改组；所有可迭代对象也应该同时迭代以重组，完成w/ zip
使用Python的原生random
也可以转换为 list，但 tuple 效率更高
恢复原始类型和return作为要解压为x, y = ordered_shuffle(a, b)

Lambda 解决方案：（来源：Gloweye）

ordered_shuffle = lambda *args:[type(args[i])(arg) for i, arg in enumerate(map(tuple, 
            zip(*random.sample([y for y in zip(*[x if type(x)!=dict else x.items() 
            for x in args])], min(len(z) for z in args)))))]

最短和最快的解决方案：（信用：GZ0）（缩短变量名称使其最短）

def ordered_shuffle(*args):
    zipped_args = list(zip(*(a.items() if isinstance(a, dict) else a for a in args)))
    random.shuffle(zipped_args)
    return [cls(elements) for cls, elements in zip(map(type, args), zip(*zipped_args))]

如何以相同的顺序打乱多个迭代？

How to shuffle multiple iterables in same order?

python

random

iterable

python-3.7