Python 如何根据枚举数合并两个列表？

如何合并两个枚举列表，如果第一个数字相同，将元素从“列表1”更改为元素从“列表2”

清单1：

[(1, 'a’), (2, ['00:00:00,280', '00:00:09,680\n']), (3, 'b), (4, 'c'), (5, 'd’),
 (6, ['00:00:08,760', '00:00:13,309\n']), (7, 'e'), (8, 'f'), (9, 'g'),
 (10, ['00:00:09,680', '00:00:15,630\n'])]

清单2：

[(2, ['00:00:00,280', '00:00:08,760']), (6, ['00:00:08,760', '00:00:09,680']),
 (10, ['00:00:09,680', '00:00:13,309'])]

结果:

[(1, 'a’), (2, ['00:00:00,280', '00:00:08,760']), (3, 'b), (4, 'c'), (5, 'd’),
 (6, ['00:00:08,760', '00:00:09,680']),  (7, 'e'), (8, 'f'), (9, 'g'),
 (10, ['00:00:09,680', '00:00:13,309'])]

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尝试将关键字与sorted（）一起使用

---

键应该是一个确定如何从数据结构中检索可比较元素的函数。在您的例子中，它是元组的第一个元素，因此我们访问[0]。

将列表转换为dict

使用第二个dict更新第一个dict

将第一个列表转换回列表

排序列表

a = [(1, 'a'), (2, ['00:00:00,280', '00:00:09,680\n']), (3, 'b'), (4, 'c'), (5, 'd'), (6, ['00:00:08,760', '00:00:13,309\n']), (7, 'e'), (8, 'f'), (9, 'g'), (10, ['00:00:09,680', '00:00:15,630\n'])]
b = [(2, ['00:00:00,280', '00:00:08,760']), (6, ['00:00:08,760', '00:00:09,680']), (10, ['00:00:09,680', '00:00:13,309'])]
c = [(1, 'a'), (2, ['00:00:00,280', '00:00:08,760']), (3, 'b'), (4, 'c'), (5, 'd'), (6, ['00:00:08,760', '00:00:09,680']),  (7, 'e'), (8, 'f'), (9, 'g'), (10, ['00:00:09,680', '00:00:13,309'])]

da = {k:v for k,v in a}
db = {k:v for k,v in b}
da.update(db)

x = sorted((k,v) for k,v in da.items())

print(x == c) # True

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编辑：正如@jferard在评论中指出的，我可以写得更简单一些

da = dict(a)
db = dict(b)
da.update(db)

x = sorted(da.items())

print(x == c) # True

或

甚至

x = sorted(dict(a+b).items())

print(x == c) # True

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我的解决方案与这里的其他答案略有不同，但我认为它获得了所需的功能，并且少了一次字典转换

我将第二个列表更改为dict，然后进行列表理解以创建一个新列表。我保留了元组中的第一个条目，并执行了

dict.get（）

尝试从第二个列表中获取一个值：

dict_b = {i: j for i,j in b}
new_a = [(i, dict_b.get(i, j)) for i, j in a]

您也不必对列表重新排序。这将导致运行时间稍好一些：

def func_1():
    dict_b = {i: j for i,j in b}
    return [(i, dict_b.get(i, j)) for i, j in a]

%timeit func_1()

2.48 µs ± 506 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)


def func_2():
    da = {k:v for k,v in a}
    db = {k:v for k,v in b}
    da.update(db)

    return sorted((k,v) for k,v in da.items())

%timeit func_2()

3.07 µs ± 71.4 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

另一个解决方案：

>>> a = [(1, 'a'), (2, ['00:00:00,280', '00:00:09,680\n']), (3, 'b'), (4, 'c'), (5, 'd'), (6, ['00:00:08,760', '00:00:13,309\n']), (7, 'e'), (8, 'f'), (9, 'g'), (10, ['00:00:09,680', '00:00:15,630\n'])]
>>> b = [(2, ['00:00:00,280', '00:00:08,760']), (6, ['00:00:08,760', '00:00:09,680']), (10, ['00:00:09,680', '00:00:13,309'])]
>>> seen = set()
>>> sorted(v for v in b+a if not (v[0] in seen or seen.add(v[0])))
[(1, 'a'), (2, ['00:00:00,280', '00:00:08,760']), (3, 'b'), (4, 'c'), (5, 'd'), (6, ['00:00:08,760', '00:00:09,680']), (7, 'e'), (8, 'f'), (9, 'g'), (10, ['00:00:09,680', '00:00:13,309'])]

<>代码>元素> < <代码> >元素> < < />代码>（<代码> >在B+A < /CODE >中），并使用该技巧获得鬼影顺序中的唯一元素：如果<代码> V [ 0 ]在可见< /代码>中，我们产生<代码> V< /代码>，否则我们添加<代码> v[ 0 ] < /代码> < <代码>参见<代码>（<代码>参见.Advd（0））。执行副作用，并且始终评估为

False

）

如果您对结果顺序不感兴趣：

>>> seen = set()
>>> [v for v in b+a if not (v[0] in seen or seen.add(v[0]))]
[(2, ['00:00:00,280', '00:00:08,760']), (6, ['00:00:08,760', '00:00:09,680']), (10, ['00:00:09,680', '00:00:13,309']), (1, 'a'), (3, 'b'), (4, 'c'), (5, 'd'), (7, 'e'), (8, 'f'), (9, 'g')]

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到目前为止，您尝试了什么？将列表转换为dicts；使用第二dict的第一dict更新；将第一个列表转换回列表；排序列表。您得到两个元素

（2，['00:00:00280'，'00:00:09680\n']），（2，['00:00:00280'，'00:00:08760']）

，但结果应该只有第二个值。是否有理由使用

{k:v表示k，v表示a}

而不是

dict（a）

？顺便说一句，

d={k:v代表k，a+b中的v}

应该有用。@jferard这是一种习惯-我通常使用它来创建dict并一步更改值-就像

{k:v.strip（）.split（，”）代表k，v在a}

：）现在我可以简单地使用

dict（a）

和

list（da.item（）

）不能进行两次投票，但

排序（dict（a+b）.items（））

就是答案。在Python>=3.6中，

sorted

可能是不必要的。@jferard我也希望在>=3.6中，它应该在没有

sorted（）

的情况下工作，但我在3.7上进行了测试，需要

sorted（）

很好地了解它。我对插入顺序的dicts不是很有信心。

>>> a = [(1, 'a'), (2, ['00:00:00,280', '00:00:09,680\n']), (3, 'b'), (4, 'c'), (5, 'd'), (6, ['00:00:08,760', '00:00:13,309\n']), (7, 'e'), (8, 'f'), (9, 'g'), (10, ['00:00:09,680', '00:00:15,630\n'])]
>>> b = [(2, ['00:00:00,280', '00:00:08,760']), (6, ['00:00:08,760', '00:00:09,680']), (10, ['00:00:09,680', '00:00:13,309'])]
>>> seen = set()
>>> sorted(v for v in b+a if not (v[0] in seen or seen.add(v[0])))
[(1, 'a'), (2, ['00:00:00,280', '00:00:08,760']), (3, 'b'), (4, 'c'), (5, 'd'), (6, ['00:00:08,760', '00:00:09,680']), (7, 'e'), (8, 'f'), (9, 'g'), (10, ['00:00:09,680', '00:00:13,309'])]

>>> seen = set()
>>> [v for v in b+a if not (v[0] in seen or seen.add(v[0]))]
[(2, ['00:00:00,280', '00:00:08,760']), (6, ['00:00:08,760', '00:00:09,680']), (10, ['00:00:09,680', '00:00:13,309']), (1, 'a'), (3, 'b'), (4, 'c'), (5, 'd'), (7, 'e'), (8, 'f'), (9, 'g')]