>>> l = [['40', '20', '10', '30'], ['20', '20', '20', '20', '20', '30', '20'], ['30', '20', '30', '50', '10', '30', '20', '20', '20'], ['100', '100'], ['100', '100', '100', '100', '100'], ['100', '100', '100', '100']]
>>> new_list = [float(x) for xs in l for x in xs]
>>> new_list
[40.0, 20.0, 10.0, 30.0, 20.0, 20.0, 20.0, 20.0, 20.0, 30.0, 20.0, 30.0, 20.0, 30.0, 50.0, 10.0, 30.0, 20.0, 20.0, 20.0, 100.0, 100.0, 100.0, 100.0, 100.0, 100.0, 100.0, 100.0, 100.0, 100.0, 100.0]

下面是如何使用嵌套的列表理解来做到这一点:

[[float(y) for y in x] for x in l]

这将给你一个列表的列表，类似于你开始的，只是用浮点数而不是字符串。

如果你想要一个平面列表，那么你可以使用

[float(y) for x in l for y in x]

注意这个循环顺序，l中的x在这个循环中排在前面。

是的，你可以用这样的代码:

l = [[float(y) for y in x] for x in l]

如果你不喜欢嵌套的列表推导式，你也可以使用map函数，

>>> from pprint import pprint

>>> l = l = [['40', '20', '10', '30'], ['20', '20', '20', '20', '20', '30', '20'], ['30', '20', '30', '50', '10', '30', '20', '20', '20'], ['100', '100'], ['100', '100', '100', '100', '100'], ['100', '100', '100', '100']] 

>>> pprint(l)
[['40', '20', '10', '30'],
['20', '20', '20', '20', '20', '30', '20'],
['30', '20', '30', '50', '10', '30', '20', '20', '20'],
['100', '100'],
['100', '100', '100', '100', '100'],
['100', '100', '100', '100']]

>>> float_l = [map(float, nested_list) for nested_list in l]

>>> pprint(float_l)
[[40.0, 20.0, 10.0, 30.0],
[20.0, 20.0, 20.0, 20.0, 20.0, 30.0, 20.0],
[30.0, 20.0, 30.0, 50.0, 10.0, 30.0, 20.0, 20.0, 20.0],
[100.0, 100.0],
[100.0, 100.0, 100.0, 100.0, 100.0],
[100.0, 100.0, 100.0, 100.0]]

在我看来，最好的方法是使用python的itertools包。

>>>import itertools
>>>l1 = [1,2,3]
>>>l2 = [10,20,30]
>>>[l*2 for l in itertools.chain(*[l1,l2])]
[2, 4, 6, 20, 40, 60]

不确定您想要的输出是什么，但如果您使用的是列表推导式，则顺序遵循嵌套循环的顺序，这是向后的。我拿到了你想要的东西

[float(y) for x in l for y in x]

其原则是:使用与编写嵌套for循环相同的顺序。

这个问题可以不使用for循环来解决。单行代码就足够了。使用带有lambda函数的嵌套映射也可以在这里工作。

l = [['40', '20', '10', '30'], ['20', '20', '20', '20', '20', '30', '20'], ['30', '20', '30', '50', '10', '30', '20', '20', '20'], ['100', '100'], ['100', '100', '100', '100', '100'], ['100', '100', '100', '100']]

map(lambda x:map(lambda y:float(y),x),l)

输出列表如下:

[[40.0, 20.0, 10.0, 30.0], [20.0, 20.0, 20.0, 20.0, 20.0, 30.0, 20.0], [30.0, 20.0, 30.0, 50.0, 10.0, 30.0, 20.0, 20.0, 20.0], [100.0, 100.0], [100.0, 100.0, 100.0, 100.0, 100.0], [100.0, 100.0, 100.0, 100.0]]

下面是如何将嵌套for循环转换为嵌套列表推导式:

下面是嵌套列表理解的工作原理:

            l a b c d e f
            ↓ ↓ ↓ ↓ ↓ ↓ ↓
In [1]: l = [ [ [ [ [ [ 1 ] ] ] ] ] ]
In [2]: for a in l:
   ...:     for b in a:
   ...:         for c in b:
   ...:             for d in c:
   ...:                 for e in d:
   ...:                     for f in e:
   ...:                         print(float(f))
   ...:                         
1.0

In [3]: [float(f)
         for a in l
   ...:     for b in a
   ...:         for c in b
   ...:             for d in c
   ...:                 for e in d
   ...:                     for f in e]
Out[3]: [1.0]

对于你的例子，如果你想要一个平面列表，它会是这样的。

In [4]: new_list = [float(y) for x in l for y in x]

我想分享一下列表推导式实际上是如何工作的，特别是对于嵌套的列表推导式:

new_list= [float(x) for x in l]

其实和:

new_list=[]
for x in l:
    new_list.append(float(x))

现在对于嵌套列表的理解:

[[float(y) for y in x] for x in l]

等于:

new_list=[]
for x in l:
    sub_list=[]
    for y in x:
        sub_list.append(float(y))

    new_list.append(sub_list)

print(new_list)

输出:

[[40.0, 20.0, 10.0, 30.0], [20.0, 20.0, 20.0, 20.0, 20.0, 30.0, 20.0], [30.0, 20.0, 30.0, 50.0, 10.0, 30.0, 20.0, 20.0, 20.0], [100.0, 100.0], [100.0, 100.0, 100.0, 100.0, 100.0], [100.0, 100.0, 100.0, 100.0]]

我有一个类似的问题要解决，所以我遇到了这个问题。我对Andrew Clark和narayan的答案做了一个性能比较，我想和大家分享一下。

两个答案之间的主要区别是它们如何遍历内部列表。其中一个使用内置映射，而另一个使用列表理解。如果Map函数不需要使用lambdas，则它的等效列表理解具有轻微的性能优势。所以在这个问题的背景下，地图的表现应该比列表理解略好。

让我们做一个性能基准测试，看看它是否真实。我使用python 3.5.0版本来执行所有这些测试。在第一组测试中，我想保持每个列表的元素为10，并改变列表的数量从10-100,000

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,10))]*10]"
>>> 100000 loops, best of 3: 15.2 usec per loop   
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,10))]*10]"
>>> 10000 loops, best of 3: 19.6 usec per loop 

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,10))]*100]"
>>> 100000 loops, best of 3: 15.2 usec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,10))]*100]"
>>> 10000 loops, best of 3: 19.6 usec per loop 

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,10))]*1000]"
>>> 1000 loops, best of 3: 1.43 msec per loop   
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,10))]*1000]"
>>> 100 loops, best of 3: 1.91 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,10))]*10000]"
>>> 100 loops, best of 3: 13.6 msec per loop   
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,10))]*10000]"
>>> 10 loops, best of 3: 19.1 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,10))]*100000]"
>>> 10 loops, best of 3: 164 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,10))]*100000]"
>>> 10 loops, best of 3: 216 msec per loop

在下一组测试中，我想将每个列表的元素数量提高到100。

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,100))]*10]"
>>> 10000 loops, best of 3: 110 usec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,100))]*10]"
>>> 10000 loops, best of 3: 151 usec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,100))]*100]"
>>> 1000 loops, best of 3: 1.11 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,100))]*100]"
>>> 1000 loops, best of 3: 1.5 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,100))]*1000]"
>>> 100 loops, best of 3: 11.2 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,100))]*1000]"
>>> 100 loops, best of 3: 16.7 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,100))]*10000]"
>>> 10 loops, best of 3: 134 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,100))]*10000]"
>>> 10 loops, best of 3: 171 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,100))]*100000]"
>>> 10 loops, best of 3: 1.32 sec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,100))]*100000]"
>>> 10 loops, best of 3: 1.7 sec per loop

让我们迈出勇敢的一步，将列表中的元素数量修改为1000

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,1000))]*10]"
>>> 1000 loops, best of 3: 800 usec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,1000))]*10]"
>>> 1000 loops, best of 3: 1.16 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,1000))]*100]"
>>> 100 loops, best of 3: 8.26 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,1000))]*100]"
>>> 100 loops, best of 3: 11.7 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,1000))]*1000]"
>>> 10 loops, best of 3: 83.8 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,1000))]*1000]"
>>> 10 loops, best of 3: 118 msec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,1000))]*10000]"
>>> 10 loops, best of 3: 868 msec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,1000))]*10000]"
>>> 10 loops, best of 3: 1.23 sec per loop

>>> python -m timeit "[list(map(float,k)) for k in [list(range(0,1000))]*100000]"
>>> 10 loops, best of 3: 9.2 sec per loop
>>> python -m timeit "[[float(y) for y in x] for x in [list(range(0,1000))]*100000]"
>>> 10 loops, best of 3: 12.7 sec per loop

从这些测试中，我们可以得出这样的结论:在这种情况下，map比列表理解具有更好的性能。这也适用于试图强制转换为int或str的情况。对于数量较少且每个列表的元素较少的列表，差异可以忽略不计。对于每个列表有更多元素的更大的列表，人们可能喜欢使用map而不是列表解析，但这完全取决于应用程序的需要。

然而，我个人认为列表理解比映射更具有可读性和惯用性。它是python中事实上的标准。通常人们更熟练和舒适(特别是初学者)使用列表理解比地图。

    deck = [] 
    for rank in ranks:
        for suit in suits:
            deck.append(('%s%s')%(rank, suit))

这可以通过列表理解来实现:

[deck.append((rank,suit)) for suit in suits for rank in ranks ]

是的，你可以这样做。

[[float(y) for y in x] for x in l]

如果需要一个扁平的列表:

[y for x in l for y in x]

如果需要一个嵌套列表(list In list):

[[float(y) for y in x] for x in l]