我有这样一个嵌套列表:
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']]
我想把l中的每个元素都转换为float。我有这样的代码:
newList = []
for x in l:
for y in x:
newList.append(float(y))
如何用嵌套的列表理解来解决这个问题?
另请参阅:如何从列表理解而不是嵌套列表中获得平坦的结果?
当前回答
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 ]
其他回答
>>> 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]
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]
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]
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。
是的，你可以这样做。
[[float(y) for y in x] for x in l]
