如何编写列表理解，包括 If Else 条件

Question

什么是简单的列表理解（不使用任何新模块或字典）以获得如下输出：

[1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]  

这里0每个 s 之间的1s 数量从 0 增加到 10 倍。

在不使用列表理解的情况下，我使用简单的 for 循环得到了输出：

for i in range(12):
    if i==0:
        print(1, end=", ")
    else:
        for j in range(i):
            if i==j+1:
                print(1, end=", ")
            else:
                print(0, end=", ")

只是想看看如何将这些循环转换为列表理解

Answer 1

您可以使用以下列表推导式生成1后面跟越来越多的0' 的列表：

[[1] + [0] * i for i in range(n)]

对于n = 4，这将产生：

[[1], [1, 0], [1, 0, 0], [1, 0, 0, 0]]

您可以通过将其嵌套在另一个理解中来展平该列表，然后添加尾随1：

res = [i for sub in [[1] + [0] * i for i in range(n)] for i in sub] + [1]

Again, for n = 4 this produces:

[1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1]

If you can use libraries, you can use itertools.chain.from_iterable to flatten the list:

res = list(itertools.chain.from_iterable([1] + [0] * i for i in range(n))) + [1]

The output will be the same.

As pointed out by @KellyBundy in the comments, the need for the trailing 1 can be removed by changing the innermost comprehension in the above code to

[0] * i + [1] for i in range(-1, n)

This makes use of the fact that [0] * n = [] for n <= 0. For n = 4 this directly produces

[1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1]