将多个列表压缩减少为单个语句

Question

寻找减少列表的压缩,减少循环和内存使用,有一些方法可以减少两个循环来构建最终路径,将其转换为单个列表压缩？

def build_paths(domains):

    http_paths  = ["http://%s"   % d for d in domains]
    https_paths = ["https://%s" % d for d in domains] 

    paths = []                                         
    paths.extend(http_paths)                           
    paths.extend(https_paths)

    return paths                          

在这种情况下,预期的结果是一个优化的列表comprehesion,从三个列表的引用(减少http_paths,https_paths,paths)成一个单一的一条线,如以下示例所示的结构:

def build_paths(domains):
    return [<reduced list comprehesion> for d in domains]

在这两种情况下,运行以下测试:

domains = ["www.ippssus.com",                      
           "www.example.com",                      
           "www.mararao.com"]                      

print(build_paths(domains))

预期输出,与列表顺序无关:

< ['http://www.ippssus.com', 'http://www.example.com', 'http://www.tetsest.com', 'https://www.ippssus.com', 'https://www.example.com', 'https://www.tetsest.com']

Answer 1

添加第二个循环:

['%s://%s' % (scheme, domain) for scheme in ('http', 'https') for domain in domains]

这将http首先构建所有URL,然后构建httpsURL,就像您的原始代码一样.