Question

下面，我有一个id列表和一个为每个id创建的URL数量列表，使用url_str作为创建的URL的基础。因此，对于身份id2，我预计会有三个网址。我用list comprehension完成了这项工作，这似乎是解决问题的最灵活方式。

但是，我现在正在学习python，并且我对map等同于列表理解感到好奇。

url_str = "www.amazon.com/gp/cdp/member-reviews/"
ids = ['id' + str(x) for x in xrange(1, 5)]
revs = [1, 3, 1, 8]

# ids output
['id1', 'id2', 'id3', 'id4']


# list comprehension
urls_compr = [url_str + id_r + "page=" + str(x)
              for id_r, page in zip(ids, revs)
              for x in xrange(1, page + 1)] 



# output of list comprehension
['www.amazon.com/gp/cdp/member-reviews/id1page=1',
 'www.amazon.com/gp/cdp/member-reviews/id2page=1',
 'www.amazon.com/gp/cdp/member-reviews/id2page=2',
 'www.amazon.com/gp/cdp/member-reviews/id2page=3',
 'www.amazon.com/gp/cdp/member-reviews/id3page=1',
 'www.amazon.com/gp/cdp/member-reviews/id4page=1',
 'www.amazon.com/gp/cdp/member-reviews/id4page=2',
 'www.amazon.com/gp/cdp/member-reviews/id4page=3',
 'www.amazon.com/gp/cdp/member-reviews/id4page=4',
 'www.amazon.com/gp/cdp/member-reviews/id4page=5',
 'www.amazon.com/gp/cdp/member-reviews/id4page=6',
 'www.amazon.com/gp/cdp/member-reviews/id4page=7',
 'www.amazon.com/gp/cdp/member-reviews/id4page=8']

理想情况下，我有一些效果：

urls_map = map(func, ids, revs)

urls_map
['www.amazon.com/gp/cdp/member-reviews/id1page=1',
 'www.amazon.com/gp/cdp/member-reviews/id2page=1',
 'www.amazon.com/gp/cdp/member-reviews/id2page=2',
 'www.amazon.com/gp/cdp/member-reviews/id2page=3',
 'www.amazon.com/gp/cdp/member-reviews/id3page=1',
 'www.amazon.com/gp/cdp/member-reviews/id4page=1',
 'www.amazon.com/gp/cdp/member-reviews/id4page=2',
 'www.amazon.com/gp/cdp/member-reviews/id4page=3',
 'www.amazon.com/gp/cdp/member-reviews/id4page=4',
 'www.amazon.com/gp/cdp/member-reviews/id4page=5',
 'www.amazon.com/gp/cdp/member-reviews/id4page=6',
 'www.amazon.com/gp/cdp/member-reviews/id4page=7',
 'www.amazon.com/gp/cdp/member-reviews/id4page=8']

我只是不确定func部分。

Answer 1

请记住，如果您希望将函数映射到序列的每个元素，则map在列表理解存在之前就已存在，例如：

map(func, seq)

相当于：

[func(x) for x in seq]

尽管在python 3中map返回一个迭代器，所以它更靠近(func(x) for x in seq)，用圆括号表示生成器。

列表理解还带来了额外的好处，即能够使用表达式以及/而不是callables。

例如，如果你想为列表的每个元素添加一个（+1），你可以这样做：

[x + 1 for x in seq]

使用map重写它，你需要创建一个可以在参数中添加一个的callable：

map(lambda x:x+1, seq)

更进一步，在列表推导中使用多个变量非常简单：

[x+y for x,y in itertools.product(range(5), range(7))]

但是map的可调用只需要一个参数，所以你需要使用元组将这些元素打包在一起：

map((lambda pair: pair[0] + pair[1]), itertools.product(range(5),range(7)))

你的情况更容易引起混淆，因为序列只是用嵌套的for循环构建的，所以你很可能最终会使用生成器表达式来完成序列：

map((lambda x: x[0] + x[1] + "page=" + str(x[2])), ((url_str, id_r, x) for id_r, page in zip(ids, revs) for x in xrange(1, page + 1)))

这显然不是map的目的，我建议你坚持使用列表理解，因为它是适合你工作的工具。

Answer 2

我不认为map会在这里以一种很好的方式发挥作用，但在我放弃它之后，我至少能够将列表组合简化为更具可读性和效率的东西： / p>

In [31]: ids = xrange(1, 5)

In [32]: pages = (1, 3, 1, 8)

In [33]: pages_per_id = zip(ids, pages)

In [34]: url_str
Out[34]: 'www.amazon.com/gp/cdp/member-reviews/id%s?page=%s'

In [35]: [url_str % (id, page_no) for id, pages in pages_per_id for page_no in range(1, pages+1)]
Out[35]:
['www.amazon.com/gp/cdp/member-reviews/id1?page=1',
 'www.amazon.com/gp/cdp/member-reviews/id2?page=1',
 'www.amazon.com/gp/cdp/member-reviews/id2?page=2',
 'www.amazon.com/gp/cdp/member-reviews/id2?page=3',
 'www.amazon.com/gp/cdp/member-reviews/id3?page=1',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=1',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=2',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=3',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=4',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=5',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=6',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=7',
 'www.amazon.com/gp/cdp/member-reviews/id4?page=8']

Answer 3

我不想删除这个问题，因为有人能够在上面的评论中回答我的问题。

 map((lambda x: x[0] + x[1] + "page=" + str(x[2])), ((url_str, id_r, x) for id_r, page in zip(ids, revs) for x in xrange(1, page + 1)))

使用zip映射等效的嵌套列表理解

3 个答案: