Question

我有我国家的报纸网站的网址和标题列表。作为一般示例：

x = ['URL1','news1','news2','news3','URL2','news1','news2','URL3','news1']

每个URL元素都有一个对应的'news'元素序列，它们的长度可以不同。在上面的示例中，URL1有3条对应的新闻，URL3只有一条。

有时，URL没有相应的“新闻”元素：

y = ['URL4','news1','news2','URL5','URL6','news1']

我可以轻松找到每个URL索引以及每个URL的“新闻”元素。

我的问题是：是否可以将此列表转换为以URL元素为键而“ news”元素为列表/元组值的字典？

预期产量

z = {'URL1':('news1', 'news2', 'news3'),
     'URL2':('news1', 'news2'),
     'URL3':('news1'),
     'URL4':('news1', 'news2'),
     'URL5':(),
     'URL6':('news1')}

我在此post中看到了类似的问题，但并不能解决我的问题。

Answer 1

您可以这样做：

>>> y = ['URL4','news1','news2','URL5','URL6','news1']
>>> result = {}
>>> current_url = None
>>> for entry in y:
...     if entry.startswith('URL'):
...         current_url = entry
...         result[current_url] = ()
...     else:
...         result[current_url] += (entry, )
...         
>>> result
{'URL4': ('news1', 'news2'), 'URL5': (), 'URL6': ('news1',)}

Answer 2

您可以将itertools.groupby与key函数一起使用来标识URL：

from itertools import groupby
def _key(url):
    return url.startswith("URL") #in the body of _key, write code to identify a URL

data = ['URL1','news1','news2','news3','URL2','news1','news2','URL3','news1', 'URL4','news1','news2','URL5','URL6','news1']
new_d = [list(b) for _, b in groupby(data, key=_key)]
grouped = [[new_d[i], tuple(new_d[i+1])] for i in range(0, len(new_d), 2)]
result = dict([i for [*c, a], b in grouped for i in [(i, ()) for i in c]+[(a, b)]])

输出：

{
 'URL1': ('news1', 'news2', 'news3'), 
 'URL2': ('news1', 'news2'), 
 'URL3': ('news1',), 
 'URL4': ('news1', 'news2'), 
 'URL5': (), 
 'URL6': ('news1',)
}

Answer 3

您可以只使用列表中URL密钥的索引并获取索引之间的内容并分配给第一个

赞：

x = ['URL1','news1','news2','news3','URL2','news1','news2','URL3','news1']
urls = [x.index(y) for y in x if 'URL' in y]
adict = {}
for i in range(0, len(urls)):
    if i == len(urls)-1:
        adict[x[urls[i]]] = x[urls[i]+1:len(x)]
    else:
        adict[x[urls[i]]] = x[urls[i]+1:urls[i+1]]
print(adict)

输出：

{'URL1': ['news1', 'news2', 'news3'], 'URL2': ['news1', 'news2'], 'URL3': ['news1']}

Answer 4

more-itertools library包含一个函数split_before()，为此目的，它非常方便：

{s[0]: tuple(s[1:]) for s in mt.split_before(x, lambda e: e.startswith('URL'))}

我认为这比在此之前发布的答案中的任何其他方法都更干净，但是它确实引入了外部依赖关系（除非您重新实现该功能），这使其不适用于每种情况。

如果您的实际用例涉及真实的URL或其他内容，而不是URL#形式的字符串，则只需用lambda e: e.startswith('URL')替换为可以用来选择键值之外的任何键值的任何函数元素。

Answer 5

使用groupby（单线）的另一种解决方案：

x = ['URL1','news1','news2','news3','URL2','news1','news2','URL3','news1', 'URL4','news1','news2','URL5','URL6','news1']

from itertools import groupby

out = {k: tuple(v) for _, (k, *v) in groupby(x, lambda k, d={'g':0}: (d.update(g=d['g']+1), d['g']) if k.startswith('URL') else (None, d['g']))}

from pprint import pprint
pprint(out)

打印：

{'URL1': ('news1', 'news2', 'news3'),
 'URL2': ('news1', 'news2'),
 'URL3': ('news1',),
 'URL4': ('news1', 'news2'),
 'URL5': (),
 'URL6': ('news1',)}

如何使用单个列表创建字典？

5 个答案: