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I have a list of url's and headers from a newspaper site in my country. As a general example:

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

Each URL element has a corresponding sequence of 'news' elements, which can differ in length. In the example above, URL1 has 3 corresponding news and URL3 has only one.

Sometimes a URL has no corresponding "news" element:

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

I can easily find every URL index and the "news" elements of each URL.

My question is: Is it possible to transform this list into a dictionary in which the URL element is the key and the "news" elements are a tuple-value?

Expected Output

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

I've seen a similar question in this post, but it doesn't solve my problem.

You can do it like this:

>>> 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',)

You can use itertools.groupby with a key function to identify a 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)]])

Output:

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

You can just use the indices of the URL keys in the list and grab what is between the indices and assign to the first

Like this:

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)

output:

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

Another solution using groupby, one-liner:

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)

Prints:

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

The more-itertools library contains a function split_before() which comes in very handy for this purpose:

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

I think this is cleaner than any of the other approaches in answers posted before this one, but it does introduce an external dependency (unless you reimplement the function), which makes it not appropriate for every situation.

If your actual use case involves real URLs or something else, rather than strings of the form URL#, then just replace lambda e: e.startswith('URL') with whatever function you can use to select the key elements apart from the value elements.