如何计算字符串列表中的元素

Question

If a have a list like this:

[['welcome','a1'],['welcome','a1'],['hello','a2'],['hello','a3']]

and I want to return something like this:

[['welcome','a1', 2],['hello','a2', 1],['hello','a3', 1]]

If the same pair of strings in a sublist is encountered, increment the count

What I have so far:

counter = 0
for i in mylist:
  counter += 1 
  if i[0]== i[0]:
    if i[1] == i[1]:
        counter -= 1
 ouptut.append([mylist, counter])

I'm new at this and I appreciate your help!

Answer 1

Use a set here to get only unique items:

>>> lis = [['welcome','a1'],['welcome','a1'],['hello','a2'],['hello','a3']] 
>>> [list(x) + [1] for x in set(map(tuple, lis))]
>>> [['welcome', 'a1', 1], ['hello', 'a3', 1], ['hello', 'a2', 1]]

Explanation:

Set always returns unique items from an iterable or iterator, but as sets can only contain immutable item so you should convert them to a tuple first. A verbose version of the above code, only difference is that will also preserve the original or

>>> lis = [['welcome','a1'],['welcome','a1'],['hello','a2'],['hello','a3']] 
>>> s = set()
>>> for item in lis:
...     tup = tuple(item)  #covert to tuple
...     s.add(tup)
>>> s
set([('welcome', 'a1'), ('hello', 'a3'), ('hello', 'a2')])

Now use a list comprehension to get the expected output:

>>> [list(item) + [1] for item in s]
[['welcome', 'a1', 1], ['hello', 'a3', 1], ['hello', 'a2', 1]]

If the order of items matter( sets don't preserve order), then use this:

>>> seen = set()
>>> ans = []
>>> for item in lis:
...     tup = tuple(item)
...     if tup not in seen:
...         ans.append(item + [1])
...         seen.add(tup)
...         
>>> ans
[['welcome', 'a1', 1], ['hello', 'a2', 1], ['hello', 'a3', 1]]

I am not sure what's the point of using 1 here.