CODEWITHSUNDEEP
pythonlist
PDavis
PDavis

Reputation: 25

Combining tuple elements from a list of tuples. Tuple elements are tuple and list

I'm relatively new with Python and am having a hard time coming up with a way to make the following work. Currently, I have a list of tuples (larger than list_one. The elements inside these tuples are 1) a tuple, and 2) a string (these strings are always one of two values: in the example 'A' or 'B' (can also be 'D' or 'Z', or any other two strings). The inner tuples include an 'id' and a list of values (strings):

list_one = [(('id1', ['v1', 'v2']), 'A'), (('id2', ['v3', 'v4', 'v5', 'v6']), 'A'), (('id3', ['v11']), 'B'), (('id4', ['v12', 'v13']), 'B'), (('id5', ['v14', 'v16']), 'B'), (('id6', ['v17', 'v18', 'v21']), 'A')]

I'm trying, unsuccessfully, to create a new list that looks as follows:

new_list1 = [('id1',['v1', 'v2', 'v3', 'v4', 'v5', 'v6']), ('id3',['v11', 'v12', 'v13', 'v14', 'v16']), ('id6',['v17', 'v18', 'v21'])]

When the string value (A, B) changes it should create a new tuple with the first element being the id of the initial tuple where the A,B changes, and the second element being a list of the values ("v's") inside the initial lists. A list if dictionaries would also be fine, e.g:

new_list2 = [{'id1': ['v1','v2', 'v3', 'v4', 'v5', 'v6']}, {'id3':['v11', 'v12', 'v13', 'v14', 'v16']}, {'id6':['v17', 'v18', 'v21']}]

I appreciate any guidance.

edit:

s = list_one[0][1]
for tup in list_one:
    if s in tup:
        new_list.extend(tup[0][1])
    else:
        new_list.append(tup[0][1])

gives:

 ['v1', 'v2', 'v3', 'v4', 'v5', 'v6', ['v11'], ['v12', 'v13'], ['v14', 'v16'], 'v17', 'v18', 'v21']

which is not really what I'm looking for, as this creates a list around one the 'A' value, with lists of 'B' inside.

Upvotes: 2

Views: 340

Answers (2)

Miro Rodozov
Miro Rodozov

Reputation: 307

Edit

This solves a related question, whereby all the lists corresponding to 'A' and 'B' values are aggregated, and the id is chose of the total first id for each group.

Here you go 

letterKeyToFirstIDMap = {}
result_dict = {}

for tpl in list_one:

  id = tpl[0][0]
  lst = tpl[0][1]
  letter_id = tpl[1]
  if not letter_id in letterKeyToFirstIDMap: 
    letterKeyToFirstIDMap.update({letter_id: id})
    result_dict.update({id:[]})

  result_dict[letterKeyToFirstIDMap[letter_id]] += lst

print result_dict

What is done here is that for each letter key (say A) you want only the first id that is met (say id1) to be the key for the following list. You create a map to store the letterKey and its idKey the first time the pair appears and then just check if the letter key already exists in that map, using it's corresponding idKey if it does, and update the list.

Upvotes: 1

Ami Tavory
Ami Tavory

Reputation: 76297

You can basically loop along, remembering what is the current key (initialized to None). 

key = None # This is 'A' or 'B' in your example, but it starts off as None
new_list = [] # This holds the final result
for r, l in [(r, l) for (l, r) in list_one]: 
    if r != key: # If the current is not the key, time to switch to a new one
        id_ = l[0]
        new_list.append({id_: []})
    new_list[-1][id_].extend(l[1]) # Extend the current list
    key = r # Make it the current key in any case.

>>> new_list
[{'id1': ['v1', 'v2', 'v3', 'v4', 'v5', 'v6']},
 {'id3': ['v11', 'v12', 'v13', 'v14', 'v16']},
 {'id6': ['v17', 'v18', 'v21']}]

Upvotes: 1

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