Two-keys dictionary into one key dictionary of lists-CodePudding

I am trying to implement a simple task. I have a dictionary with keys (ti, wi)

y={('t1', 'w1'): 1, ('t2', 'w1'): 2, ('t3', 'w1'): 3, ('t1', 'w2'): 4, ('t2', 'w2'): 5, ('t3', 'w2'): 6}

I want to create a new dictionary where keys will be wi, and value is a list of all ti. So I want to have an output dictionary like:

{'w1': [1, 2, 3], 'w2': [4, 5, 6]}

I wrote the following code:

y={('t1', 'w1'): 1, ('t2', 'w1'): 2, ('t3', 'w1'): 3, ('t1', 'w2'): 4, ('t2', 'w2'): 5, ('t3', 'w2'): 6} 
y_w={}  
y_t=[]    
for w in range(1,3):
    y_t.clear()
    for t in range(1,4):
        print('t= ', t, 'w= ', w, 'y=' , y['t{0}'.format(t), 'w{0}'.format(w)])
        y_t.append(y['t{0}'.format(t), 'w{0}'.format(w)])            
    print(y_t)
    y_w['w{0}'.format(w)]=y_t
print(y_w)

But the result I am getting is

{'w1': [4, 5, 6], 'w2': [4, 5, 6]}

I can not understand where the first list disappeared? Can someone help me explain where I am wrong? Is there a nicer way to do it, maybe without for lops?

CodePudding user response：

Your problem lies in the assumption that setting the value in the dictionary somehow freezes the list.

It's no accident the lists have the same values: They are identical, two pointers to the same list. Observe:

>>> a_dict = {}
>>> a_list = []
>>> a_list.append(23)
>>> a_dict["a"] = a_list
>>> a_list.clear()
>>> a_list.append(42)
>>> a_dict["b"] = a_list
>>> a_dict
{'a': [42], 'b': [42]}

You could fix your solution by replacing y_t.clear() with y_t = [], which does create a new list:

y = {('t1', 'w1'): 1, ('t2', 'w1'): 2, ('t3', 'w1'): 3, ('t1', 'w2'): 4, ('t2', 'w2'): 5, ('t3', 'w2'): 6} 
y_w = {}      
for w in range(1,3):
    y_t = []
    for t in range(1,4):
        print('t= ', t, 'w= ', w, 'y=' , y['t{0}'.format(t), 'w{0}'.format(w)])
        y_t.append(y['t{0}'.format(t), 'w{0}'.format(w)])            
    print(y_t)
    y_w['w{0}'.format(w)]=y_t
print(y_w)

But there are, as you suspect, easier ways of doing this, for example the defaultdict solution shown by Riccardo Bucco.

CodePudding user response：

Try this:

from collections import defaultdict

d = defaultdict(list)
for k, v in y.items():
    d[k[1]].append(v)
d = dict(d)

CodePudding user response：

The line number 10 is causing the problem, if you replace it with y_t = [] it will work as you expect

CodePudding user response：

You could first find all unique keys:

unique_keys = set(list(zip(*k))[1])

and then create the dict with list-values using those:

{u: [v for k, v in y.items() if k[1] == u] for u in unique_keys}

CodePudding user response：

According to your output here's what you can try:

y = {('t1', 'w1'): 1, ('t2', 'w1'): 2, ('t3', 'w1'): 3, ('t1', 'w2'): 4, ('t2', 'w2'): 5, ('t3', 'w2'): 6}

def new_dict_with_keys(dictionary):
    new_dictionary = dict()

    # Go through the dictionary keys to read each key's value
    for tuple_key in dictionary: 
        if "w1" in tuple_key or "w2" in tuple_key:
            
            # Determine which key to use
            if "w1" in tuple_key:
                key = "w1"  
            else:
                key = "w2"

            # Check if the new dictionary has the "w1" or "w2" as a an item
            # If it does not, create a new list
            if new_dictionary.get(key) is None:
                new_dictionary[key] = list()
            
            # Append the value in the respective key
            new_dictionary[key].append(dictionary[tuple_key])
    
    # Return the dictionary with the items
    return new_dictionary

print(new_dict_with_keys(y))
# Prints: {'w1': [1, 2, 3], 'w2': [4, 5, 6]}

CodePudding user response：

Here's a solution using itertools.groupby:

import itertools as it
from operator import itemgetter

items = sorted((k, v) for (_, k), v in y.items())
groups = it.groupby(items, key=itemgetter(0))
result = {k: [v for _, v in vs] for k, vs in groups}

# {'w1': [1, 2, 3], 'w2': [4, 5, 6]}