0

```
l1 = [('x', [1]), ('y', [1]) ('z', [2]]
l2 = ['1', '2', '3']
```

Desired output, assuming that '1' can be done by 'x' and 'y', and 'z' can do '2':

```
l3 = [[['1'['x', 'y']], ['2'['z']], ['3'[]]
```

So then i can do an untie between 'x' and 'y' using aditional paramenters that for the sake of redability i wont be putting here unless asked.

This is what i've come to so far:

```
x = []
for i in l2:
for j in l1:
if i in j[1][0]:
x.append(j[0])
```

This gives me the possible l1, but i'm having no luck in adding the l2 part in a way i can put 'x' and 'y' toghether. Since this a school project, i can't use tools like zip.

# Answer

1

Build a `dict`

with the keys being the elements of `l2`

, initializing their values as a list, then iterate over `l1`

elements appending `element[0]`

to the list of the key that matches `element[1]`

.

```
l3 = {}
for elem in l2:
l3[elem] = []
# Look for matches in l1
for letter, tasks in l2:
for task in tasks:
l3[str(task)].append(letter)
```

In this algorithm, `l3`

will be a `dict`

instead of `list`

which makes it easier to access the values in each task. If you really need to be a `list`

, just convert it to one.

Answer author Lucasnadalutti

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