Python Group List of Lists

Question

I have a list of lists in the following format:

[['Sarah', '12', 'Chocolate'],
 ['Anders', '11', 'Vanilla'],
 ['Sarah', '13', 'Strawberry'],
 ['John', '11', 'None'],
 # ...
]

And I want to group the sublists as follows:

[['Sarah', '12', 'Chocolate', '13', 'Strawberry'],
 ['Anders', '11', 'Vanilla'],
 ['John', '11', 'None'],
 # ...
]

Where I group by the first item of the sublists and order by the second (so Sarahs with age 12 come before Sarahs with age 13).

How to do this?

You probably want to preserve the (name, age, flavor) tuples, and put the tuples in a list, rather than concatenating the lists together. I'm guessing you actually want to sort the list using sorted — Filip Haglund, Oct 09 '17 at 08:08

Eric Duminil · Answer 1 · 2017-10-09T10:16:58.983

2

You didn't show any code, so I won't give a complete solution.

One good data structure would be to use a dict, with names as keys and a list of tuples as values:

data =[['Sarah', '12', 'Chocolate'],
    ['Anders', '11', 'Vanilla'],
    ['Sarah', '13', 'Strawberry'],
    ['John', '11', 'None']]

grouped = {}

for name, x, y in data:
    grouped.setdefault(name, []).append((x,y))

print(grouped)
# {'Sarah': [('12', 'Chocolate'), ('13', 'Strawberry')], 'Anders': [('11', 'Vanilla')], 'John': [('11', 'None')]}

You'd just need to sort the values.

edited Oct 09 '17 at 10:16

answered Oct 09 '17 at 08:09

Eric Duminil

52,989
9
71
124

Rather than use `not dict.get(key)`, use `key not in dict`. And for adding a list, instead of `if : ... else: ...` you can use `dict.setdefault(key, []).append(...)`. – Martijn Pieters Oct 09 '17 at 09:20
@MartijnPieters: Excellent comment, thanks. You're also correct about `setdefault`. Since OP didn't show any code, I didn't want to use `defaultdict`, `groupby` or `dict.setdefault` in order to keep the example as simple as possible. – Eric Duminil Oct 09 '17 at 10:14
I don't see why `setdefault` is not simple. If you are going to answer, use best practices. Why dumb things down? – Martijn Pieters Oct 09 '17 at 10:15
@MartijnPieters: Modified. What bothers me about about `dict.setdefault` is that it's used as a getter, which can be surpising. – Eric Duminil Oct 09 '17 at 10:20
It's no more surprising than using `defaultdict`, in my opinion. – Martijn Pieters Oct 09 '17 at 10:22
@MartijnPieters: I probably shouldn't argue with you about Python :). But here goes : `d = defaultdict(list); d['not_here']` It's pretty clear that it returns `[]`. Instead, knowing that assignments return `None` in Python, I'd expect `dict.setdefault('not_here', [])` to also return `None`. – Eric Duminil Oct 09 '17 at 10:34
Assignments don't return anything in Python, because assignments are statements, not expressions. `dict.setdefault(...)` is an expression, so returns *something*, always. It is documented as returning the current value for the key, using the second value as a default if there is no such key (setting the key-value pair in the process). – Martijn Pieters Oct 09 '17 at 11:55
@MartijnPieters: Yes, it's documented. All I'm saying is that the method name isn't very consistent with its behaviour. – Eric Duminil Oct 09 '17 at 11:59

score 0 · Answer 2 · answered Oct 09 '17 at 08:38

A few possible solutions jump to mind, but here's one. Read through the comments and feel free to ask questions:

from collections import defaultdict

l = [
        ['Sarah', '12', 'Chocolate'],
        ['Anders', '11', 'Vanilla'],
        ['Sarah', '13', 'Strawberry'],
        ['John', '11', 'None'],
]

d = defaultdict(list)

for entry in l:
    d[entry[0]].append(entry[1:])

# Here d looks something like this:
# {'Sarah': [['12', 'Chocolate'], ['13', 'Strawberry']], ...}

result = [
    # 1. Sort the list by the first element of each sublist (parsed as an int).
    # 2. Flatten the result as in https://stackoverflow.com/questions/952914/making-a-flat-list-out-of-list-of-lists-in-python/952952#952952.
    # 3. Prepend the key (name).
    [k] + [item for sublist in sorted(v, key=lambda x: int(x[0])) for item in sublist]
    for k,v in d.items()
]

print(result)

# Output: [['Sarah', '12', 'Chocolate', '13', 'Strawberry'], ['Anders', '11', 'Vanilla'], ['John', '11', 'None']]

score 0 · Answer 3 · answered Oct 09 '17 at 09:57

You can try this :

import itertools


list_1=[['Sarah', '12', 'Chocolate'],
       ['Anders', '11', 'Vanilla'],
       ['Sarah', '13', 'Strawberry'],
       ['John', '11', 'None']]

repeated_list=[]
unique_list=[]
for item_1,item_2 in itertools.combinations(list_1,2):
    if item_1[0]==item_2[0]:
        repeated_list.append(item_1+item_2)

    else:

        for item_3 in repeated_list:
            if item_2[0] not in item_3:
                if item_2 not in unique_list:
                    unique_list.append(item_2)

            elif item_1[0] not in item_3:
                if item_1 not in unique_list:
                    unique_list.append(item_1)


print(unique_list+repeated_list)

Output:

[['John', '11', 'None'], ['Anders', '11', 'Vanilla'], ['Sarah', '12', 'Chocolate', 'Sarah', '13', 'Strawberry']]

Python Group List of Lists

3 Answers3