Grouping two lists in python

Question

I have two lists which I want to group on the basis of the first element of the lists.

list1 = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]

list2 = [['1','rofl','pole'],['2','sole','pop'],['3','lmao','wtf']]

Here the first elements in the list inside the list are '1' , '2' and '3'.

I want my final list to be like :-

Final_List = [['1', 'abc', 'zef', 'rofl', 'pole'], ['3', 'lol', 'pop', 'lmao', 'wtf'], ['2', 'qwerty', 'opo', 'sole', 'pop']]

I have tried this using below code.

#!/usr/bin/python
list1 = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]
list2 = [['1','rofl','pole'],['2','sole','pop'],['3','lmao','wtf']]
d = {}
for i in list1:
    d[i[0]] = i[1:]
for i in list2:
    d[i[0]].extend(i[1:])
Final_List = []
for key, value in d.iteritems():
    value.insert(0,key)
   Final_List.append(value)

This code works but i was wondering if there was an easy and cleaner way to do it

Any help?

are the first two elements always the exact same side by side like your example? — jamylak
– jamylak, Commented Dec 18, 2014 at 14:40
your data structure is strange. why is there what appears to be an index or a key as the first element of each list item? — njzk2
– njzk2, Commented Dec 18, 2014 at 14:42
the way you do it using by changing the data structure using dicts is way cleaner than what is suggested in the posted answers. — njzk2
– njzk2, Commented Dec 18, 2014 at 14:43
the thing you can shorten in your version is using list comprehensions — njzk2
– njzk2, Commented Dec 18, 2014 at 14:44

thefourtheye · Accepted Answer · 2014-12-18 14:58:09Z

5

I would have written like you have written with a little modification, like this

Prepare a dictionary with all the elements from the second position gathered corresponding to the first element.

d = {}
for items in (list1, list2):
    for item in items:
        d.setdefault(item[0], [item[0]]).extend(item[1:])

And then just get all the values from the dictionary (Thanks @jamylak) :-)
```
print(d.values())
```

Output

[['3', 'lol', 'pop', 'lmao', 'wtf'],
 ['1', 'abc', 'zef', 'rofl', 'pole'],
 ['2', 'qwerty', 'opo', 'sole', 'pop']]

edited Dec 18, 2014 at 14:58

answered Dec 18, 2014 at 14:47

thefourtheye

241k53 gold badges466 silver badges505 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

thefourtheye Over a year ago

I would love to hear how I can improve this post.

jamylak Over a year ago

I didn't downvote but one improvement is d.setdefault(item[0], [item[0]]).extend(item[1:]) and then to get the list all it is, is d.values()

salmanwahed · Accepted Answer · 2014-12-18 14:58:39Z

1

If item sequence in the lists inside of the Final_List is not important then this can be used,

[list(set(sum(itm, []))) for itm in zip(list1, list2)]

answered Dec 18, 2014 at 14:58

salmanwahed

9,6878 gold badges37 silver badges56 bronze badges

Comments

Shahriar · Accepted Answer · 2014-12-18 14:46:29Z

0

Your code seems correct. Just modify the following portion:

Final_List = []    
for key in d:
    L = [key] + [x for x in d[key]]
    Final_List.append(L)

answered Dec 18, 2014 at 14:46

Shahriar

13.9k11 gold badges83 silver badges97 bronze badges

Comments

markcial · Accepted Answer · 2014-12-18 14:47:02Z

0

Yes, with list comprehension and enumerate

list1 = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]

list2 = [['1','rofl','pole'],['2','sole','pop'],['3','lmao','wtf']]

print [set(v + list2[k]) for k,v in enumerate(list1)]

[['1', 'abc', 'zef', 'rofl', 'pole'], ['2', 'qwerty', 'opo', 'sole', 'pop'], ['3', 'lol', 'pop', 'lmao', 'wtf']]

EDIT

With index relation

list1 = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]
list2 = [['1','rofl','pole'],['3','lmao','wtf'],['2','sole','pop']]

d1 = {a[0]:a for a in list1}
d2 = {a[0]:a for a in list2}
print [set(v + d2[k]) for k, v in d1.items()]

edited Dec 18, 2014 at 14:47

answered Dec 18, 2014 at 14:41

markcial

9,3634 gold badges33 silver badges41 bronze badges

Comments

njzk2 · Accepted Answer · 2014-12-18 14:50:08Z

0

Using default dict and list comprehensions you can shorten your code

from collections import defaultdict

list1 = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]
list2 = [['1','rofl','pole'],['2','sole','pop'],['3','lmao','wtf']]

d = defaultdict(list)

for i in list1 + list2:
    d[i[0]].extend(i[1:])

Final_List = [[key] + value for key, value in d.iteritems()]

print Final_List

answered Dec 18, 2014 at 14:50

njzk2

39.4k7 gold badges72 silver badges111 bronze badges

Comments

Epoch · Accepted Answer · 2014-12-18 14:58:45Z

0

list3 = []
for i in xrange(0,max(len(list1[0]), len(list2[0]))):
    list3.append(list(list1[i]))
    list3[i].extend(x for x in list2[i] if x not in list3[i])

with a xrange, you can iterate only once through the list.

answered Dec 18, 2014 at 14:58

Epoch

13 bronze badges

Comments

wwii · Accepted Answer · 2014-12-18 15:21:15Z

A bit of functional style:

import operator, itertools
from pprint import pprint
one = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]
two = [['1','rofl','pole'],['2','sole','pop'],['3','lmao','wtf']]

A few helpers:

zero = operator.itemgetter(0)
all_but_the_first = operator.itemgetter(slice(1, None))
data = (one, two)

def foo(group):
    # group is (key, iterator) from itertools.groupby
    key = group[0]
    lists = group[1]
    result = list(key)
    for item in lists:
        result.extend(all_but_the_first(item))
    return result

Function to process the daa

def process(data, func = foo):
    # concatenate all the sublists
    new = itertools.chain(*data)
    # group by item zero
    three = sorted(new, key = zero)
    groups = itertools.groupby(three, zero)
    # iterator that builds the new lists
    return itertools.imap(foo, groups)

Usage

>>> pprint(list(process(data)))

[['1', 'abc', 'zef', 'rofl', 'pole'],
 ['2', 'qwerty', 'opo', 'sole', 'pop'],
 ['3', 'lol', 'pop', 'lmao', 'wtf']]
>>>
>>> for thing in process(data):
    print thing

['1', 'abc', 'zef', 'rofl', 'pole']
['2', 'qwerty', 'opo', 'sole', 'pop']
['3', 'lol', 'pop', 'lmao', 'wtf']
>>>

hariK · Accepted Answer · 2014-12-18 16:46:44Z

0

list1 = [['1','abc','zef'],['2','qwerty','opo'],['3','lol','pop']]

list2 = [['1','rofl','pole'],['2','sole','pop'],['3','lmao','wtf']]

Final_List = []

for i in range(0, len(list1)):

    Final_List.append(list1[i] + list2[i])

    del Final_List[i][3]

print Final_List

Output

[['1', 'abc', 'zef', 'rofl', 'pole'], ['2', 'qwerty', 'opo', 'sole', 'pop'], ['3', 'lol', 'pop', 'lmao', 'wtf']]

answered Dec 18, 2014 at 16:46

hariK

3,09915 silver badges19 bronze badges

Collectives™ on Stack Overflow

Grouping two lists in python

8 Answers 8

2 Comments

Comments

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

8 Answers 8

2 Comments

Comments

Comments

Comments

Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related