如何在python中合并多个列表

Question

I have multiple lists like below

[u'a', 11, u'P']
[u'a', 11, u'A']
[u'b', 2, u'P']
[u'c', 1, u'P']
[u'c', 2, u'P']
[u'd', 1, u'P']
[u'e', 3, u'P']
[u'f', 2, u'P']
[u'a', 1, u'P']
[u'a', 2, u'P']
[u'b', 1, u'P']
[u'b', 11, u'P']

how to merge above lists,to loop the list to add up like below

[u'a', 11, u'P'] + [u'a', 2, u'P'] + [u'a', 11, u'A'] = ['a',('P' : 13) ,('A': 11)]

[u'b', 2, u'P'] + [u'b', 1, u'P'] + [u'b', 11, u'P'] = ['b',14,p]

output should like below :

['a',('P' : 13) ,('A': 11)]
['b',14,'p']

Answer 1

You could consider using collections.defaultdict and iterating through the list of dicts in the values.

import collections
d = collections.defaultdict(list)
l = [[u'a', 11, u'P'],[u'a', 11, u'A'],[u'a', 3, u'P'],[u'b', 2, u'P'],[u'c', 1, u'P'],[u'c', 2, u'P'],[u'd', 1, u'P'],[u'e', 3, u'P']]
for k1, v, k2 in l:
    if k1 in d:
            d[k1].append({k2:v})
    else: 
        d[k1] = [{k2:v}]

newdict = {}
for key,value in d.items():
    newvalue = {}
    for valuedict in value:
        for key2,value2 in valuedict.items():
            if key2 in newvalue:
                newvalue[key2] += value2
            else:
                newvalue[key2] = value2
    newdict[key] = newvalue

print newdict

And this would give you

{u'a': {u'A': 11, u'P': 14}, u'c': {u'P': 3}, u'b': {u'P': 2}, u'e': {u'P': 3}, u'd': {u'P': 1}}

Answer 2

The output that you want looks slightly strange due to the inconsistency between the two cases. You could trivially change this example to get whatever output you want, however:

lists = [
 [u'a', 11, u'P'],
 [u'a', 11, u'A'],
 [u'b', 2, u'P'],
 [u'c', 1, u'P'],
 [u'c', 2, u'P'],
 [u'd', 1, u'P'],
 [u'e', 3, u'P'],
 [u'f', 2, u'P'],
 [u'a', 1, u'P'],
 [u'a', 2, u'P'],
 [u'b', 1, u'P'],
 [u'b', 11, u'P']]

# Each key in this dictionary will be one of the first elements
# from the lists shown above.  The values will be dictionaries
# mapping a letter (one of the third elements in each list) to
# their total count (i.e. the sum of the second elements matching
# the other two columns)
from collections import defaultdict
results = defaultdict(dict)

for main_key, count, subkey in lists:
    d = results[main_key]
    d[subkey] = d.get(subkey,0) + count

for main_key, values in results.items():
    print main_key, "=>", values

The output is:

a => {u'A': 11, u'P': 14}
c => {u'P': 3}
b => {u'P': 14}
e => {u'P': 3}
d => {u'P': 1}
f => {u'P': 2}

Update: Thanks to sharjeel for suggesting in the comment below that I remove the setdefault by using defaultdict instead.

---

Update 2: In your further question in the comments below, you indicate that instead you want as output "[a] set of lists like [[u'a', 11, u'P'], [u'a', 11, u'A'] ". (I'm assuming for the moment that you mean a list of lists rather than a set, but that's almost as easy.) In order to construct such a list of lists , you could replace the loop that prints the values with the following:

lists_output = []

for main_key, values in results.items():
    for subkey, count in values.items():
       lists_output.append([main_key,count,subkey])

print lists_output

... which will give the output:

[[u'a', 11, u'A'], [u'a', 14, u'P'], [u'c', 3, u'P'], [u'b', 14, u'P'], [u'e', 3, u'P'],
 [u'd', 1, u'P'], [u'f', 2, u'P']]

Answer 3

Its a one liner solution if you use groupby from itertools.

Put all the lists in one list say lst.

lst = [
    [u'a', 11, u'P']
    [u'a', 11, u'A']
    [u'b', 2, u'P']
    [u'c', 1, u'P']
    [u'c', 2, u'P']
    [u'd', 1, u'P']
    [u'e', 3, u'P']
    [u'f', 2, u'P']
    [u'a', 1, u'P']
    [u'a', 2, u'P']
    [u'b', 1, u'P']
    [u'b', 11, u'P']
]

Now use group by in outer elements to group over a,b,c etc. then within each grouped data group over the third element ie P,A etc. The further grouped data needs to be summed.

Here is the solution:

from itertools import groupby
result = dict(
                ( k, dict( (k1, sum([i[1] for i in g2])) for k1, g2 in groupby(g, key=lambda y: y[2] ) ) )
                for k, g in groupby(lst, key=lambda x: x[0])
            )

To have a better understanding, I would recommend you to play around with single groupby then jump to nested groupby.

Here are a few links:

http://docs.python.org/library/itertools.html#itertools.groupby

http://www.builderau.com.au/program/python/soa/Python-groupby-the-iterator-swiss-army-knife/0,2000064084,339280431,00.htm