python - タプルの最初の値に基づくPython合計タプルリスト

Question

次のリストタプルがあるとします。

myList = [(0,2),(1,3),(2,4),(0,5),(1,6)]

同じ最初のタプル値に基づいてこのリストを合計したい:

[(n,m),(n,k),(m,l),(m,z)] = m*k + l*z

為にmyList

sum = 2*5 + 3*6 = 28

どうすればこれを入手できますか？

Answer 1

使用できますcollections.defaultdict：

>>> from collections import defaultdict
>>> from operator import mul
>>> lis = [(0,2),(1,3),(2,4),(0,5),(1,6)]
>>> dic = defaultdict(list)
>>> for k,v in lis:
    dic[k].append(v)  #use the first item of the tuple as key and append second one to it
...     

#now multiply only those lists which contain more than 1 item and finally sum them.
>>> sum(reduce(mul,v) for k,v in dic.items() if len(v)>1)
 28

Answer 2

以下のプログラムを使用すると、同じキーに対して2つだけでなく、複数のエントリがある場合でも機能します

#!/usr/local/bin/python3

myList = [(0,2),(1,3),(2,4),(0,5),(1,6),(1,2)]

h = {}
c = {}
sum = 0

for k in myList:
        # if key value already present
        if k[0] in c:
                if k[0] in h:
                        sum = sum - h[k[0]]
                        h[k[0]] = h[k[0]] * k[1]
                else:
                        h[k[0]] = c[k[0]] * k[1]
                sum = sum + h[k[0]]
        else:
                # stores key and value if first time though the loop
                c[k[0]] = k[1]                
print('sum is' + str(sum))

Answer 3

from operator import itemgetter
from itertools import groupby

def mul(args): # will work with more than 2 arguments, e.g. 2*3*4
    return reduce(lambda acc, x: acc*x, args, 1)

myList = [(0,2),(1,3),(2,4),(0,5),(1,6)]
sorted_ = sorted(myList, key=itemgetter(0))
grouped = groupby(sorted_, key=itemgetter(0))
numbers = [[t[1] for t in items] for _, items in grouped]
muls = [mul(items) for items in numbers if len(items) > 1]
print sum(muls)