蟒蛇和元組列表基於元組的第一個值

Question

假設我有以下列表中的元組：

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

我想基於相同的第一個元組值來總結這個名單：

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

對於myList

sum = 2*5 + 3*6 = 28 

我怎麼能得到這個？

Answer 1

可以使用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

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) 

Answer 3

將該溶液做它在相對於所述多個可讀defaultdict版本採用兩個通行證和可能會佔用更多空間的單程：

myList = [(0,2),(1,3),(2,4),(0,5),(1,6)] 
sum_ = 0 
once, twice = {}, {} 
for x, y in myList: 
    if x in once: 
     sum_ -= twice.get(x, 0) 
     twice[x] = twice.get(x, once[x]) * y 
     sum_ += twice[x] 
    else: 
     once[x] = y 


>>> sum_ 
28 

Answer 4

即使您有多個輸入，並且不只有兩個輸入相同的密鑰，也可以使用以下程序

#!/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))