Question

假设我有以下列表元组：

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

我想根据相同的第一个元组值对此列表求和：

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

myList

sum = 2*5 + 3*6 = 28

我怎么能得到这个？

Answer 1

您可以使用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

这个解决方案只需一次通过，而不是更具可读性的defaultdict版本，它需要两次通过并可能占用更多空间：

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

使用以下程序，即使您有多个条目而不只有两个相同的密钥，它也可以使用

#!/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))

基于元组第一个值的python sum元组列表

4 个答案: