Question

如何在按第一个值分组的元组列表中获取第二个和第三个值的总和？

即：

list_of_tuples = [(1, 3, 1), (1, 2, 4), (2, 1, 0), (2, 2, 0)]

expected_output = [(1, 5, 5), (2, 3, 0)]

我在StackOverflow上发现了几个great answers，发现这个元组有两个值，但是无法弄清楚如何调整它们来对第二个和第三个值求和。

第二个值的一个好答案是：

def sum_pairs(pairs):
sums = {}
for pair in pairs:
    sums.setdefault(pair[0], 0)
    sums[pair[0]] += pair[1]
return sums.items()

Answer 1

你也可以这样做：

list_of_tuples = [(1, 3, 1), (1, 2, 4), (2, 1, 0), (2, 2, 0)]

# create empty dictionary to store data
sums = {}

# iterate over list of typles
for pair in list_of_tuples:

  # create new item in dictionary if it didnt exist
  if pair[0] not in sums: sums[pair[0]] = [pair[0], 0 ,0]

  # sum the values
  sums[pair[0]][1] += pair[1]
  sums[pair[0]][2] += pair[2]

#print resulting tuple   
print(tuple(sums.values()))

Answer 2

您可以使用itertools.groupby根据第一项进行分组，然后获取每组中所有最后两项的累计总和：

from itertools import groupby

list_of_tuples = [(1, 3, 1), (1, 2, 4), (2, 1, 0), (2, 2, 0)]
lst = [(k,)+tuple(sum(x) for x in zip(*g))[1:] 
                         for k, g in groupby(list_of_tuples, lambda x: x[0])]
print(lst)
# [(1, 5, 5), (2, 3, 0)]

Answer 3

使用defaultdict作为石斑鱼：

>>> from collections import defaultdict
>>> grouper = defaultdict(lambda: (0,0))
>>> list_of_tuples = [(1, 3, 1), (1, 2, 4), (2, 1, 0), (2, 2, 0)]
>>> for a, b, c in list_of_tuples:
...     x, y = grouper[a]
...     grouper[a] = (x + b, y + c)
...
>>> grouper
defaultdict(<function <lambda> at 0x102b240d0>, {1: (5, 5), 2: (3, 0)})

现在，您可以随时获取这样的元组列表：

>>> [(k, a, b) for k, (a, b) in grouper.items()]
[(1, 5, 5), (2, 3, 0)]

在第一个分组的元组列表中元组的第二个和第三个元素的Python和

3 个答案: