Question

如果前两个元素相同，我怎样才能在元组中添加第三个元素？在列表中，我们有三个元组，前两个相同的元素6220和586。因此，我们需要创建一个新的单个元组，如[(6220,586,(5+5+5))]。

list = [(6220, 586, 5), (7032, 621, 0), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5)]

预期产出：

[(6220, 586, 15),(7032, 621, 0),(7030, 589, 0)]

到目前为止，我已经提出以下建议：

a_b=[]
for x in lst:
    for y in lst:
        if len(a_b) == 0:
            a_b.append((x[0],x[1],x[2]))
        if x[0]==y[0] and x[1]==y[1]:
                if (x[0],x[1],x[2]+y[2]) not in a_b:
                    t=(x[0],x[1],x[2]+y[2])
                    a_b.append(t)

Answer 1

您可以使用简单的整数聚合（defaultdict可以缩短整数）：

from collections import defaultdict                        
agg = defaultdict(int)                                     
for item in lst:                                           
    agg[item[:2]] += item[2]   

result = [(k[0], k[1], v,) for k, v in agg.iteritems()]

结果：

[(6220, 586, 15), (7032, 621, 0), (7030, 589, 0)]

Answer 2

如果订单很重要，我会让OrderedCounter对货币对进行求和，然后将货币对与价值组合起来：

>>> from collections import OrderedDict, Counter
>>> l1= [(6220, 586, 5), (7032, 621, 0), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5)]
>>> class OrderedCounter(Counter, OrderedDict):
...     pass
...
>>> oc = OrderedCounter()
>>> for a,b,c in l1:
...     oc[a,b] += c
...
>>> oc
OrderedCounter([((6220, 586), 15), ((7032, 621), 0), ((7030, 589), 0)])

最后：

>>> result = [k+(v,) for k,v in oc.items()]
>>> result
[(6220, 586, 15), (7032, 621, 0), (7030, 589, 0)]
>>>

如果顺序无关紧要，那么普通的Counter就足够了。

Answer 3

from collections import defaultdict

d = defaultdict(int)
list_ = [(6220, 586, 5), (7032, 621, 0), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5)]

for l in list_:
    d[l[0:2]] += l[-1]
#defaultdict(int, {(6220, 586): 15, (7030, 589): 0, (7032, 621): 0})
[tuple(list(k) + [v]) for k, v in d.items()]

出：

    [(6220, 586, 15), (7032, 621, 0), (7030, 589, 0)]

Answer 4

这是一个功能性编程经典！这是：

from itertools import groupby
from functools import reduce

lst = [(6220, 586, 5), (7032, 621, 0), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5), (7030, 589, 0), (7032, 621, 0), (6220, 586, 5)]

ab = []
for k, v in groupby(sorted(lst), lambda t: (t[0], t[1])):
    ab.append(reduce(lambda x, y: (x[0], x[1], x[2] + y[2]), v))

print(ab)

>>>
[(6220, 586, 15), (7030, 589, 0), (7032, 621, 0)]

如果前两个元素是相同的，则在元组列表中添加第三个元素，每个元素包含三个元素

4 个答案: