Python 列表——去重和添加子列表元素

Question

輸入：

a = [[['a','b'],3],[['b','a'],9],[['b','z'],4]]

所需的輸出（重復數據刪除不考慮順序並添加整數）

[[['a','b'],12],[['b','z'],4]]

這當然不起作用：

a2 = [sorted(list) for list in [i[0] for i in a]]

print(a2)

[['a', 'b'], ['a', 'b'], ['b', 'z']]

當然可以刪除：

a3= [] 

for i in a2: 
    if i not in a3: 
       a3.append(i)

print(a3)

[['a', 'b'], ['b', 'z']]

但當然，我正在丟失被重復數據刪除的子列表的計數。 有什么建議嗎？
謝謝。

Answer 1

我想這就是你想要的：

a = [[['a','b'],3],[['b','a'],9],[['b','z'],4]]

# we use a dict to do the job.
# we sort the list of char and change to tuple. So i can be use as key.
dct = {}

for i in a:
    srt = tuple(sorted(i[0]))
    if srt in dct:
        # the tuple of char is already existing as key, just sum int.
        dct[srt]+= i[1]
    else:
        # if not we had a new key
        dct[srt] = i[1]
# Convert the dict to list of list
new_list = [[list(k),v] for k,v in dct.items()]
print(new_list) # [[['a', 'b'], 12], [['b', 'z'], 4]]

Answer 2

您可以使用：

from collections import defaultdict

a = [[['a','b'],3],[['b','a'],9],[['b','z'],4]]
d = defaultdict(lambda : 0)

for e, n in a:
    d[frozenset(e)] += n

a =  [[list(k), v]  for k, v in d.items()]
a

輸出：

[[['a', 'b'], 12], [['b', 'z'], 4]]

---

或者你可以使用：

from functools import reduce
from operator import add
from collections import Counter

f = ({frozenset(k): v} for k, v in a) 
d = reduce(add, map(Counter, f)) # sum unique data
result = [[list(k), v]  for k, v in d.items()]
result

輸出：

[[['a', 'b'], 12], [['b', 'z'], 4]]

---

如果您喜歡單線解決方案：

 [[list(k), v] for k, v in reduce(add, map(Counter, ({frozenset(k): v} for k, v in a))).items()]

輸出：

[[['a', 'b'], 12], [['b', 'z'], 4]]

Answer 3

您可以先將子列表轉換為frozensets，以便您可以將它們視為鍵並使用collections.Counter將整數添加為每個不同鍵的計數，然后在迭代結果項時將frozensets轉換回列表：

from collections import Counter
counter = Counter()
for t, c in a:
    counter.update({frozenset(t): c})
print([[list(t), c] for t, c in counter.items()])

Answer 4

這將解決問題：

from itertools import groupby

a = [[['a','b'],3],[['b','a'],9],[['b','z'],4]]

# this will ensure a is ordered by the merging key:
a=sorted(a, key=lambda x: sorted(x[0]))

# this will group by merging key, extract the value with index 1, and sum it:
a=[[k, sum(map(lambda z: z[1], v))] for k,v in groupby(a, key=lambda x: sorted(x[0]))]

輸出：

[[['a', 'b'], 12], [['b', 'z'], 4]]