TypeError：“元組”對象不支持項目分配在非元組對象上

Question

此代碼采用有序的 （從最高分到最低分）元組列表，並收集最高，第二高和第三高得分者的姓名和得分。 如果他們的關系相同，則兩個名稱都將附加到同一列表中。

myresults=[('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Otter', '2')]


FirstScore=myresults[0][1]
SecondHighestScore=0
ThirdHighestScore=0
for i in myresults:
    if i[1]==FirstScore:
        FirstPlacePatrols.append(i[0])
for i in myresults:
    print(i[1])
    print(repr(i[1]))
    if int(i[1])<int(FirstScore):
        if int(i[1])>=SecondHighestScore:
            print(i[1])
            i[1]=SecondHighestScore
            SecondPlacePatrols.append(i[0])
for i in myresults:
    if int(i[1])<SecondHighestScore:
        if int(i[1])>=ThirdHighestScore:
            i[0]=ThirdHighestScore
            ThirdPlacePatrols.append(i[0])
print(FirstPlacePatrols)
print(FirstScore)
print(SecondPlacePatrols)
print(SecondHighestScore)
print(ThirdPlacePatrols)
print(ThirdHighestScore)

然而，

i[1]=SecondHighestScore

產量

TypeError: 'tuple' object does not support item assignment

盡管，

print(repr(i[1]))

屈服

'18'

顯然這不是元組。

Answer 1

您不能更改tuple（） -它們是不可變的。 您可以創建一個新的。 或者，您可以使用itertools.groupby將元組分組在一起，並執行一些選擇性輸出：

myresults=[('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Swine', '6'), ('Otter', '2')]

from itertools import groupby

grped = groupby(myresults, lambda x: int(x[1])) 

# create a dict for all results
result_dict = {}
for key in grped :
    result_dict[key[0]] = [value for value,_ in key[1]] 

# print top 3 results:
for k in sorted(result_dict,reverse=True):
    print(k)
    print(result_dict[k])

# whole dict
print(result_dict)

輸出：

18
['Raven']
8
['Cobra']
6
['Lion', 'Swine']

# whole dict
{18: ['Raven'], 8: ['Cobra'], 6: ['Lion', 'Swine'], 2: ['Otter']}            

---

通過使用collections.defaultdict解決此問題的第二種方法：

myresults=[('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Swine', '6'), ('Otter', '2')]

from collections import defaultdict

result_dict = defaultdict(list)

for value,key in myresults:
    result_dict[int(key)].append(value)

for k in sorted(result_dict,reverse=True):
    print(k)
    print(result_dict[k])

print(result_dict)


18
['Raven']
8
['Cobra']
6
['Lion', 'Swine']
2
['Otter']

# whole dict
defaultdict(<class 'list'>, {18: ['Raven'], 8: ['Cobra'], 
                              6: ['Lion', 'Swine'], 2: ['Otter']})

Doku：

• itertools.groupby（）
• collections.defaultdict（）
• 排序（iterable [，key = ...]）

Answer 2

這是我的解決方案：

from collections import defaultdict
given_list = [('Raven', '18'), ('Cobra', '8'), ('Lion', '6'), ('Python', '6'),('Otter', '2')]
reversed_dict = defaultdict(list)
for key,value in given_list:
    reversed_dict[int(value)].append(key)

for k in reversed(sorted(reversed_dict)[-3:]):
     print(k,reversed_dict[k])

輸出：

18 ['Raven']
8 ['Cobra']
6 ['Lion', 'Python']