Python：迭代列表中的元组

现在，我有一段代码，它提供了以下输出：

[('KOR', [3, 1, 0]), ('ITA', [1, 0, 0]), ('TPE', [0, 1, 1]), ('CHN', [0, 1, 0]), ('JPN', [0, 1, 0]), ('AUS', [0, 0, 1]), ('GBR', [0, 0, 1]), ('UKR', [0, 0, 1])]

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提前谢谢你

提供相同的结构，您可以通过连接每个内部列表的两个索引，在列表理解中轻松地将其展开：

data = [('KOR', [3, 1, 0]), ('ITA', [1, 0, 0]), ('TPE', [0, 1, 1]), ('CHN', [0, 1, 0]),
        ('JPN', [0, 1, 0]), ('AUS', [0, 0, 1]), ('GBR', [0, 0, 1]), ('UKR', [0, 0, 1])]

flattened = [" ".join(str(e) for e in [item[0]] + item[1]) for item in data]
# ['KOR 3 1 0', 'ITA 1 0 0', 'TPE 0 1 1', 'CHN 0 1 0', 'JPN 0 1 0', 'AUS 0 0 1',
#  'GBR 0 0 1', 'UKR 0 0 1']

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如果不想使用列表理解，可以按如下方式更改原始代码：

answer = []
for item in data:
    result = item[0]
    for x in item[1]:
        result = result + ' ' + str(x)
    answer.append(result)
return answer

---

我把正确的输出格式弄错了

for item in alist:
    for x in item:
        answer.append(x)

for item in answer:
    if type(item)==str:
        answer2.append(item)
    else:
        for f in item:
            answer2.append(f)
return answer2

data = [('KOR', [3, 1, 0]), ('ITA', [1, 0, 0]), ('TPE', [0, 1, 1]), ('CHN', [0, 1, 0]),
        ('JPN', [0, 1, 0]), ('AUS', [0, 0, 1]), ('GBR', [0, 0, 1]), ('UKR', [0, 0, 1])]

flattened = [" ".join(str(e) for e in [item[0]] + item[1]) for item in data]
# ['KOR 3 1 0', 'ITA 1 0 0', 'TPE 0 1 1', 'CHN 0 1 0', 'JPN 0 1 0', 'AUS 0 0 1',
#  'GBR 0 0 1', 'UKR 0 0 1']

answer = []
for item in data:
    result = item[0]
    for x in item[1]:
        result = result + ' ' + str(x)
    answer.append(result)
return answer