Question

我从以下内容中获得了良好的结果，但是如何从元组中提取数据？换句话说，我该如何清理数据？

这是数据库中的数据，我用完了。

>>> policy_id = ((2309L,), (118L,), (94L,))
>>> for i in policy_id:
        print i


(2309L,)
(118L,)
(94L,)

但我希望结果如下：

2309
118
94

Answer 1

policy_id = ((2309L,), (118L,), (94L,))
for i in policy_id:
    print i[0]

Answer 2

>>> from itertools import chain
>>> policy_id = ((2309L,), (118L,), (94L,))
>>> for i in chain.from_iterable(policy_id):
        print i


2309
118
94

Answer 3

print '\n'.join(str(x[0]) for x in policy_id)

Answer 4

我喜欢的概念，可能会引起混淆，如下：

<强> Python2

let dict = ["emailOK" : "Hello"]
request.setValue("application/json", forHTTPHeaderField: "Content-Type")
do {
   request.HTTPBody = try NSJSONSerialization.dataWithJSONObject(dict, options: NSJSONWritingOptions.PrettyPrinted)
} catch _ {
   request.HTTPBody = nil
}

<强> Python3

policy_id = ((2309L,), (118L,), (94L,))
for i, in policy_id:
    print i

policy_id = ((2309,), (118,), (94,)) for i, in policy_id: print(i)之后,解包单元素元组的元素（对于元素多于一元素的元组不起作用）。

Answer 5

>>> policy_id = ((2309L,), (118L,), (94L,))
>>> print("\n".join(str(x[0]) for x in policy_id))
2309
118
94

Answer 6

使用地图的其他方式

map(lambda x: str(x[0]), policy_id)

如果你想要换行

"\n".join(map(lambda x: str(x[0]), policy_id))

如何将元组数据提取为单个元素格式

6 个答案: