Python 以列表的形式获取输出

有一个集合列表，如下所示：

expiry = [{'20180830', '20180928', '20181025'},
          {'20180830', '20180927', '20181026'},
          {'20180830', '20180927', '20181025'}]

和一个函数：

def get_dte(Expiry):
    exp_date = datetime.strptime(Expiry, '%Y%m%d')
    dte = (exp_date- datetime.now()).days
    return dte

如何以相同的输入形式获得函数的输出（集合列表）

产生：

[61, 90, 33, 61, 89, 34, 61, 89, 33]

…需要的是：

[{61, 90, 33}, {61, 89, 34}, {61, 89, 33}]

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你只需要稍微修改一下你的理解力就可以了

试试看：

from datetime import datetime

expiry = [{'20180830', '20180928', '20181025'},
          {'20180830', '20180927', '20181026'},
          {'20180830', '20180927', '20181025'}]

def get_dte(Expiry):
    exp_date = datetime.strptime(Expiry, '%Y%m%d')
    dte = (exp_date- datetime.now()).days
    return dte

print( [{get_dte(i) for i in elem } for elem in expiry] )

[{89, 33, 62}, {33, 90, 61}, {89, 33, 61}]

输出：

from datetime import datetime

expiry = [{'20180830', '20180928', '20181025'},
          {'20180830', '20180927', '20181026'},
          {'20180830', '20180927', '20181025'}]

def get_dte(Expiry):
    exp_date = datetime.strptime(Expiry, '%Y%m%d')
    dte = (exp_date- datetime.now()).days
    return dte

print( [{get_dte(i) for i in elem } for elem in expiry] )

[{89, 33, 62}, {33, 90, 61}, {89, 33, 61}]

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您需要分别迭代过期的每个元素：

>>> [{get_dte(j) for j in i} for i in expiry]
[{33, 62, 89}, {33, 61, 90}, {33, 61, 89}]

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打印（[{get_dte（i）for i in elem}for elem in expiration]）

？输出应为目录列表或列表列表？集合列表（非目录）