对python中的元组列表进行排序

Question

我有这样的元组

[('AVAX', '070122'), ('AVAX', '201221'), ('AVAX', '211221'), ('AVAX', '241221'), ('AVAX', '311221'), ('BNB', '070122'), ('BNB', '201221'), ('BNB', '211221'), ('BNB', '241221'), ('BNB', '280122'), ('BNB', '311221'), ('BTC', '070122'), ('BTC', '201221'), ('BTC', '211221'), ('BTC', '241221'), ('BTC', '250222'), ('BTC', '250322'), ('BTC', '280122'), ('BTC', '311221'), ('ETH', '070122'), ('ETH', '201221'), ('ETH', '211221'), ('ETH', '241221'), ('ETH', '250222'), ('ETH', '250322'), ('ETH', '280122'), ('ETH', '311221'), ('MATIC', '070122'), ('MATIC', '201221'), ('MATIC', '211221'), ('MATIC', '241221'), ('MATIC', '311221'), ('SOL', '070122'), ('SOL', '201221'), ('SOL', '211221'), ('SOL', '241221'), ('SOL', '280122'), ('SOL', '311221')]

这些是硬币和它的报应。日期是字符串格式，所以安排是错误的。

因此，我已经改变了格式到目前为止，并试图安排。我试过下面的代码。

filtered_final_product_list = [list(ele) for ele in filtered_final_product_list]
new_list=list()
for i in filtered_final_product_list:
    i[1]=datetime.strptime(i[1],'%d%m%y')
    new_list.append(i)
    
print(sorted(new_list))

结果和预期的一样。

[['AVAX', datetime.datetime(2021, 12, 20, 0, 0)], ['AVAX', datetime.datetime(2021, 12, 21, 0, 0)], ['AVAX', datetime.datetime(2021, 12, 24, 0, 0)], ['AVAX', datetime.datetime(2021, 12, 31, 0, 0)], ['AVAX', datetime.datetime(2022, 1, 7, 0, 0)], ['BNB', datetime.datetime(2021, 12, 20, 0, 0)], ['BNB', datetime.datetime(2021, 12, 21, 0, 0)], ['BNB', datetime.datetime(2021, 12, 24, 0, 0)], ['BNB', datetime.datetime(2021, 12, 31, 0, 0)], ['BNB', datetime.datetime(2022, 1, 7, 0, 0)], ['BNB', datetime.datetime(2022, 1, 28, 0, 0)], ['BTC', datetime.datetime(2021, 12, 20, 0, 0)], ['BTC', datetime.datetime(2021, 12, 21, 0, 0)], ['BTC', datetime.datetime(2021, 12, 24, 0, 0)], ['BTC', datetime.datetime(2021, 12, 31, 0, 0)], ['BTC', datetime.datetime(2022, 1, 7, 0, 0)], ['BTC', datetime.datetime(2022, 1, 28, 0, 0)], ['BTC', datetime.datetime(2022, 2, 25, 0, 0)], ['BTC', datetime.datetime(2022, 3, 25, 0, 0)], ['ETH', datetime.datetime(2021, 12, 20, 0, 0)], ['ETH', datetime.datetime(2021, 12, 21, 0, 0)], ['ETH', datetime.datetime(2021, 12, 24, 0, 0)], ['ETH', datetime.datetime(2021, 12, 31, 0, 0)], ['ETH', datetime.datetime(2022, 1, 7, 0, 0)], ['ETH', datetime.datetime(2022, 1, 28, 0, 0)], ['ETH', datetime.datetime(2022, 2, 25, 0, 0)], ['ETH', datetime.datetime(2022, 3, 25, 0, 0)], ['MATIC', datetime.datetime(2021, 12, 20, 0, 0)], ['MATIC', datetime.datetime(2021, 12, 21, 0, 0)], ['MATIC', datetime.datetime(2021, 12, 24, 0, 0)], ['MATIC', datetime.datetime(2021, 12, 31, 0, 0)], ['MATIC', datetime.datetime(2022, 1, 7, 0, 0)], ['SOL', datetime.datetime(2021, 12, 20, 0, 0)], ['SOL', datetime.datetime(2021, 12, 21, 0, 0)], ['SOL', datetime.datetime(2021, 12, 24, 0, 0)], ['SOL', datetime.datetime(2021, 12, 31, 0, 0)], ['SOL', datetime.datetime(2022, 1, 7, 0, 0)], ['SOL', datetime.datetime(2022, 1, 28, 0, 0)]]

现在，当我想要回到我在字符串中提到的旧格式时，它的顺序就不一样了。它进入了混乱的状态。需要帮助，使排序的方式，在字符串格式的结尾。

Answer 1

使用datetime.datetime按硬币和日期排序

from datetime import datetime
sorted(lst, key=lambda x:(x[0], datetime.strptime(x[1], '%d%m%y')))

输出：

[('AVAX', '201221'), ('AVAX', '211221'), ('AVAX', '241221'), 
 ('AVAX', '311221'), ('AVAX', '070122'), ('BNB', '201221'), 
 ('BNB', '211221'), ('BNB', '241221'), ('BNB', '311221'), 
 ('BNB', '070122'), ('BNB', '280122'), ('BTC', '201221'),
 ('BTC', '211221'), ('BTC', '241221'), ('BTC', '311221'), 
 ('BTC', '070122'), ('BTC', '280122'), ('BTC', '250222'), 
 ('BTC', '250322'), ('ETH', '201221'), ('ETH', '211221'), 
 ('ETH', '241221'), ('ETH', '311221'), ('ETH', '070122'), 
 ('ETH', '280122'), ('ETH', '250222'), ('ETH', '250322'), 
 ('MATIC', '201221'), ('MATIC', '211221'), ('MATIC', '241221'), 
 ('MATIC', '311221'), ('MATIC', '070122'), ('SOL', '201221'),
 ('SOL', '211221'), ('SOL', '241221'), ('SOL', '311221'), 
 ('SOL', '070122'), ('SOL', '280122')]

Answer 2

这将满足您的要求，但正如我所说的，最好先以正确的格式生成这些带有日期的元组。

product_list = [('AVAX', '070122'), ('AVAX', '201221'), ('AVAX', '211221'), ('AVAX', '241221'), ('AVAX', '311221'), ('BNB', '070122'), ('BNB', '201221'), ('BNB', '211221'), ('BNB', '241221'), ('BNB', '280122'), ('BNB', '311221'), ('BTC', '070122'), ('BTC', '201221'), ('BTC', '211221'), ('BTC', '241221'), ('BTC', '250222'), ('BTC', '250322'), ('BTC', '280122'), ('BTC', '311221'), ('ETH', '070122'), ('ETH', '201221'), ('ETH', '211221'), ('ETH', '241221'), ('ETH', '250222'), ('ETH', '250322'), ('ETH', '280122'), ('ETH', '311221'), ('MATIC', '070122'), ('MATIC', '201221'), ('MATIC', '211221'), ('MATIC', '241221'), ('MATIC', '311221'), ('SOL', '070122'), ('SOL', '201221'), ('SOL', '211221'), ('SOL', '241221'), ('SOL', '280122'), ('SOL', '311221')]

product_list.sort( key=lambda k: (k[0],k[1][4:6]+k[1][2:4]+k[1][0:2]))
print(product_list)

Answer 3

当您想要创建new_list并希望备份现有列表时

注意: sorted() :->创建一个新列表

从@manlai A借来的代码:谢谢您分享答案

   from datetime import datetime
    new_list = sorted(lst, key=lambda x:(x[0], datetime.strptime(x[1], '%d%m%y')))
    print(new_list)

当您想要修改现有的列表时

注: given_list_obj.sort(key=function) :->对现有列表进行排序

new_list1 = lst.copy()
new_list1.sort(key=lambda x:(x[0], datetime.strptime(x[1], '%d%m%y')))
print(new_list)

输出

    [('AVAX', '201221'), ('AVAX', '211221'), ('AVAX', '241221'), 
 ('AVAX', '311221'), ('AVAX', '070122'), ('BNB', '201221'), 
 ('BNB', '211221'), ('BNB', '241221'), ('BNB', '311221'), 
 ('BNB', '070122'), ('BNB', '280122'), ('BTC', '201221'),
 ('BTC', '211221'), ('BTC', '241221'), ('BTC', '311221'), 
 ('BTC', '070122'), ('BTC', '280122'), ('BTC', '250222'), 
 ('BTC', '250322'), ('ETH', '201221'), ('ETH', '211221'), 
 ('ETH', '241221'), ('ETH', '311221'), ('ETH', '070122'), 
 ('ETH', '280122'), ('ETH', '250222'), ('ETH', '250322'), 
 ('MATIC', '201221'), ('MATIC', '211221'), ('MATIC', '241221'), 
 ('MATIC', '311221'), ('MATIC', '070122'), ('SOL', '201221'),
 ('SOL', '211221'), ('SOL', '241221'), ('SOL', '311221'), 
 ('SOL', '070122'), ('SOL', '280122')]