用日期字符串對列表進行排序python

Question

我想用我的排序名單的名單a.sort(key=lambda x: x[0])或a.sort() 我約會的第一個要素是日期，我需要對其進行排序。

a = [['1/8/2018', 71], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['2/8/2018', 81], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77]]

我需要的結果：

[['1/8/2018', 71],['2/8/2018', 81], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12],['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79]]

Answer 1

使用datetime.datetime.strptime

例如：

import datetime
a = [['1/8/2018', 71], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['2/8/2018', 81], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77]]
a.sort(key=lambda x: datetime.datetime.strptime(x[0], "%d/%m/%Y"))
print(a)

輸出：

[['1/8/2018', 71], ['2/8/2018', 81], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79]]

Answer 2

您可以使用str.split ：

a = [['1/8/2018', 71], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['2/8/2018', 81], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77]]
new_a = sorted(a, key=lambda x:[int(i) for i in x[0].split('/')])

輸出：

[['1/8/2018', 71], ['2/8/2018', 81], ['3/8/2018', 8], ['4/8/2018', 64], ['5/8/2018', 64], ['6/8/2018', 12], ['7/8/2018', 8], ['8/8/2018', 83], ['9/8/2018', 77], ['10/8/2018', 76], ['11/8/2018', 6], ['12/8/2018', 60], ['13/8/2018', 81], ['14/8/2018', 74], ['15/8/2018', 7], ['16/8/2018', 77], ['17/8/2018', 81], ['18/8/2018', 69], ['19/8/2018', 66], ['20/8/2018', 74], ['21/8/2018', 88], ['22/8/2018', 92], ['23/8/2018', 90], ['24/8/2018', 84], ['25/8/2018', 66], ['26/8/2018', 64], ['27/8/2018', 82], ['28/8/2018', 82], ['29/8/2018', 79]]