Question

如果我有这样的列表：

lst = [['00:10:00', '00:08:00'], ["00:05:00', '00:04:00']]

如何减去每个列表的第一个和第二个数字，并将结果附加到每个列表的末尾，这样我就可以得到：

lst = [['00:10:00', '00:08:00', '00:02:00'], ['00:05:00', '00:04:00', '00:01:00']]

我当前的代码：

end_lst = []
lst = [['00:10:00', '00:08:00'], ['00:05:00', '00:04:00']]
for t1, t2 in lst:
    d1 = datetime.datetime.strptime(t1, "%H:%M:%S")
    d2 = datetime.datetime.strptime(t2, "%H:%M:%S")
    diff = (d2 - d1)
    end_lst = [item + [diff] for item in lst]
print(end_lst)

，但每个列表的输出均不正确。有什么想法吗？

输出：

[['00:10:00', '00:08:00', datetime.timedelta(-1, 86160)], ['00:05:00', '00:04:00', datetime.timedelta(-1, 86160)]]

Answer 1

以下内容如何？

导入日期时间

end_lst = []
lst = [['00:10:00', '00:08:00'], ['00:05:00', '00:04:00']]
for tup in lst:
    d1 = datetime.datetime.strptime(tup[0], "%H:%M:%S")
    d2 = datetime.datetime.strptime(tup[1], "%H:%M:%S")
    diff = (d1 - d2)
    end_lst = [item + [str(diff)] for item in lst]

print(end_lst)

Answer 2

如果您将每个元素都视为一个列表，而又没有拆包，那将非常简单：

lst = [['00:10:00', '00:08:00'], ['00:05:00', '00:04:00']]
for internal_list in lst:
    d1 = datetime.datetime.strptime(internal_list[0], "%H:%M:%S")
    d2 = datetime.datetime.strptime(internal_list[1], "%H:%M:%S")
    diff = (d2 - d1)
    internal_list.append(diff)
print(lst)

Answer 3

只需将datetime.timedelta对象转换为字符串，同时确保从d1中减去d2。

lst = [['00:10:00', '00:08:00'], ['00:05:00', '00:04:00']]
for t1, t2 in lst:
    d1 = datetime.datetime.strptime(t1, "%H:%M:%S")
    d2 = datetime.datetime.strptime(t2, "%H:%M:%S")
    diff = str(d1 - d2) # Only change to your current source code
    end_lst = [item + [diff] for item in lst]
print(end_lst) # [['00:10:00', '00:08:00', '0:02:00'], ['00:05:00', '00:04:00', '0:01:00']]

受@ferhen的启发，尽管可以说可读性被牺牲了，但是您也可以使用列表理解来使其更加简洁。

end_lst = [[t1, t2, str(datetime.datetime.strptime(t1, "%H:%M:%S") - datetime.datetime.strptime(t2, "%H:%M:%S"))] for t1, t2 in lst]
print(end_lst) # [['00:10:00', '00:08:00', '0:02:00'], ['00:05:00', '00:04:00', '0:01:00']]

现在，这是假设t1表示晚于t2的时间。如果没有，您的嵌套列表中可能会得到诸如'-1 day, 23:59:00'之类的结果。

Answer 4

一支班轮

end_lst = [[t1, t2, str(datetime.datetime.strptime(t1, "%H:%M:%S") - datetime.datetime.strptime(t2, "%H:%M:%S"))] for (t1, t2) in lst]

Answer 5

您的lst不正确，否则您的代码就可以了。

import datetime
lst = [['00:10:00', '00:08:00'], ["00:05:00", '00:04:00']]

end_lst = []
lst = [['00:10:00', '00:08:00'], ['00:05:00', '00:04:00']]
for t1, t2 in lst:
    d1 = datetime.datetime.strptime(t1, "%H:%M:%S")
    d2 = datetime.datetime.strptime(t2, "%H:%M:%S")
    diff = (d2 - d1)
    end_lst = [item + [diff] for item in lst]
print(end_lst)

输出为：

[['00:10:00', '00:08:00', datetime.timedelta(-1, 86340)], ['00:05:00', '00:04:00', datetime.timedelta(-1, 86340)]]

那么我猜的问题是您的timedelta对象。您可以在这里找到文档：datetime documentation。

这是我根据您的示例和正确的编辑创建的Google Colab笔记本。 Colab Notebook

将项目追加到列表中的每个列表

5 个答案: