Question

我想根据旧字典来创建新字典。我要包括的项目是从特定的索引顺序到另一个特定的索引顺序。

我看到有很多关于基于值或键偷看项目的答案，这不是我想要的。我有这个字典位于特定行中的pandas DataFrame中，该字典包含用于键的元组和用于值的Id：

{(24, Timestamp('1900-01-01 08:20:48')): '77034', (25, Timestamp('1900-01-01 08:21:47')): '77034', (26, Timestamp('1900-01-01 08:31:25')): '77034', (27, Timestamp('1900-01-01 08:32:11')): '77034', (53, Timestamp('1900-01-01 13:22:51')): '77034', (54, Timestamp('1900-01-01 13:30:40')): '77034', (57, Timestamp('1900-01-01 15:28:25')): '77034', (58, Timestamp('1900-01-01 15:29:32')): '77034', (59, Timestamp('1900-01-01 15:43:55')): '77034', (60, Timestamp('1900-01-01 15:44:46')): '77034', (70, Timestamp('1900-01-01 16:56:44')): '77034', (71, Timestamp('1900-01-01 16:57:41')): '77034', (72, Timestamp('1900-01-01 17:02:57')): '77034', (73, Timestamp('1900-01-01 17:33:29')): '19975', (74, Timestamp('1900-01-01 17:43:56')): '14145', (77, Timestamp('1900-01-01 19:23:26')): '77034'}

到目前为止，我使用的dict理解什么都没有产生。

我希望能够得到这样的东西：

for row in df.itertuples():
    df.at[row.Index, 'dict_by_index'] = row.full_dict[13:14]

这将在数据框的新列中产生一个新字典： {(73, Timestamp('1900-01-01 17:33:29')): '19975', (74, Timestamp('1900-01-01 17:43:56')): '14145'}

Answer 1

您可以改用它：

df = [{(24, Timestamp('1900-01-01 08:20:48')): '77034'}, {(25, Timestamp('1900-01-01 08:21:47')): '77034'}, {(26, Timestamp('1900-01-01 08:31:25')): '77034'}, {(27, Timestamp('1900-01-01 08:32:11')): '77034'}, {(53, Timestamp('1900-01-01 13:22:51')): '77034'}, {(54, Timestamp('1900-01-01 13:30:40')): '77034'}, {(57, Timestamp('1900-01-01 15:28:25')): '77034'}, {(58, Timestamp('1900-01-01 15:29:32')): '77034'}, {(59, Timestamp('1900-01-01 15:43:55')): '77034'}, {(60, Timestamp('1900-01-01 15:44:46')): '77034'}, {(70, Timestamp('1900-01-01 16:56:44')): '77034'}, {(71, Timestamp('1900-01-01 16:57:41')): '77034'} , {(72, Timestamp('1900-01-01 17:02:57')): '77034'} , {(73, Timestamp('1900-01-01 17:33:29')): '19975'} , {(74, Timestamp('1900-01-01 17:43:56')): '14145'} , {(77, Timestamp('1900-01-01 19:23:26')): '77034'}]

基本上，我将所有这些内容都放在[]内，然后将每个对象都分成一个独立的对象，所以现在，它不再是字典，而是列表，只要它是列表，它就可以与索引

现在尝试做：

print(df[0])

它将打印第一个对象

{(24, Timestamp('1900-01-01 08:20:48')): '77034'}

您可以执行df[1]或df[2]等等，既然已经可以使用索引了，那么您应该会很高兴。

希望这会有所帮助。

是否可以像列表（基于索引）那样切片字典？

1 个答案: