Question

我需要使用python：

在字典中嵌套三个数据

ID（可以重复）
日期（行动 - 可以重复）
操作（与ID和日期关联的单词列表）

数据示例（制表符分隔）：

UID DATE    ACTIONS
abc123  12/25/2016  break, pullover
abc123  12/25/2016  stop
abc123  10/15/2015  break, pullover, turn
def456  6/14/2015   turn, wash, skid
def456  11/24/2016  stop, wash, pullover, break
ghi789  2/12/2015   pullover, stop

代码 - 使用@moogle评论修改

from collections import defaultdict

date = ['12/25/16','12/25/16','10/15/2015','6/14/2015','11/24/2016','2/12/2015']
uid = ['abc123','abc123', 'abc123','def456', 'def456', 'ghi789']
action = [['break', 'pullover'],['stop'],['break','pullover','turn'],['turn','wash','skid'],['stop','wash','pullover','break'],['pullover','stop']]

d = defaultdict(list)
for uid, date, action in zip(uid, date, action):
    d[id].append((date,action))

    print dict(d)

渴望输出 所需的输出是列表的嵌套字典。其中父key是ID，父value是嵌套字典，其中嵌套key是日期，嵌套value是列表（操作）

当前实际输出

{'ghi789': [('2/12/2015', ['pullover', 'stop'])], 'def456': [('6/14/2015', ['turn', 'wash', 'skid']), ('11/24/2016', ['stop', 'wash', 'pullover', 'break'])], 'abc123': [('12/25/16', ['break', 'pullover']), ('12/25/16', ['stop']), ('10/15/2015', ['break', 'pullover', 'turn'])]}

**desired output**
{'abc123':[{'12/25/2016':[['break', 'pullover'],['stop']]}, {'10/15/2015':[['break','pullover','turn']]}],'def456':[{'6/14/2015':[['turn','wash','skid'],['stop','wash','pullover','break']},'ghi789':{'2/12/2915':[['pullover','stop']]}]}

我尝试使用上面的代码获取上述输出，我从HERE改编并查找HERE。但是，我一直在收到错误。我认为这与我试图在列表列表中嵌套的事实有关，我不确定要修复它的方向。

Answer 1

我认为基于对象的方法对这些数据要好得多。

您可以执行以下操作：

class Event:
    def __init__(self, ID, date, actions):
        self.ID=ID
        self.date=date
        self.actions=actions

    def __repr__(self):
        return 'ID: {} date: {} actions: {}'.format(self.ID, self.date, self.actions)

然后创建一个对象列表，如下所示：

 >>> objs=[Event(id_, d, actions) for id_, d, actions in zip(uid, date, action)]
 >>> objs
 [ID: abc123 date: 12/25/16 actions: ['break', 'pullover'], ID: abc123 date: 12/25/16 actions: ['stop'], ID: abc123 date: 10/15/2015 actions: ['break', 'pullover', 'turn'], ID: def456 date: 6/14/2015 actions: ['turn', 'wash', 'skid'], ID: def456 date: 11/24/2016 actions: ['stop', 'wash', 'pullover', 'break'], ID: ghi789 date: 2/12/2015 actions: ['pullover', 'stop']]

然后可以根据需要对动作/事件列表进行排序，分析和保存。

按日期排序：

>>> sorted(objs, key=lambda o: o.date)
[ID: abc123 date: 10/15/2015 actions: ['break', 'pullover', 'turn'], ID: def456 date: 11/24/2016 actions: ['stop', 'wash', 'pullover', 'break'], ID: abc123 date: 12/25/16 actions: ['break', 'pullover'], ID: abc123 date: 12/25/16 actions: ['stop'], ID: ghi789 date: 2/12/2015 actions: ['pullover', 'stop'], ID: def456 date: 6/14/2015 actions: ['turn', 'wash', 'skid']]

按事件：

>>> [o for o in objs if 'stop' in o.actions]
[ID: abc123 date: 12/25/16 actions: ['stop'], ID: def456 date: 11/24/2016 actions: ['stop', 'wash', 'pullover', 'break'], ID: ghi789 date: 2/12/2015 actions: ['pullover', 'stop']]

然后创建一个类似于你想要的dict（尽管那个例子不是合法的Python dict ......）是相当明显的：

di={o.ID:[] for o in objs}
for user in di:
    di[user].append({o.date:o.actions for o in objs if o.ID==user})

>>> di
{'ghi789': [{'2/12/2015': ['pullover', 'stop']}], 'def456': [{'6/14/2015': ['turn', 'wash', 'skid'], '11/24/2016': ['stop', 'wash', 'pullover', 'break']}], 'abc123': [{'10/15/2015': ['break', 'pullover', 'turn'], '12/25/16': ['stop']}]}

Answer 2

如果我在“操作”列表中添加了缺少的逗号，那么您的代码适用于我...

我得到了输出：

{'ghi789': [('2/12/2015', ['pullover', 'stop'])], 'def456': [('6/14/2015', ['turn', 'wash', 'skid']), ('11/24/2016', ['stop', 'wash', 'pullover', 'break'])], 'abc123': [('12/25/16', ['break', 'pullover']), ('12/25/16', ['stop']), ('10/15/2015', ['break', 'pullover', 'turn'])]}

以下解决方案如何：

from collections import defaultdict

date = ['12/25/16','12/25/16','10/15/2015','6/14/2015','11/24/2016','2/12/2015']
uid = ['abc123','abc123', 'abc123','def456', 'def456', 'ghi789']
action = [['break', 'pullover'],['stop'],['break','pullover','turn'],['turn','wash','skid'],['stop','wash','pullover','break'],['pullover','stop']]

d = defaultdict(dict)
for uid, date, action in zip(uid, date, action):
    d[uid].setdefault(date,[]).append(action)

print dict(d)

输出：

{'ghi789': {'2/12/2015': [['pullover', 'stop']]}, 'def456': {'6/14/2015': [['turn', 'wash', 'skid']], '11/24/2016': [['stop', 'wash', 'pullover', 'break']]}, 'abc123': {'10/15/2015': [['break', 'pullover', 'turn']], '12/25/16': [['break', 'pullover'], ['stop']]}}

将列表列表嵌套为嵌套字典中的值

2 个答案: