当前尝试
在下面,我已经在你的代码最后一行之前添加了几行:
d = ({'Time': ['8:03:00', '8:17:00', '8:20:00', '10:15:00', '10:15:00', '11:48:00', '12:00:00', '12:10:00'],
'Place': ['House 1', 'House 2', 'House 1', 'House 3', 'House 4', 'House 5', 'House 1', 'House 1'],
'Area': ['X', 'X', 'Y', 'X', 'X', 'X', 'X', 'X']})
df = pd.DataFrame(data=d)
def g(gps):
s = gps['Place'].unique()
d = dict(zip(s, np.arange(len(s)) // 3 + 1))
gps['Person'] = gps['Place'].map(d)
return gps
df = df.groupby('Area', sort=False).apply(g)
s = df['Person'].astype(str) + df['Area']
t = s.value_counts()
df_sub = df.loc[s[s.isin(t[t < 3].index)].index].copy()
df_sub["tag"] = df_sub["Place"] + df_sub["Area"]
tags = list(df_sub.tag.unique())
f = lambda x: f'R{int(tags.index(x) / 3) + 1}'
df_sub['reassign'] = df_sub.tag.apply(f)
s[s.isin(t[t < 3].index)] = df_sub['reassign']
df['Person'] = pd.Series(pd.factorize(s)[0] + 1).map(str).radd('Person ')
老实说,我不确定它是否适用于所有情况,但在测试案例中它能给出你想要的输出结果。
之前的尝试:
让我们看看我是否能帮助你有限的理解你所要做的事情。
你有一些连续的数据(我称之为事件),你想为每个事件分配一个“人员”标识符。你将分配到每个连续事件上的标识符取决于先前的分配,我认为它需要按以下规则逐步应用:
1. “我认识你”:如果给定标识符的“地点”和“区域”具有相同的值(时间有关系吗?),则可以重用以前的标识符。
2. “我不认识你”:如果出现了新的区域值(那么地点和区域发挥不同的角色吗?),则会创建一个新的标识符。
3. “我是否认识你?”:如果一个标识符未被分配到至少三个事件(如果这对多个标识符都发生了怎么办?我会假设我使用最旧的...),则可能重新使用以前使用过的标识符。
4. “不,我不认识”:如果前面的规则都不适用,则会创建一个新的标识符。
在假设了以上内容后,以下是一个解决方案的实现:
people = dict()
persons = list()
def i_know_you(people, now):
def conditions(now, past):
return [e for e in past if (now.Place == e.Place) and (now.Area == e.Area)]
i_do = [person for person, past in people.items() if conditions(now, past)]
if i_do:
return i_do[0]
return False
def i_do_not_know_you(people, now):
conditions = not bool([e for past in people.values() for e in past if e.Area == now.Area])
if conditions:
return f'Person {len(people) + 1}'
return False
def do_i_know_you(people, now):
i_do = [person for person, past in people.items() if len(past) < 3]
if i_do:
return i_do[0]
return False
for event in df.itertuples():
print('event:', event)
for rule in [i_know_you, i_do_not_know_you, do_i_know_you]:
person = rule(people, event)
print('\t', rule.__name__, person)
if person:
break
if not person:
person = f'Person {len(people) + 1}'
print('\t', "nah, I don't", person)
if person in people:
people[person].append(event)
else:
people[person] = [event]
persons.append(person)
df['Person'] = persons
输出:
event: Pandas(Index=0, Time='8:00:00', Place='House 1', Area='X', Person='Person 1')
i_know_you False
i_do_not_know_you Person 1
event: Pandas(Index=1, Time='8:30:00', Place='House 2', Area='X', Person='Person 1')
i_know_you False
i_do_not_know_you False
do_i_know_you Person 1
event: Pandas(Index=2, Time='9:00:00', Place='House 1', Area='Y', Person='Person 2')
i_know_you False
i_do_not_know_you Person 2
event: Pandas(Index=3, Time='9:30:00', Place='House 3', Area='X', Person='Person 1')
i_know_you False
i_do_not_know_you False
do_i_know_you Person 1
event: Pandas(Index=4, Time='10:00:00', Place='House 4', Area='X', Person='Person 2')
i_know_you False
i_do_not_know_you False
do_i_know_you Person 2
event: Pandas(Index=5, Time='10:30:00', Place='House 5', Area='X', Person='Person 2')
i_know_you False
i_do_not_know_you False
do_i_know_you Person 2
event: Pandas(Index=6, Time='11:00:00', Place='House 1', Area='X', Person='Person 1')
i_know_you Person 1
event: Pandas(Index=7, Time='11:30:00', Place='House 6', Area='X', Person='Person 3')
i_know_you False
i_do_not_know_you False
do_i_know_you False
nah, I don't Person 3
event: Pandas(Index=8, Time='12:00:00', Place='House 7', Area='X', Person='Person 3')
i_know_you False
i_do_not_know_you False
do_i_know_you Person 3
event: Pandas(Index=9, Time='12:30:00', Place='House 8', Area='X', Person='Person 3')
i_know_you False
i_do_not_know_you False
do_i_know_you Person 3
最终的数据框如您所需:
Time Place Area Person
0 8:00:00 House 1 X Person 1
1 8:30:00 House 2 X Person 1
2 9:00:00 House 1 Y Person 2
3 9:30:00 House 3 X Person 1
4 10:00:00 House 4 X Person 2
5 10:30:00 House 5 X Person 2
6 11:00:00 House 1 X Person 1
7 11:30:00 House 6 X Person 3
8 12:00:00 House 7 X Person 3
9 12:30:00 House 8 X Person 3
备注: 注意,我故意避免使用分组操作并按顺序处理数据。我认为这种复杂性(
而且不太理解你想做什么...)需要采用这种方法。此外,您可以使用上面相同的结构使规则更加复杂(
时间是否真的起到作用?)。
新数据更新的答案
查看新数据,很明显我没有理解你想做什么(特别是,该任务似乎不遵循顺序规则)。
我有一个解决方案适用于第二个数据集,但对第一个数据集会产生不同的结果。
解决方案要简单得多,将添加一列(如果您愿意,可以稍后删除):
df["tag"] = df["Place"] + df["Area"]
tags = list(df.tag.unique())
f = lambda x: f'Person {int(tags.index(x) / 3) + 1}'
df['Person'] = df.tag.apply(f)
在第二个数据集上,它会得到:
Time Place Area tag Person
0 8:00:00 House 1 X House 1X Person 1
1 8:30:00 House 2 X House 2X Person 1
2 9:00:00 House 3 X House 3X Person 1
3 9:30:00 House 1 Y House 1Y Person 2
4 10:00:00 House 1 Z House 1Z Person 2
5 10:30:00 House 1 V House 1V Person 2
在第一个数据集上,它给出:
Time Place Area tag Person
0 8:00:00 House 1 X House 1X Person 1
1 8:30:00 House 2 X House 2X Person 1
2 9:00:00 House 1 Y House 1Y Person 1
3 9:30:00 House 3 X House 3X Person 2
4 10:00:00 House 4 X House 4X Person 2
5 10:30:00 House 5 X House 5X Person 2
6 11:00:00 House 1 X House 1X Person 1
7 11:30:00 House 6 X House 6X Person 3
8 12:00:00 House 7 X House 7X Person 3
9 12:30:00 House 8 X House 8X Person 3
这与您在索引2和3上的预期输出不同。这个输出是否符合您的要求?如果不是,为什么?