【问题标题】：Re-assign unique values - pandas DataFrame重新分配唯一值 - pandas DataFrame
【发布时间】：2019-06-03 17:09:37
【问题描述】：

我正在尝试将 assign unique 中的值 pandas df 分配给特定的个人。

对于下面的df，[Area] 和[Place] 将共同构成unique 值，它们是各种工作。这些值将分配给个人，总体目标是使用尽可能少的个人。

诀窍是这些值不断地开始和结束，并持续不同的时间长度。任何时候分配给个人的unique 值最多为3。 [On] 显示当前出现的 [Place] 和 [Area] 唯一值的数量。

因此，这为我需要多少人提供了具体指南。例如3 unique 值 1 = 1 人，6 唯一值 = 2 人

我无法执行groupby 语句，其中我将assign 第一个3 unique values 到individual 1 和接下来的3 个unique 值到individual 2 等。

我的设想是，当 unique 的值大于 3 时，我想先将 [Area] 中的值分组，然后合并剩余部分。因此，请查看 assign 中 [Area] 中的相同值给个人（最多 3 个）。然后，如果有_leftover_ 值(，则应尽可能将它们组合成一组 3。

我设想这项工作的方式是：展望未来hour。对于每个新的row 值，script 应该看到有多少值是[On]（这表明需要多少个人）。如果 unique 的值 >3，则它们应该是 assigned 乘以 grouping 与 [Area] 中的相同值。如果有剩余值，则无论如何都应该将它们组合成一组 3。

对于下面的 df，[Place] 和 [Area] 出现的 unique 值的数量在 1-6 之间变化。所以我们不应该有超过 2 个人 assigned。当unique 值>3 时，应首先由[Area] 分配。剩余值应与具有少于 3 个unique 值的其他个体组合。

为大 df 道歉。这是我可以复制问题的唯一方法！

import pandas as pd
import numpy as np
from collections import Counter

d = ({   
    'Time' : ['8:03:00','8:17:00','8:20:00','8:33:00','8:47:00','8:48:00','9:03:00','9:15:00','9:18:00','9:33:00','9:45:00','9:48:00','10:03:00','10:15:00','10:15:00','10:15:00','10:18:00','10:32:00','10:33:00','10:39:00','10:43:00','10:48:00','10:50:00','11:03:00','11:03:00','11:07:00','11:25:00','11:27:00','11:42:00','11:48:00','11:51:00','11:57:00','12:00:00','12:08:00','12:15:00','12:17:00','12:25:00','12:30:00','12:35:00','12:39:00','12:47:00','12:52:00','12:55:00','13:00:00','13:03:00','13:07:00','13:12:00','13:15:00','13:22:00','13:27:00','13:27:00'],
    'Area' : ['A','A','A','A','A','A','A','A','A','A','A','A','A','A','A','B','A','B','A','A','A','A','B','A','A','B','B','A','B','C','A','B','C','C','A','B','C','C','B','A','C','B','C','C','A','C','B','C','C','A','C'],
    'Place' : ['House 1','House 2','House 3','House 1','House 3','House 2','House 1','House 3','House 2','House 1','House 3','House 2','House 1','House 3','House 4','House 1','House 2','House 1','House 1','House 4','House 3','House 2','House 1','House 1','House 4','House 1','House 1','House 4','House 1','House 1','House 4','House 1','House 2','House 1','House 4','House 1','House 1','House 2','House 1','House 4','House 1','House 1','House 3','House 2','House 4','House 1','House 2','House 4','House 1','House 4','House 2'],
    'On' : ['1','2','3','3','3','3','3','3','3','3','3','3','3','3','4','5','5','5','5','5','5','4','3','3','3','2','2','2','2','3','3','3','4','4','4','4','4','4','4','4','4','4','4','4','4','4','5','6','6','6','6'],
    'Person' : ['Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 2','Person 3','Person 1','Person 3','Person 1','Person 2','Person 1','Person 1','Person 3','Person 1','Person 2','Person 3','Person 3','Person 2','Person 3','Person 4','Person 2','Person 3','Person 4','Person 4','Person 2','Person 3','Person 4','Person 4','Person 3','Person 2','Person 4','Person 3','Person 4','Person 4','Person 2','Person 4','Person 3','Person 5','Person 4','Person 2','Person 4'],
    })

df = pd.DataFrame(data=d)

def getAssignedPeople(df, areasPerPerson):
    areas = df['Area'].values
    places = df['Place'].values
    times = pd.to_datetime(df['Time']).values
    maxPerson = np.ceil(areas.size / float(areasPerPerson)) - 1
    assignmentCount = Counter()
    assignedPeople = []
    assignedPlaces = {}
    heldPeople = {}
    heldAreas = {}
    holdAvailable = True
    person = 0

    # search for repeated areas. Mark them if the next repeat occurs within an hour
    ixrep = np.argmax(np.triu(areas.reshape(-1, 1)==areas, k=1), axis=1)
    holds = np.zeros(areas.size, dtype=bool)
    holds[ixrep.nonzero()] = (times[ixrep[ixrep.nonzero()]] - times[ixrep.nonzero()]) < np.timedelta64(1, 'h')

    for area,place,hold in zip(areas, places, holds):
        if (area, place) in assignedPlaces:
            # this unique (area, place) has already been assigned to someone
            assignedPeople.append(assignedPlaces[(area, place)])
            continue

        if assignmentCount[person] >= areasPerPerson:
            # the current person is already assigned to enough areas, move on to the next
            a = heldPeople.pop(person, None)
            heldAreas.pop(a, None)
            person += 1

        if area in heldAreas:
            # assign to the person held in this area
            p = heldAreas.pop(area)
            heldPeople.pop(p)
        else:
            # get the first non-held person. If we need to hold in this area, 
            # also make sure the person has at least 2 free assignment slots,
            # though if it's the last person assign to them anyway 
            p = person
            while p in heldPeople or (hold and holdAvailable and (areasPerPerson - assignmentCount[p] < 2)) and not p==maxPerson:
                p += 1

        assignmentCount.update([p])
        assignedPlaces[(area, place)] = p
        assignedPeople.append(p)

        if hold:
            if p==maxPerson:
                # mark that there are no more people available to perform holds
                holdAvailable = False

            # this area recurrs in an hour, mark that the person should be held here
            heldPeople[p] = area
            heldAreas[area] = p

    return assignedPeople

def allocatePeople(df, areasPerPerson=3):
    assignedPeople = getAssignedPeople(df, areasPerPerson=areasPerPerson)
    df = df.copy()
    df.loc[:,'Person'] = df['Person'].unique()[assignedPeople]
    return df

print(allocatePeople(df))

输出：

        Time Area    Place On    Person
0    8:03:00    A  House 1  1  Person 1
1    8:17:00    A  House 2  2  Person 1
2    8:20:00    A  House 3  3  Person 1
3    8:33:00    A  House 1  3  Person 1
4    8:47:00    A  House 3  3  Person 1
5    8:48:00    A  House 2  3  Person 1
6    9:03:00    A  House 1  3  Person 1
7    9:15:00    A  House 3  3  Person 1
8    9:18:00    A  House 2  3  Person 1
9    9:33:00    A  House 1  3  Person 1
10   9:45:00    A  House 3  3  Person 1
11   9:48:00    A  House 2  3  Person 1
12  10:03:00    A  House 1  3  Person 1
13  10:15:00    A  House 3  3  Person 1
14  10:15:00    A  House 4  4  Person 2
15  10:15:00    B  House 1  5  Person 2
16  10:18:00    A  House 2  5  Person 1
17  10:32:00    B  House 1  5  Person 2
18  10:33:00    A  House 1  5  Person 1
19  10:39:00    A  House 4  5  Person 2
20  10:43:00    A  House 3  5  Person 1
21  10:48:00    A  House 2  4  Person 1
22  10:50:00    B  House 1  3  Person 2
23  11:03:00    A  House 1  3  Person 1
24  11:03:00    A  House 4  3  Person 2
25  11:07:00    B  House 1  2  Person 2
26  11:25:00    B  House 1  2  Person 2
27  11:27:00    A  House 4  2  Person 2
28  11:42:00    B  House 1  2  Person 2
29  11:48:00    C  House 1  3  Person 2
30  11:51:00    A  House 4  3  Person 2
31  11:57:00    B  House 1  3  Person 2
32  12:00:00    C  House 2  4  Person 3
33  12:08:00    C  House 1  4  Person 2
34  12:15:00    A  House 4  4  Person 2
35  12:17:00    B  House 1  4  Person 2
36  12:25:00    C  House 1  4  Person 2
37  12:30:00    C  House 2  4  Person 3
38  12:35:00    B  House 1  4  Person 2
39  12:39:00    A  House 4  4  Person 2
40  12:47:00    C  House 1  4  Person 2
41  12:52:00    B  House 1  4  Person 2
42  12:55:00    C  House 3  4  Person 3
43  13:00:00    C  House 2  4  Person 3
44  13:03:00    A  House 4  4  Person 2
45  13:07:00    C  House 1  4  Person 2
46  13:12:00    B  House 2  5  Person 3
47  13:15:00    C  House 4  6  Person 4
48  13:22:00    C  House 1  6  Person 2
49  13:27:00    A  House 4  6  Person 2
50  13:27:00    C  House 2  6  Person 3

我认为应该分配它的预期输出和评论：

【问题讨论】：

这里的代码和文字太多了。请将其简化为minimal reproducible example，其中包含一些易于消化的简单数据。
您没有发布数据框df，它为您提供了您发布的输出。请补充一下，因为没有它，这个问题是无法回答的。
另外，虽然在last question 中找出汤这个词是一个有趣的挑战，但它有点像试图破译死海古卷。我将回应coldspeed 并请您阅读MCV example 文档并将其牢记在心。下次写 SO 问题时，请像圣经一样遵循它。这将帮助人们写出更好、更通用的答案，从而在未来帮助其他寻求答案的人。
到目前为止df 变量是什么？
目前您的问题无法回答，只有知道df的样子我们才会给出答案，请发帖df...

标签： python pandas numpy dataframe assign

【解决方案1】：

There's a live version of this answer online that you can try for yourself.

问题

您看到的错误是由于您的问题的（又一个）有趣的边缘案例。在6th 作业期间，代码将person 2 分配给(A, House 4)。然后它会看到A 区域在一小时内重复，因此它在该区域中保持person 2。这使得person 2 无法用于下一个工作，该工作位于区域B。

但是，没有理由为了在(A, House 1) 中发生的工作而在区域A 中持有person 2，因为区域和地点(A, House 1) 的唯一组合已经分配给person 1 .

解决办法

在决定何时将人留在某个区域时，可以通过仅考虑区域和地点的独特组合来解决此问题。只需更改几行代码。

首先，我们构造一个对应于唯一（区域、地点）对的区域列表：

unqareas = df[['Area', 'Place']].drop_duplicates()['Area'].values

然后我们只需在标识保持的代码的第一行中将unqareas 替换为areas：

ixrep = np.argmax(np.triu(unqareas.reshape(-1, 1)==unqareas, k=1), axis=1)

完整的清单/测试

import pandas as pd
import numpy as np
from collections import Counter

d = ({
     'Time' : ['8:03:00','8:07:00','8:10:00','8:23:00','8:27:00','8:30:00','8:37:00','8:40:00','8:48:00'],
     'Place' : ['House 1','House 2','House 3','House 1','House 2','House 3','House 4','House 1','House 1'],
     'Area' : ['A','A','A','A','A','A','A','B','A'],
     'Person' : ['Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 2','Person 3','Person 1'],
     'On' : ['1','2','3','3','3','3','4','5','5']
     })

df = pd.DataFrame(data=d)

def getAssignedPeople(df, areasPerPerson):
    areas = df['Area'].values
    unqareas = df[['Area', 'Place']].drop_duplicates()['Area'].values
    places = df['Place'].values
    times = pd.to_datetime(df['Time']).values

    maxPerson = np.ceil(areas.size / float(areasPerPerson)) - 1
    assignmentCount = Counter()
    assignedPeople = []
    assignedPlaces = {}
    heldPeople = {}
    heldAreas = {}
    holdAvailable = True
    person = 0

    # search for repeated areas. Mark them if the next repeat occurs within an hour
    ixrep = np.argmax(np.triu(unqareas.reshape(-1, 1)==unqareas, k=1), axis=1)
    holds = np.zeros(areas.size, dtype=bool)
    holds[ixrep.nonzero()] = (times[ixrep[ixrep.nonzero()]] - times[ixrep.nonzero()]) < np.timedelta64(1, 'h')

    for area,place,hold in zip(areas, places, holds):
        if (area, place) in assignedPlaces:
            # this unique (area, place) has already been assigned to someone
            assignedPeople.append(assignedPlaces[(area, place)])
            continue

        if assignmentCount[person] >= areasPerPerson:
            # the current person is already assigned to enough areas, move on to the next
            a = heldPeople.pop(person, None)
            heldAreas.pop(a, None)
            person += 1

        if area in heldAreas:
            # assign to the person held in this area
            p = heldAreas.pop(area)
            heldPeople.pop(p)
        else:
            # get the first non-held person. If we need to hold in this area, 
            # also make sure the person has at least 2 free assignment slots,
            # though if it's the last person assign to them anyway 
            p = person
            while p in heldPeople or (hold and holdAvailable and (areasPerPerson - assignmentCount[p] < 2)) and not p==maxPerson:
                p += 1

        assignmentCount.update([p])
        assignedPlaces[(area, place)] = p
        assignedPeople.append(p)

        if hold:
            if p==maxPerson:
                # mark that there are no more people available to perform holds
                holdAvailable = False

            # this area recurrs in an hour, mark that the person should be held here
            heldPeople[p] = area
            heldAreas[area] = p

    return assignedPeople

def allocatePeople(df, areasPerPerson=3):
    assignedPeople = getAssignedPeople(df, areasPerPerson=areasPerPerson)
    df = df.copy()
    df.loc[:,'Person'] = df['Person'].unique()[assignedPeople]
    return df

print(allocatePeople(df))

输出：

      Time    Place Area    Person On
0  8:03:00  House 1    A  Person 1  1
1  8:07:00  House 2    A  Person 1  2
2  8:10:00  House 3    A  Person 1  3
3  8:23:00  House 1    A  Person 1  3
4  8:27:00  House 2    A  Person 1  3
5  8:30:00  House 3    A  Person 1  3
6  8:37:00  House 4    A  Person 2  4
7  8:40:00  House 1    B  Person 2  5
8  8:48:00  House 1    A  Person 1  5

【讨论】：