在 Pandas df 中重新分配列值 [英] Re-assign column values in a pandas df
问题描述
这个问题与排班或人员配备有关.我正在尝试为个人(员工)分配各种工作.使用下面的 df，
This question is related to rostering or staffing. I'm trying to assign various jobs to individuals (employees). Using the df below,
`[Person]` = Individuals (employees)
`[Area]` and `[Place]` = unique jobs
`[On]` = How many unique jobs are occurring at each point in time
所以 [Area] 和 [Place] 一起构成了不同作业的 unique 值.这些值将分配给个人，总体目标是尽可能少地使用个人.分配的给任何一个人的最独特的值是3.[On]显示[Place]<的当前unique值的数量/code> 和 [Area] 正在发生.因此，这提供了有关我需要多少个人的具体指南.例如，
So [Area] and [Place] together will make up unique values that are different jobs. These values will be assigned to individuals with the overall aim to use the least amount of individuals possible. The most unique values assigned to any one individual is 3. [On] displays how many current unique values for [Place] and [Area] are occurring. So this provides a concrete guide on how many individuals I need. For example,
1-3 unique values occurring = 1 individual
4-6 unique values occurring = 2 individuals
7-9 unique values occurring = 3 individuals etc
问题:[Area] 和 [Place] 中 unique 值的数量大于 3 给我带来了麻烦.我不能做一个 groupby，在那里我 assign 第一个 3 唯一值 到 individual 1 和接下来的 3unique 值到 individual 2 等.我想将 [Area] 和 [Place] 中的唯一值分组[区域].因此，希望将 [Area] 中的 assign 相同的值分配给一个人(最多 3 个).然后，如果有 剩余 值 (<3)，则应尽可能将它们组合成一组 3.
Question:
Where the amount of unique values in [Area] and [Place] is greater than 3 is causing me trouble. I can't do a groupby where I assign the first 3 unique values to individual 1 and the next 3 unique values to individual 2 etc. I want to group unique values in [Area] and [Place] by [Area]. So look to assign same values in [Area] to an individual (up to 3). Then, if there are leftover values (<3), they should be combined to make a group of 3, where possible.
我设想的工作方式是:在小时之前看到未来.对于每个新的 row 值，script 应该看到有多少个值是 [On](这提供了有多少个人是必需的).当unique 值>3 时，它们应该通过grouping [Area] 中的相同值assigned.如果有剩余值，无论如何都应该将它们组合起来组成一组 3.
The way I envisage this working is: see into the future by an hour. For each new row of values the script should see how many values will be [On](this provides an indication of how many total individuals are required). Where unique values are >3, they should be assigned by grouping the same value in [Area]. If there are leftover values they should be combined anyhow to make up to a group of 3.
将其放入一步一步的过程中:
Putting that into a step by step process:
1) 使用 [On] Column 来确定需要多少个人，通过展望未来小时
1) Use the [On] Column to determine how many individuals are required by looking into the future for an hour
2) 如果出现超过 3 个 unique 值，则首先在 [Area] 中分配相同的值.
2) Where there are more than 3 unique values occurring assign the identical values in [Area] first.
3) 如果有任何剩余值，则尽可能组合.
3) If there are any leftover values then look to combine anyway possible.
对于下面的 df，有 9 个 unique 值出现在 [Place] 和 [Area]带有 hour.所以我们应该有3个人assigned.当 unique 值 >3 时，它应该由 [Area] 分配并查看是否出现相同的值.剩余的值应该与少于 3 个 unique 值的其他个体相结合.
For the df below, there are 9 unique values occurring for [Place] and [Area] with an hour. So we should have 3 individuals assigned. When unique values >3 it should be assigned by [Area] and seeing if the same value occurs. The leftover values should be combined with other individuals that have less than 3 unique values.
import pandas as pd
import numpy as np
d = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','08:40:00','08:42:00','08:45:00','08:50:00'],
'Place' : ['House 1','House 2','House 3','House 4','House 5','House 1','House 2','House 3','House 2'],
'Area' : ['A','B','C','D','E','D','E','F','G'],
'On' : ['1','2','3','4','5','6','7','8','9'],
'Person' : ['Person 1','Person 2','Person 3','Person 4','Person 5','Person 4','Person 5','Person 6','Person 7'],
})
df = pd.DataFrame(data=d)
这是我的尝试:
def reduce_df(df):
values = df['Area'] + df['Place']
df1 = df.loc[~values.duplicated(),:] # ignore duplicate values for this part..
person_count = df1.groupby('Person')['Person'].agg('count')
leftover_count = person_count[person_count < 3] # the 'leftovers'
# try merging pairs together
nleft = leftover_count.shape[0]
to_try = np.arange(nleft - 1)
to_merge = (leftover_count.values[to_try] +
leftover_count.values[to_try + 1]) <= 3
to_merge[1:] = to_merge[1:] & ~to_merge[:-1]
to_merge = to_try[to_merge]
merge_dict = dict(zip(leftover_count.index.values[to_merge+1],
leftover_count.index.values[to_merge]))
def change_person(p):
if p in merge_dict.keys():
return merge_dict[p]
return p
reduced_df = df.copy()
# update df with the merges you found
reduced_df['Person'] = reduced_df['Person'].apply(change_person)
return reduced_df
df1 = (reduce_df(reduce_df(df)))
这是输出:
Time Place Area On Person
0 8:03:00 House 1 A 1 Person 1
1 8:17:00 House 2 B 2 Person 1
2 8:20:00 House 3 C 3 Person 1
3 8:28:00 House 4 D 4 Person 4
4 8:35:00 House 5 E 5 Person 5
5 8:40:00 House 1 D 6 Person 4
6 8:42:00 House 2 E 7 Person 5
7 8:45:00 House 3 F 8 Person 5
8 8:50:00 House 2 G 9 Person 7
这是我的预期输出:
Time Place Area On Person
0 8:03:00 House 1 A 1 Person 1
1 8:17:00 House 2 B 2 Person 1
2 8:20:00 House 3 C 3 Person 1
3 8:28:00 House 4 D 4 Person 2
4 8:35:00 House 5 E 5 Person 3
5 8:40:00 House 6 D 6 Person 2
6 8:42:00 House 2 E 7 Person 3
7 8:45:00 House 3 F 8 Person 2
8 8:50:00 House 2 G 9 Person 3
关于我想如何获得此输出的说明:
Description on how I want to get this output:
Index 0: One `unique` value occurring. So `assign` to individual 1
Index 1: Two `unique` values occurring. So `assign` to individual 1
Index 2: Three `unique` values occurring. So `assign` to individual 1
Index 3: Four `unique` values on. So `assign` to individual 2
Index 4: Five `unique` values on. This one is a bit tricky and hard to conceptualise. But there is another `E` within an `hour`. So `assign` to a new individual so it can be combined with the other `E`
Index 5: Six `unique` values on. Should be `assigned` with the other `D`. So individual 2
Index 6: Seven `unique` values on. Should be `assigned` with other `E`. So individual 3
Index 7: Eight `unique` values on. New value in `[Area]`, which is a _leftover_. `Assign` to either individual 2 or 3
Index 8: Nine `unique` values on. New value in `[Area]`, which is a _leftover_. `Assign` to either individual 2 or 3
示例 2:
d = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','8:40:00','8:42:00','8:45:00','8:50:00'],
'Place' : ['House 1','House 2','House 3','House 1','House 2','House 3','House 1','House 2','House 3'],
'Area' : ['X','X','X','X','X','X','X','X','X'],
'On' : ['1','2','3','3','3','3','3','3','3'],
'Person' : ['Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1'],
})
df = pd.DataFrame(data=d)
我收到一个错误:
IndexError: index 1 is out of bounds for axis 1 with size 1
在这一行:
df.loc[:,'Person'] = df['Person'].unique()[assignedPeople]
但是，如果我将 Person 更改为 1,2,3 重复，它会返回以下内容:
However, if I change the Person to 1,2,3 repeating, it returns the following:
'Person' : ['Person 1','Person 2','Person 3','Person 1','Person 2','Person 3','Person 1','Person 2','Person 3'],
Time Place Area On Person
0 8:03:00 House 1 X 1 Person 1
1 8:17:00 House 2 X 2 Person 1
2 8:20:00 House 3 X 3 Person 1
3 8:28:00 House 1 X 3 Person 2
4 8:35:00 House 2 X 3 Person 2
5 8:40:00 House 3 X 3 Person 2
6 8:42:00 House 1 X 3 Person 3
7 8:45:00 House 2 X 3 Person 3
8 8:50:00 House 3 X 3 Person 3
预期输出:
Time Place Area On Person
0 8:03:00 House 1 X 1 Person 1
1 8:17:00 House 2 X 2 Person 1
2 8:20:00 House 3 X 3 Person 1
3 8:28:00 House 1 X 3 Person 1
4 8:35:00 House 2 X 3 Person 1
5 8:40:00 House 3 X 3 Person 1
6 8:42:00 House 1 X 3 Person 1
7 8:45:00 House 2 X 3 Person 1
8 8:50:00 House 3 X 3 Person 1
示例 2 的主要内容是:
The main takeaway from Example 2 is:
1) There are <3 unique values on so assign to individual 1
推荐答案
更新
这是 allocatePeople 函数形式的答案.它基于预先计算区域在一小时内重复的所有索引:
Here's an answer in the form of the allocatePeople function. It's based around precomputing all of the indices where the areas repeat within an hour:
from collections import Counter
import numpy as np
import pandas as pd
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
注意 df['Person'].unique() 在 allocatePeople 中的使用.这处理了人们在输入中重复的情况.假设输入中人员的顺序是分配这些人员时所需的顺序.
Note the use of df['Person'].unique() in allocatePeople. That handles the case where people are repeated in the input. It is assumed that the order of people in the input is the desired order in which those people should be assigned.
我针对 OP 的示例输入(example1 和 example2)测试了 allocatePeople，还针对我想出的几个边缘情况进行了测试我认为(?)匹配 OP 所需的算法:
I tested allocatePeople against the OP's example input (example1 and example2) and also against a couple of edge cases I came up with that I think(?) match the OP's desired algorithm:
ds = dict(
example1 = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','08:40:00','08:42:00','08:45:00','08:50:00'],
'Place' : ['House 1','House 2','House 3','House 4','House 5','House 1','House 2','House 3','House 2'],
'Area' : ['A','B','C','D','E','D','E','F','G'],
'On' : ['1','2','3','4','5','6','7','8','9'],
'Person' : ['Person 1','Person 2','Person 3','Person 4','Person 5','Person 4','Person 5','Person 6','Person 7'],
}),
example2 = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','8:40:00','8:42:00','8:45:00','8:50:00'],
'Place' : ['House 1','House 2','House 3','House 1','House 2','House 3','House 1','House 2','House 3'],
'Area' : ['X','X','X','X','X','X','X','X','X'],
'On' : ['1','2','3','3','3','3','3','3','3'],
'Person' : ['Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1','Person 1'],
}),
long_repeats = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:25:00','8:30:00','8:31:00','8:35:00','8:45:00','8:50:00'],
'Place' : ['House 1','House 2','House 3','House 4','House 1','House 1','House 2','House 3','House 2'],
'Area' : ['A','A','A','A','B','C','C','C','B'],
'Person' : ['Person 1','Person 1','Person 1','Person 2','Person 3','Person 4','Person 4','Person 4','Person 3'],
'On' : ['1','2','3','4','5','6','7','8','9'],
}),
many_repeats = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','08:40:00','08:42:00','08:45:00','08:50:00'],
'Place' : ['House 1','House 2','House 3','House 4','House 1','House 1','House 2','House 1','House 2'],
'Area' : ['A', 'B', 'C', 'D', 'D', 'E', 'E', 'F', 'F'],
'On' : ['1','2','3','4','5','6','7','8','9'],
'Person' : ['Person 1','Person 1','Person 1','Person 2','Person 3','Person 4','Person 3','Person 5','Person 6'],
}),
large_gap = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','08:40:00','08:42:00','08:45:00','08:50:00'],
'Place' : ['House 1','House 2','House 3','House 4','House 1','House 1','House 2','House 1','House 3'],
'Area' : ['A', 'B', 'C', 'D', 'E', 'F', 'D', 'D', 'D'],
'On' : ['1','2','3','4','5','6','7','8','9'],
'Person' : ['Person 1','Person 1','Person 1','Person 2','Person 3','Person 4','Person 3','Person 5','Person 6'],
}),
different_times = ({
'Time' : ['8:03:00','8:17:00','8:20:00','8:28:00','8:35:00','08:40:00','09:42:00','09:45:00','09:50:00'],
'Place' : ['House 1','House 2','House 3','House 4','House 1','House 1','House 2','House 1','House 1'],
'Area' : ['A', 'B', 'C', 'D', 'D', 'E', 'E', 'F', 'G'],
'On' : ['1','2','3','4','5','6','7','8','9'],
'Person' : ['Person 1','Person 1','Person 1','Person 2','Person 3','Person 4','Person 3','Person 5','Person 6'],
})
)
expectedPeoples = dict(
example1 = [1,1,1,2,3,2,3,2,3],
example2 = [1,1,1,1,1,1,1,1,1],
long_repeats = [1,1,1,2,2,3,3,3,2],
many_repeats = [1,1,1,2,2,3,3,2,3],
large_gap = [1,1,1,2,3,3,2,2,3],
different_times = [1,1,1,2,2,2,3,3,3],
)
for name,d in ds.items():
df = pd.DataFrame(d)
expected = ['Person %d' % i for i in expectedPeoples[name]]
ap = allocatePeople(df)
print(name, ap, sep='
', end='
')
np.testing.assert_array_equal(ap['Person'], expected)
assert_array_equal 语句通过，并且输出与 OP 的预期输出匹配:
The assert_array_equal statements pass, and the output matches OP's expected output:
example1
Time Place Area On Person
0 8:03:00 House 1 A 1 Person 1
1 8:17:00 House 2 B 2 Person 1
2 8:20:00 House 3 C 3 Person 1
3 8:28:00 House 4 D 4 Person 2
4 8:35:00 House 5 E 5 Person 3
5 08:40:00 House 1 D 6 Person 2
6 08:42:00 House 2 E 7 Person 3
7 08:45:00 House 3 F 8 Person 2
8 08:50:00 House 2 G 9 Person 3
example2
Time Place Area On Person
0 8:03:00 House 1 X 1 Person 1
1 8:17:00 House 2 X 2 Person 1
2 8:20:00 House 3 X 3 Person 1
3 8:28:00 House 1 X 3 Person 1
4 8:35:00 House 2 X 3 Person 1
5 8:40:00 House 3 X 3 Person 1
6 8:42:00 House 1 X 3 Person 1
7 8:45:00 House 2 X 3 Person 1
8 8:50:00 House 3 X 3 Person 1
我的测试用例的输出也符合我的期望:
The output for my test cases matches my expectations as well:
long_repeats
Time Place Area Person On
0 8:03:00 House 1 A Person 1 1
1 8:17:00 House 2 A Person 1 2
2 8:20:00 House 3 A Person 1 3
3 8:25:00 House 4 A Person 2 4
4 8:30:00 House 1 B Person 2 5
5 8:31:00 House 1 C Person 3 6
6 8:35:00 House 2 C Person 3 7
7 8:45:00 House 3 C Person 3 8
8 8:50:00 House 2 B Person 2 9
many_repeats
Time Place Area On Person
0 8:03:00 House 1 A 1 Person 1
1 8:17:00 House 2 B 2 Person 1
2 8:20:00 House 3 C 3 Person 1
3 8:28:00 House 4 D 4 Person 2
4 8:35:00 House 1 D 5 Person 2
5 08:40:00 House 1 E 6 Person 3
6 08:42:00 House 2 E 7 Person 3
7 08:45:00 House 1 F 8 Person 2
8 08:50:00 House 2 F 9 Person 3
large_gap
Time Place Area On Person
0 8:03:00 House 1 A 1 Person 1
1 8:17:00 House 2 B 2 Person 1
2 8:20:00 House 3 C 3 Person 1
3 8:28:00 House 4 D 4 Person 2
4 8:35:00 House 1 E 5 Person 3
5 08:40:00 House 1 F 6 Person 3
6 08:42:00 House 2 D 7 Person 2
7 08:45:00 House 1 D 8 Person 2
8 08:50:00 House 3 D 9 Person 3
different_times
Time Place Area On Person
0 8:03:00 House 1 A 1 Person 1
1 8:17:00 House 2 B 2 Person 1
2 8:20:00 House 3 C 3 Person 1
3 8:28:00 House 4 D 4 Person 2
4 8:35:00 House 1 D 5 Person 2
5 08:40:00 House 1 E 6 Person 2
6 09:42:00 House 2 E 7 Person 3
7 09:45:00 House 1 F 8 Person 3
8 09:50:00 House 1 G 9 Person 3
让我知道它是否可以满足您的所有需求，或者是否仍需要进行一些调整.我想每个人都渴望看到你实现自己的愿景.
Let me know if it does everything you wanted, or if it still needs some tweaks. I think everyone is eager to see you fulfill your vision.
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