Re-assign column values in a pandas df
This question is related to rostering or staffing. I\'m trying to assign various jobs to individuals (employees). Using the df below,
`[Person]` =
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In writing my other answer, I slowly came around to the idea that the OP's algorithm might be easier to implement with an approach that focuses on the jobs (which can be different), instead of the people (which are all the same). Here's a solution that uses the job-centric approach:
from collections import Counter import numpy as np import pandas as pd def assignJob(job, assignedix, areasPerPerson): for i in range(len(assignedix)): if (areasPerPerson - len(assignedix[i])) >= len(job): assignedix[i].extend(job) return True else: return False def allocatePeople(df, areasPerPerson=3): areas = df['Area'].values times = pd.to_datetime(df['Time']).values peopleUniq = df['Person'].unique() npeople = int(np.ceil(areas.size / float(areasPerPerson))) # 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') jobs =[] _jobdict = {} for i,(area,hold) in enumerate(zip(areas, holds)): if hold: _jobdict[area] = job = _jobdict.get(area, []) + [i] if len(job)==areasPerPerson: jobs.append(_jobdict.pop(area)) elif area in _jobdict: jobs.append(_jobdict.pop(area) + [i]) else: jobs.append([i]) jobs.sort() assignedix = [[] for i in range(npeople)] for job in jobs: if not assignJob(job, assignedix, areasPerPerson): # break the job up and try again for subjob in ([sj] for sj in job): assignJob(subjob, assignedix, areasPerPerson) df = df.copy() for i,aix in enumerate(assignedix): df.loc[aix, 'Person'] = peopleUniq[i] return dfThis version of
allocatePeoplehas also been extensively tested and passes all of the same checks described in my other answer.It does have more looping than my other solution, so it is likely to be slightly less efficient (though it'll only matter if your dataframe is very large, say
1e6rows and up). On the other hand, it is somewhat shorter and, I think, more straightforward and easy to understand.讨论(0) -
Ok, before we delve into the logic of the problem it is worthwhile to do some housekeeping to tidy-up the data and bring it into a more useful format:
#Create table of unique people unique_people = df[['Person']].drop_duplicates().sort_values(['Person']).reset_index(drop=True) #Reformat time column df['Time'] = pd.to_datetime(df['Time'])Now, getting to the logic of the problem, it is useful to break the problem down in to stages. Firstly, we will want to create individual jobs (with job numbers) based on the 'Area' and the time between them. i.e. jobs in the same area, within an hour can share the same job number.
#Assign jobs df= df.sort_values(['Area','Time']).reset_index(drop=True) df['Job no'] = 0 current_job = 1 df.loc[0,'Job no'] = current_job for i in range(rows-1): prev_row = df.loc[i] row = df.loc[i+1] time_diff = (row['Time'] - prev_row['Time']).seconds //3600 if (row['Area'] == prev_row['Area']) & (time_diff == 0): pass else: current_job +=1 df.loc[i+1,'Job no'] = current_jobWith this step now out of the way, it is a simple matter of assigning 'Persons' to individual jobs:
df= df.sort_values(['Job no']).reset_index(drop=True) df['Person'] = "" df_groups = df.groupby('Job no') for group in df_groups: group_size = group[1].count()['Time'] for person_idx in range(len(unique_people)): person = unique_people.loc[person_idx]['Person'] person_count = df[df['Person']==person]['Person'].count() if group_size <= (3-person_count): idx = group[1].index.values df.loc[idx,'Person'] = person breakAnd finally,
df= df.sort_values(['Time']).reset_index(drop=True) print(df)I've attempted to code this in a way that is easier to unpick, so there may well be efficiencies to be made here. The aim however was to set out the logic used.
This code gives the expected results on both data sets, so I hope it answers your question.
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Update
There's a live version of this answer online that you can try for yourself.
Here's an answer in the form of the
allocatePeoplefunction. 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 dfNote the use of
df['Person'].unique()inallocatePeople. 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.I tested
allocatePeopleagainst the OP's example input (example1andexample2) 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='\n', end='\n\n') np.testing.assert_array_equal(ap['Person'], expected)The
assert_array_equalstatements 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 1The 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 3Let 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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