动态分配唯一值 - Python
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【中文标题】动态分配唯一值 - Python【英文标题】：Dynamic assignment of unique values - Python 【发布时间】：2020-02-13 19:13:28 【问题描述】：我正在尝试为特定分配或组分配唯一值。棘手的是这些独特的值是动态开始和结束的。因此，这些群体将保持先前看到的价值，并在不同的时间段采用新的独特价值。关于df，唯一值位于Place 中，可供选择的组位于Available Group 中，每个时间段Period。
我试图遵守的广泛准则是：
1) 每个Group在任何时候都不能超过 3 个唯一的Places
2) 当前唯一的Places 应均匀分布在每个Group 中
3) 一旦Places 被分配给Group，保持直到Group 完成。 除非Group 变为 NA 或会议分配不均
为了了解当前出现了多少Places，我已包含Total，它基于Place 值是否再次出现。我正在满足我的前两个准则和部分第三个准则。当 Place 分配给 Group 时，它会保持在相同的 Group 直到 Place 完成（不会再次出现）。
但是，我没有引用 Available Group 来了解 Group 是否可用。当Group 变得不可用时，我想将那些Places 重新排列在另一个Available Group 之间。使用下面的df，可以很好地分配位置，从而增加独特的位置。但是一旦他们开始完成并且第 2 组变得不可用，这些位置就不会重新分配给第 1 组。此时只有 3 个位置出现。
df = pd.DataFrame(
'Period' : [1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6],
'Place' : ['CLUB','CLUB','CLUB','HOME','HOME','AWAY','AWAY','WORK','WORK','AWAY','AWAY','GOLF','GOLF','CLUB','CLUB','POOL','POOL','HOME','HOME','WORK','WORK','AWAY','AWAY','POOL','POOL','TENNIS','TENNIS'],
'Total' : [1,1,1,2,2,3,3,4,4,4,4,5,5,4,4,4,4,4,4,4,4,4,4,4,4,5,5],
'Available Group' : ['1','2','1','2','1','2','1','2','1','1','2','1','2','2','1','2','1','2','1','2','1','1','2','1','2','2','1'],
)
尝试：
# df to store all unique places
uniquePlaces = pd.DataFrame(df["Place"].unique(), columns=["Place"])
# Start stores index of df where the place appears 1st
uniquePlaces["Start"] = -1
# End stores index of df where the place appears last
uniquePlaces["End"] = -1
def assign_place_label(group):
''' Create a label column that calculates the amount of unique meetings
throughout the racing schedule '''
label = uniquePlaces[uniquePlaces["Place"] == group.name].index[0]
group["Place Label"] = label
uniquePlaces.loc[label, "Start"] = group.index.min()
uniquePlaces.loc[label, "End"] = group.index.max()
return group
# Based on Start and End of each place, assign index to each place.
# when 'freed' the index is reused to new place appearing after that
def Assign_Meetings_group(up):
up["Index"] = 0
up["Freed"] = False
max_ind=0
free_indx = []
for i in range(len(up)):
ind_freed = up.index[(up["End"]<up.iloc[i]["Start"]) & (~up["Freed"])]
free = list(up.loc[ind_freed, "Index"])
free_indx += free
up.loc[ind_freed, "Freed"] = True
if len(free_indx)>0:
m = min(free_indx)
up.loc[i, "Index"] = m
free_indx.remove(m)
else:
up.loc[i, "Index"] = max_ind
max_ind+=1
up["Group"] = up["Index"]//3+1
return up
df2 = df.groupby("Place").apply(assign_place_label)
uniquePlaces = Assign_Meetings_group(uniquePlaces)
df3 = df2[df2['Period']!=0].drop_duplicates(subset = ['Period','Place'])
result = df3.merge(uniquePlaces[["Group"]], how="left", left_on="Place Label", right_index=True, sort=False)
输出：
Period Place Total Available Group Place Label Group
0 1 CLUB 1 1 0 1
1 2 CLUB 1 1 0 1
3 2 HOME 2 1 1 1
5 2 AWAY 3 1 2 1
7 3 WORK 4 1 3 2
9 3 AWAY 4 1 2 1
11 3 GOLF 5 1 4 2
13 3 HOME 5 1 1 1
15 4 CLUB 4 1 0 1
17 4 AWAY 3 1 2 1
19 4 POOL 3 1 5 1
21 5 WORK 3 1 3 2
23 5 POOL 2 1 5 1
25 6 GOLF 1 1 4 2
预期输出：
Period Place Total Available Group Place Label Group
0 1 CLUB 1 1 0 1
1 2 CLUB 1 1 0 1
3 2 HOME 2 1 1 1
5 2 AWAY 3 1 2 1
7 3 WORK 4 1 3 2
9 3 AWAY 4 1 2 1
11 3 GOLF 5 1 4 2
13 3 HOME 5 1 1 1
15 4 CLUB 4 1 0 1
17 4 AWAY 3 1 2 1
19 4 POOL 3 1 5 1
21 5 WORK 3 1 3 1
23 5 POOL 2 1 5 1
25 6 GOLF 1 1 4 1
【问题讨论】：
【参考方案1】：这是我的问题的解决方案，请在评论中找到详细信息
import pandas as pd
import numpy as np
df = pd.DataFrame(
'Period' : [1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,6,6],
'Place' : ['CLUB','CLUB','CLUB','HOME','HOME','AWAY','AWAY','WORK','WORK','AWAY','AWAY','GOLF','GOLF','CLUB','CLUB','POOL','POOL','HOME','HOME','WORK','WORK','AWAY','AWAY','POOL','POOL','TENNIS','TENNIS'],
'Total' : [1,1,1,2,2,3,3,4,4,4,4,5,5,4,4,4,4,4,4,4,4,4,4,4,4,5,5],
'Available Group' : ['1','2','1','2','1','2','1','2','1','1','2','1','2','2','1','2','1','2','1','2','1','1','2','1','2','2','1'],
)
# df to store all unique places
uniquePlaces = pd.DataFrame(df["Place"].unique(), columns=["Place"])
# Start stores index of df where the place appears 1st
uniquePlaces["Start"] = -1
# End stores index of df where the place appears last
uniquePlaces["End"] = -1
def assign_place_label(group):
''' Create a label column that calculates the amount of unique meetings
throughout the racing schedule '''
label = uniquePlaces[uniquePlaces["Place"] == group.name].index[0]
group["Place Label"] = label
uniquePlaces.loc[label, "Start"] = group.index.min()
uniquePlaces.loc[label, "End"] = group.index.max()+1
return group
df2 = df.groupby("Place").apply(assign_place_label)
def calc_groups(uniquePlaces, df2):
## group need to be changed only when a group starts or finishes
change_points = np.sort(uniquePlaces[["Start", "End"]].values.ravel()).reshape(-1,1)
## for each change points find boolean indxes for places (True if place is in use at that point)
inds = (change_points>=uniquePlaces["Start"].values) & (change_points<uniquePlaces["End"].values)
## all available indexes for place
all_ind = set(uniquePlaces.index.values+1)
prev_ind = np.array([0]*len(all_ind))
result = []
for ind in inds:
## copy prev_ind where place exists
new_ind = prev_ind * ind
## mark places with index greater than available places with -1
new_ind[new_ind>sum(ind)] = -1
## mark existing places with index 0 with -1
new_ind[(new_ind==0) & ind] = -1
available_ind = all_ind - set(new_ind[new_ind>0])
## replace indxes marked by -1 with minimum values from available_ind
for i in range(len(new_ind)):
if new_ind[i]==-1:
new_ind[i] = min(available_ind)
available_ind.remove(new_ind[i])
result.append(new_ind)
prev_ind = new_ind
result = np.r_[result]
repeats = np.r_[change_points[1:] - change_points[:-1], [[0]]].ravel()
## place index calculated only for change points, now fill the gap between change points
## by repeating index in the gap
result = np.repeat(result, repeats, axis=0)
df2["group"] = (result[np.arange(len(result)), df2["Place Label"].values]-1)//3 + 1
return df2
df2 = calc_groups(uniquePlaces, df2)
df2.drop_duplicates(subset=['Period','Place'])
结果
Period Place Total Available Group Place Label group
0 1 CLUB 1 1 0 1
1 2 CLUB 1 2 0 1
3 2 HOME 2 2 1 1
5 2 AWAY 3 2 2 1
7 3 WORK 4 2 3 2
9 3 AWAY 4 1 2 1
11 3 GOLF 5 1 4 2
13 4 CLUB 4 2 0 1
15 4 POOL 4 2 5 1
17 4 HOME 4 2 1 1
19 5 WORK 4 2 3 1
21 5 AWAY 4 1 2 1
23 5 POOL 4 1 5 1
25 6 TENNIS 5 2 6 1
【讨论】：
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