以 3 个一组分配列值 - Python

Question

我的目标是assignpandasdf 中的人们。具体来说，我使用下面的df 来确定当前开启的地方有多少。我想以 3 个一组的形式使用这些值和 assign。

例如，出现少于 3 个的总位置应分配给 P1。 3-6的地方应该分配给P2等。

注意：一次出现的地点总数最多可达到 20 个，因此分配的组数需要适应此情况。

这是我的尝试。

import pandas as pd
import numpy as np

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 4','House 5','House 1','House 2','House 3','House 4'],                                     
     })

df = pd.DataFrame(data=d)

df['u'] = df[::-1].groupby('Place').Place.cumcount()
ids = [1]
seen = set([df.iloc[0].Place])
dec = False
for val, u in zip(df.Place[1:], df.u[1:]):
    ids.append(ids[-1] + (val not in seen) - dec)
    seen.add(val)
    dec = u == 0
df['Places On'] = ids

df = df.drop(df[['u']], axis=1)

def g(gps):
        s = gps['Place'].unique()
        d = dict(zip(s, np.arange(len(s)) // 3 + 1))
        gps['P'] = gps['Place'].map(d)
        return gps

df = df.groupby('Place', sort=False).apply(g)

输出：

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

预期输出：

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

Answer 1

import pandas as pd
import numpy as np

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 4','House 5','House 1','House 2','House 3','House 4'],                                     
     })

df = pd.DataFrame(data=d)

df['u'] = df[::-1].groupby('Place').Place.cumcount()
ids = [1]
seen = set([df.iloc[0].Place])
dec = False
for val, u in zip(df.Place[1:], df.u[1:]):
    ids.append(ids[-1] + (val not in seen) - dec)
    seen.add(val)
    dec = u == 0
df['Places On'] = ids

df = df.drop(df[['u']], axis=1)

def g(gps):
        s = gps['Place'].unique()
        d = dict(zip(s, np.arange(len(s)) // 3 + 1))
        gps['P'] = gps['Place'].map(d)
        return gps

df = df.groupby('Place', sort=False).apply(g)

for i in range(df.shape[0]):
    if(df['Places On'][i]<=3):
        df['P'][i]=1
    else:
        df['P'][i]=2
print(df)

这应该基于对 df['Places On'] 的排序工作。