以3组为单位分配列值-Python

时间:2019-02-20 23:55:45

标签: python pandas numpy dataframe assign

我的目标是将assign pandas中的人放置在df中。具体来说,使用下面的df,我确定当前在有多少个地方。我想使用这些值并将它们assign以3为一组。

例如,应该将出现的总地点少于3个分配给P1。应该将3-6的位置分配给P2等。

注意:一次发生的地点总数可以增加到20个,因此为此需要容纳分配的组数。

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

1 个答案:

答案 0 :(得分:0)

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']进行排序。