Question

我有一个包含以下列的数据框：

账号、通讯日期、开户日期

对于每个开设的账户，我被要求回顾其中发生的所有通信该帐户的开放日期为 30 天，然后将积分分配如下：

Forty-twenty-forty: Attribute 40% (0.4 points) of the attribution to the first touch,
40% to the last touch, and divide the remaining 20% between all touches in between

所以我知道按功能申请和分组，但这超出了我的工资等级。我必须按帐户分组，条件是基于 2 列相互之间的比较，我必须这样做才能获得对应的总数，我想它们也必须进行排序，因为为对应分配点的以下步骤取决于它们出现的顺序。

我想有效地做到这一点，因为我有很多行，我知道 apply() 可以很快，但是当我在行级操作时，我很不擅长应用我试图做的甚至有点复杂。

感谢您的帮助，因为我不擅长熊猫。

编辑根据要求

Acct, ContactDate, OpenDate, Points (what I need to calculate)
123, 1/1/2018, 1/1/2021, 0 (because correspondance not within 30 days of open)
123, 12/10/2020, 1/1/2021, 0.4 (first touch gets 0.4)
123, 12/11/2020, 1/1/2021, 0.2 (other 'touches' get 0.2/(num of touches-2) 'points')
123, 12/12/2020, 1/1/2021, 0.4 (last touch gets 0.4)
456, 1/1/2018, 1/1/2021, 0 (again, because correspondance not within 30 days of open)
456, 12/10/2020, 1/1/2021, 0.4 (first touch gets 0.4)
456, 12/11/2020, 1/1/2021, 0.1 (other 'touches' get 0.2/(num of touches-2) 'points')
456, 12/11/2020, 1/1/2021, 0.1 (other 'touches' get 0.2/(num of touches-2) 'points')
456, 12/12/2020, 1/1/2021, 0.4 (last touch gets 0.4)

Answer 1

这将返回一个减少的数据帧，因为它排除了超过 30 天的时间帧，然后将原始 df 合并到其中，将所有数据合并到一个 df 中。这假设您的日期排序是正确的，否则，您可能需要在应用下面的函数之前预先执行此操作。

df['Points'] = 0 #add column to dataframe before analysis

#df.columns
#Index(['Acct', 'ContactDate', 'OpenDate', 'Points'], dtype='object')

def points(x):
    newx = x.loc[(x['OpenDate'] - x['ContactDate']) <= timedelta(days=30)] # reduce for wide > 30 days
    # print(newx.Acct)
    if newx.Acct.count() > 2: # check more than two dates exist
        newx['Points'].iloc[0] = .4 # first row
        newx['Points'].iloc[-1] = .4 # last row
        newx['Points'].iloc[1:-1] = .2 / newx['Points'].iloc[1:-1].count() # middle rows / by count of those rows
        return newx
    elif newx.Acct.count() == 2: # placeholder for later
        #edge case logic here for two occurences
        return newx
    elif newx.Acct.count() == 1: # placeholder for later
        #edge case logic here one onccurence
        return newx

# groupby Acct then clean up the indices so it can be merged back into original df
dft = df.groupby('Acct', as_index=False).apply(points).reset_index().set_index('level_1').drop('level_0', axis=1)

# merge on index
df_points = df[['Acct', 'ContactDate', 'OpenDate']].merge(dft['Points'], how='left', left_index=True, right_index=True).fillna(0)

输出：

   Acct ContactDate   OpenDate  Points
0   123  2018-01-01 2021-01-01     0.0
1   123  2020-12-10 2021-01-01     0.4
2   123  2020-12-11 2021-01-01     0.2
3   123  2020-12-12 2021-01-01     0.4
4   456  2018-01-01 2021-01-01     0.0
5   456  2020-12-10 2021-01-01     0.4
6   456  2020-12-11 2021-01-01     0.1
7   456  2020-12-11 2021-01-01     0.1
8   456  2020-12-12 2021-01-01     0.4

将多个条件 groupby + sort + sum 应用于 pandas 数据框行

1 个答案: