Question

输入

假设我有一个具有以下结构的数据框：

   transaction_code   transaction_time   amount reversed_transaction_code
0       TX051       2019-01-01 13:00:00   150   
1       TX002       2019-01-01 14:00:00   250           TX004
2       TX113       2019-01-01 15:00:00   100   
3       TX004       2019-01-01 16:00:00    80           TX002
4       TX805       2019-01-01 17:00:00    30

可以使用以下代码来复制它：

eg = {'transaction_code': ['TX051','TX002','TX113','TX004','TX805'],
      'transaction_time': pd.to_datetime(['1 Jan 2019 1pm','1 Jan 2019 2pm','1 Jan 2019 3pm','1 Jan 2019 4pm','1 Jan 2019 5pm']),
      'amount': [150,250,100,80,30],
      'reversed_transaction_code': ['','TX004','','TX002','']}
df = pd.DataFrame(eg)

在df中，每一行对应于在我的商店进行的一笔交易。返回项目后，将添加新交易并记录在reversed_transaction_code列中。

问题

例如，TX004中返还了TX002中的$ 80物品。我该如何匹配这些交易，记录时间和退货金额，然后删除反向交易的ROWS？

预期产量

新列应如下所示：

   reversed_amount  reversed_transaction_time
0        NaN                   NaT
1        80            2019-01-01 16:00:00
2        NaN                   NaT
4        NaN                   NaT

可以使用以下代码来复制它：

da = df[df.index!=3]
da['reversed_amount'] = [None, 80, None, None]
da['reversed_transaction_time'] =  pd.to_datetime([None, '1 Jan 2019 4pm', None, None])

Answer 1

我已经修改了您的原始数据，使其变得更加复杂。

解决方案-

eg = {'transaction_code': ['TX051','TX002','TX113','TX004','TX805'],
          'transaction_time': pd.to_datetime(['1 Jan 2019 1pm','1 Jan 2019 2pm','1 Jan 2019 3pm','1 Jan 2019 4pm','1 Jan 2019 5pm']),
          'amount': [150,250,100,80,30],
          'reversed_transaction_code': ['','TX004','TX805','TX002','TX113']}
df = pd.DataFrame(eg)
df


+---+--------+---------------------------+------------------+---------------------+
|   | amount | reversed_transaction_code | transaction_code |  transaction_time   |
+---+--------+---------------------------+------------------+---------------------+
| 0 |    150 |                           | TX051            | 2019-01-01 13:00:00 |
| 1 |    250 | TX004                     | TX002            | 2019-01-01 14:00:00 |
| 2 |    100 | TX805                     | TX113            | 2019-01-01 15:00:00 |
| 3 |     80 | TX002                     | TX004            | 2019-01-01 16:00:00 |
| 4 |     30 | TX113                     | TX805            | 2019-01-01 17:00:00 |
+---+--------+---------------------------+------------------+---------------------+

# Fetching the index where there's an entry on reversed_transaction_code
idx_ = df[df.reversed_transaction_code.str.startswith('T')].index
idx_
# Int64Index([1, 2], dtype='int64')

# Creating blank columns
df['reversed_amount'] = np.NaN
df['reversed_transaction_time'] = None

# Reverse transaction index
idxR_ = df.iloc[idx_, :][df.loc[idx_, 'reversed_transaction_code'].str.split('TX', expand=True).iloc[:, 1] < df.loc[idx_, 'transaction_code'].str.split('TX', expand=True).iloc[:, 1]].index
idxR_
# Int64Index([3, 4], dtype='int64')

# Fetching valid reversed transaction code from reversed_transaction_code column
val = df.loc[idxR_, 'reversed_transaction_code'] 
val
# 3    TX002
# 4    TX113
# Name: reversed_transaction_code, dtype: object

# Fetching transaction code from transaction_code column  
code_idx_ = df[np.where(df.transaction_code.isin(val), True , False)].index
code_idx_
# Int64Index([1, 2], dtype='int64')

# checking where does transaction code lies and adding corresponding results to new columns
# Below code can be made shorter or more efficient (say using merge/join, etc)
for i in range(len(val)):
    for j in range(len(code_idx_)):        
        if val.iloc[i] == df.loc[code_idx_[j], 'transaction_code']: 
            df.loc[code_idx_[j], 'reversed_transaction_time'] = df.loc[val.index[i], 'transaction_time']
            df.loc[code_idx_[j], 'reversed_amount'] = df.loc[val.index[i], 'amount']

# Removing the rows with reversed transactions
df.drop(val.index, inplace=True)
df            

+---+--------+---------------------------+------------------+---------------------+-----------------+---------------------------+
|   | amount | reversed_transaction_code | transaction_code |  transaction_time   | reversed_amount | reversed_transaction_time |
+---+--------+---------------------------+------------------+---------------------+-----------------+---------------------------+
| 0 |    150 |                           | TX051            | 2019-01-01 13:00:00 | NaN             | None                      |
| 1 |    250 | TX004                     | TX002            | 2019-01-01 14:00:00 | 80.0            | 2019-01-01 16:00:00       |
| 2 |    100 | TX805                     | TX113            | 2019-01-01 15:00:00 | 30.0            | 2019-01-01 17:00:00       |
+---+--------+---------------------------+------------------+---------------------+-----------------+---------------------------+

根据值匹配熊猫中的行是不同的列

1 个答案:

解决方案-