Question

给出以下数据框：

data = [['2019-06-20 12:28:00', '05123', 2, 8888],
        ['2019-06-20 13:28:00', '55874', 6, 8888],
        ['2019-06-20 13:35:00', '12345', 1, 8888],
        ['2019-06-20 13:35:00', '35478', 2, 1234],
        ['2019-06-20 13:35:00', '12345', 2, 8888],
        ['2019-06-20 14:22:00', '98765', 1, 8888]]

columns = ['pdate', 'station', 'ptype', 'train']
df = pd.DataFrame(data, columns = columns)

其中“ pdate” =乘车时间，“ station” =乘车代号，“ ptype” =乘车类型，“ train” =火车号

'ptype'可以具有以下值（1 =到达，2 =出发，6 =通过）

这是结果：

                 pdate station  ptype  train
0  2019-06-20 12:28:00   05123      2   8888
1  2019-06-20 13:28:00   55874      6   8888
2  2019-06-20 13:35:00   12345      1   8888
3  2019-06-20 13:35:00   35478      2   1234
4  2019-06-20 13:35:00   12345      2   8888
5  2019-06-20 14:22:00   98765      1   8888

不幸的是，有时错误地在车站而不是注册“ ptype” = 6（通过），而是输入了“ ptype” = 1（到达）并且“ ptype” = 2（出发）发生在同一时间，所以这2条记录我必须考虑只是一张通行证记录

我必须从数据帧中删除具有ptype = 6 OR的每一行（ptype = 1，并且同一站点和同一列车编号的ptype = 2的下一条记录恰好同时发生）

因此，从给定的示例中，我必须删除以下行（1、2、4）

我可以毫无疑问地删除ptype = 6的所有行

df = df.drop(df[(df['ptype']==6)].index)

但是我不知道如何删除其他对。有想法吗？

Answer 1

IIUC，您可以执行groupby和nunique：

# convert to datetime. Skip if already is.
df.pdate = pd.to_datetime(df.pdate)

# drop all the 6 records:
df = df[df.ptype.ne(6)]

(df[df.groupby(['pdate','train'])
      .ptype.transform('nunique').eq(1)]
)

输出：

                pdate station  ptype  train
0 2019-06-20 12:28:00   05123      2   8888
3 2019-06-20 13:35:00   35478      2   1234
5 2019-06-20 14:22:00   98765      1   8888

Answer 2

这是您的方法：

# We look at the problematic ptypes
# We groupby station train and pdate to  identify the problematic rows
test = df[(df['ptype'] == 1) | (df['ptype'] == 2)].groupby(['station', 'train', 'pdate']).size().reset_index()

# If there is more than one row that means there is a duplicate 
errors = test[test[0] >1][['station', 'train', 'pdate']]
# We create a column to_remove to later identify the problematic rows
errors['to_remove'] = 1

df = df.merge(errors, on=['station', 'train', 'pdate'], how='left')

#We drop the problematic rows
df = df.drop(index = df[df['to_remove'] == 1].index)

# We drop the column to_remove which is no longer necessary
df.drop(columns='to_remove', inplace = True)

输出：

                 pdate station  ptype  train
0  2019-06-20 12:28:00   05123      2   8888
1  2019-06-20 13:28:00   55874      6   8888
3  2019-06-20 13:35:00   35478      2   1234
5  2019-06-20 14:22:00   98765      1   8888

Answer 3

这不是熊猫式的方法，但是如果我正确地理解了你的追求，就会真正得到想要的结果

# a dict for unique filtered records
filtered_records = {}

def unique_key(row):
    return '%s-%s-%d' % (row[columns[0]],row[columns[1]],row[columns[3]])

# populate a map of unique dt, train, station records
for index, row in df.iterrows():
    key = unique_key(row)
    val = filtered_records.get(key,None)
    if val is None:
        filtered_records[key] = row[columns[2]]
    else:
        # is there's a 1 and 2 record, declare the record a 6
        if val * row[columns[2]] == 2:
            filtered_records[key] = 6

# helper function for apply
def update_row_ptype(row):
    val = filtered_records[unique_key(row)]
    return val if val == 6 else row[columns[2]]

# update the dataframe with invalid detected entries from the dict
df[columns[2]] = df.apply(lambda row: update_row_ptype(row), axis = 1)
# drop em
df.drop(df[(df[columns[2]]==6)].index,inplace=True)

print df

输出

                 pdate station  ptype  train
0  2019-06-20 12:28:00   05123      2   8888
3  2019-06-20 13:35:00   35478      2   1234
5  2019-06-20 14:22:00   98765      1   8888

熊猫根据“邻居”拖放行

3 个答案: