Question

我正在尝试将Dataframe的单行拆分为两行。在数据框的开始和结束列中。我想根据情况拆分行。

我有一个如下所示的数据框：

symbol,start,end,size
ABC,2015-08-27 18:00:00,2015-08-28 05:00:00,12
ABC,2015-11-20 02:00:00,2015-11-20 06:00:00,5
ABC,2016-01-22 03:00:00,2016-01-22 06:00:00,4
PQR,2016-02-12 02:00:00,2016-02-12 06:00:00,5
PQR,2016-02-12 22:00:00,2016-02-13 03:00:00,6
PQR,2016-02-12 02:00:00,2016-02-12 07:00:00,6

条件：

如果开始和结束是同一天，则无需执行任何操作。
如果开始日期和结束日期不同，则需要将其分为两行。

示例：让我们考虑这样的行：

PQR,2016-02-12 22:00:00,2016-02-13 03:00:00,6

在上面的行中，开始包含第12天，结束包含第13天，因此，需要将其分成两行，如下所示：

PQR,2016-02-12 22:00:00,2016-02-12 23:00:00,2
PQR,2016-02-12 00:00:00,2016-02-13 03:00:00,4

如果该行包含第12天开始和第14天结束之间的三天，则需要将其分成三行。

预期输出为：

symbol,start,end,size
ABC,2015-08-27 18:00:00,2015-08-27 23:00:00,6
ABC,2015-08-28 00:00:00,2015-08-28 05:00:00,6
ABC,2015-11-20 02:00:00,2015-11-20 06:00:00,5
ABC,2016-01-22 03:00:00,2016-01-22 06:00:00,4
PQR,2016-02-12 02:00:00,2016-02-12 06:00:00,5
PQR,2016-02-12 22:00:00,2016-02-12 23:00:00,2
PQR,2016-02-12 00:00:00,2016-02-13 03:00:00,4
PQR,2016-02-12 02:00:00,2016-02-12 07:00:00,6

Answer 1

选项1

遍历行，并逐行附加一个新的DataFrame。

import pandas as pd
import datetime

df2 = pd.DataFrame(columns=df.columns)

for (_,r) in df.iterrows():

    while r['start'].date()<r['end'].date():
        # create new row
        newR = r.copy()
        newR['end']=newR['start']
        newR['end']=newR['end'].replace(hour=23)

        newSize = 24-newR['start'].hour
        newR['size']=newSize

        # update row to process 
        r['start']=r['start']+datetime.timedelta(days=1)
        r['start']=r['start'].replace(hour=0)

        r['size'] = r['size'] - newSize

        df2 = df2.append(newR)

    df2 = df2.append(r)

df2.reset_index(drop=True, inplace=True)

选项2

如果原始Dataframe中的行要在两天内拆分，请使用掩码和递归调用Dataframe进行明智的操作。

import pandas as pd
import numpy as np
import datetime


def splitMultiDayRows(df):
    mask = df['end'].dt.day>df['start'].dt.day

    if np.any(mask):
        df_new = df.loc[mask]

        newSizes = 24-df.loc[mask,'start'].dt.hour

        df.loc[mask,'end'] = df.loc[mask,'start']
        df.loc[mask,'end'] = df.loc[mask,
                                    'end'].apply(lambda x:
                                                 x.replace(hour=23))
        df.loc[mask,'size'] = newSizes

        df_new.loc[:,'start'] = df_new['start']+datetime.timedelta(days=1)
        df_new.loc[:,'start'] = df_new['start'].apply(lambda x:
                                                      x.replace(hour=0))

        df_new.loc[:,'size'] = df_new['size'] - newSizes

        return pd.concat([df,splitMultiDayRows(df_new)])
    else:
        return df

与通话配合使用：

splitMultiDayRows(df.copy()).\
sort_values(['symbol','start']).\
reset_index(drop=True)

Answer 2

此答案避免重复，并且不会复制不必要的行，因此可以节省时间和空间。

df['start'] = pd.to_datetime(df['start'])
df['end'] = pd.to_datetime(df['end'])

df2 = pd.DataFrame(columns=df.columns)

mask_to_change = df.apply(lambda x: x['end'].day > x['start'].day, axis=1)

for (_,r) in df[mask_to_change].iterrows():

    while r['start'].date()<r['end'].date():
        # create new row
        newR = r.copy()
        newR['end']=newR['start']
        newR['end']=newR['end'].replace(hour=23)

        newSize = 24-newR['start'].hour
        newR['size']=newSize

        # update row to process 
        r['start']=r['start']+datetime.timedelta(days=1)
        r['start']=r['start'].replace(hour=0)

        r['size'] = r['size'] - newSize

        df2 = df2.append(newR)

    df2 = df2.append(r)

df = pd.concat([df[~mask_to_change], df2])
df.sort_values(['symbol', 'start'], inplace=True)

如何将熊猫数据框单行分为两行？

2 个答案: