Question

我试图根据Column D中的地点连续返回每个项目。

目前我通过df = df.loc[df['D'] == '#specific place']这样做。

这样可以正常工作，但如果我有50个不同的地方，这个代码会变得非常低效。我必须改变df = df.loc[df['D'] == '#another specific place'] 50次。如果地方保持不变但每个数据集包含不同的地方，我会这样做。

是否有更有效的方法来返回每个地方的行？我在每个地方都有一个单独的df，我可以返回list。我可以使用此列表中的值来返回

中的函数

import pandas as pd

places = ['Home','Away','Shops']

d = ({
    'C' : ['08:00:00','XX','08:10:00','XX','08:41:42','XX','08:50:00','XX', '09:00:00', 'XX','09:15:00','XX','09:21:00','XX','09:30:00','XX','09:40:00','XX'],
    'D' : ['Home','','Home','','Away','','Shops','','Away','','Shops','','Home','','Away','','Home',''],
    'E' : ['Num:','','Num:','','Num:','','Num:','','Num:', '','Num:','','Num:','','Num:', '','Num:', ''],
    'F' : ['1','','1','','1','','1','','1', '','2','','2','','1', '','2',''],   
    'A' : ['A','','A','','A','','A','','A','','A','','A','','A','','A',''],           
    'B' : ['Stop','','Res','','Stop','','Start','','Res','','Stop','','Res','','Start','','Start','']
    })

df = pd.DataFrame(data=d)

#Select desired place
Home = df.loc[df['D'] == 'Home']
Shops = df.loc[df['D'] == 'Shops']
Away = df.loc[df['D'] == 'Away']

预期输出：

    A      B         C     D     E  F
0   A   Stop  08:00:00  Home  Num:  1
2   A    Res  08:10:00  Home  Num:  1
12  A    Res  09:21:00  Home  Num:  2
16  A  Start  09:40:00  Home  Num:  2
6   A  Start  08:50:00  Shops  Num:  1
10  A   Stop  09:15:00  Shops  Num:  2
4   A   Stop  08:41:42  Away  Num:  1
8   A    Res  09:00:00  Away  Num:  1
14  A  Start  09:30:00  Away  Num:  1

Answer 1

您可以通过循环显示df ['B']来识别您的位置，并识别非空字符串并将其保存在列表中，例如

places = []
for i in df['B']:
    if i!="":places.append(i)
###now you can create a dict to save the differences
diff_result = {}
for i in places:
    Stop = df.loc[df['B'] == i].reset_index()['C']
    Start = df.loc[df['B'] == i].reset_index()['C']
    Res = df.loc[df['B'] == i].reset_index()['C']
    diff_result[i+"_diff"] = Res

现在你有一个填满想要结果的词典

Answer 2

您只需要一个groupby（），然后可能是“D”列中排除/包含的唯一项目列表：

df = pd.DataFrame(data=d)

df['C'] = pd.to_timedelta(df['C'], errors="coerce").dt.total_seconds()

places = ['Home','Away','Shops']

for d, dfg in df[df['D'].isin(places)].groupby('D'):
    # print out the processing place
    print('group:{}\n{}'.format(d, dfg))
    # do whatever you want on `dfg` and `d`
    # In your example, when d is 'Home', dfg is df.loc[df['D'] == 'Home']
    # below your code, just change `df` to `dfg`
    #Return start-stop times
    Stop = dfg.loc[df['B'] == 'Stop'].reset_index()['C']
    Start = dfg.loc[df['B'] == 'Start'].reset_index()['C']
    Res = dfg.loc[df['B'] == 'Res'].reset_index()['C']

    #return difference
    Start_diff = Start - Stop
    Res_diff = Res - Start

有效返回pandas df中的特定行

2 个答案: