通过Python基于if条件添加新行

Question

如果列door的两个连续单元格相同，并且列time的两个相邻单元格之间的差异超过5分钟，则需要添加新行。df:

Time        door             name
09:10:00    RDC_OUT-1        alex
09:10:00    RDC_OUT-1        alex
11:23:00    RDC_IN-1         alex
12:13:00    RDC_IN-1         alex
12:39:00    RDC_OUT-1        alex
15:23:00    RDC_IN-1         alex

代码：

import pandas as pd
import numpy as np
file_name='test.xlsx'
from datetime import timedelta
import datetime
df = pd.read_excel(file_name, header=0, index= False)
df= df.sort_values(by='Time')
df.reset_index(inplace = True) 
print(df)
idx=[]
for i in range (0,len(df)):
if i == 0:
print ('Door Name '+str(i)+' ok')
elif df['Door Name'][i] != df['Door Name'][i-1]:
print('index '+str(i)+' ok')
elif ((df['Door Name'][i] == df['Door Name'][i-1]) & ((df['Time'].iloc[i] - df['Time'].iloc[i-1]) > datetime.timedelta(minutes=5))):
print('index '+str(i)+' ok')
df.iloc[i] = [i,'RDC_OUT-1', str('12:00:00'), 'ARYA'] 

elif ((df['Door Name'][i] == df['Door Name'][i-1]) & ((df['Time'].iloc[i] - df['Time'].iloc[i-1]) < datetime.timedelta(minutes=5))):
print('index '+str(i)+' nok')
idx.append(i)
print('idxn',idx)
df.drop(df.index[[idx]],inplace=True)
print('n',df) 

所需输出：

Time        door              name
Time        door             name
09:10:00    RDC_OUT-1        alex
11:23:00    RDC_IN-1         alex
12:00:00    RDC_OUT-1        ARYA
12:13:00    RDC_IN-1         alex
12:39:00    RDC_OUT-1        alex
15:23:00    RDC_IN-1         alex

输出

0      4  09:10:00    RDC_OUT-1   alex
2      3  11:23:00    RDC_IN-1    alex
3      2  12:13:00    RDC_IN-1    alex
4      3  12:00:00    RDC_OUT-1   ARYA
5      0  15:23:00    RDC_IN-1    alex

Answer 1

因此，首先，我强烈建议您始终提供一个工作示例，用于复制粘贴！

import pandas as pd
import numpy as np
import datetime as dt
df= pd.DataFrame({'Time':['17:01:10', '13:23:00', '11:23:00', '10:01:10','09:01:10','09:01:10'],
'door':['RDC_OUT-1', 'RDC_IN-1','RDC_IN-1','RDC_OUT-1','RDC_IN-1','RDC_IN-1'],
'name':['alex','alex','alex','alex','alex','alex']})

然后，转换你的时间戳和功能，这样你就可以在上面做数学运算：

# replace door with bin value
df['door']= df['door'].map({'RDC_IN-1': 0, 'RDC_OUT-1': 1})
# convert time stamp
df['Time'] = pd.to_datetime(df['Time'], format="%H:%M:%S")

现在你可以释放熊猫数据帧的力量了；(

# sort by time stamp
df= df.sort_values(by='Time')
# calculate difference to next row per column
df_diff = df[['Time', 'door']].diff(periods=-1)
# select and copy relevant rows 
df_add = df[(df_diff.Time < dt.timedelta(minutes=-5))& (df_diff.door ==0)].copy()
# change the time stamp of copied rows
df_add.loc[df_add.door == 0, 'Time'] =  pd.to_datetime('12:00:00', format="%H:%M:%S")
df_add.loc[df_add.door == 1, 'Time'] =  pd.to_datetime('14:00:00', format="%H:%M:%S")

# switch the label of copied rows
df_add['door']= -(df['door']-1)
# change name to mark the new
df_add['name']= 'new_alex'
# append existing data frame with new rows and sort by time stamp
df = df.append(df_add ).sort_values(by='Time')
# remap the door featuere
df['door']= df['door'].map({0:'RDC_IN-1', 1:'RDC_OUT-1'})

这应该会给你输出：

Time       door      name
4 1900-01-01 09:01:10   RDC_IN-1      alex
5 1900-01-01 09:01:10   RDC_IN-1      alex
3 1900-01-01 10:01:10  RDC_OUT-1      alex
2 1900-01-01 11:23:00   RDC_IN-1      alex
2 1900-01-01 12:00:00  RDC_OUT-1  new_alex
1 1900-01-01 13:23:00   RDC_IN-1      alex
0 1900-01-01 17:01:10  RDC_OUT-1      alex