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Python 我试图在dataframe新列中输入一个列表_Python_Pandas - Fatal编程技术网

Python 我试图在dataframe新列中输入一个列表

Python 我试图在dataframe新列中输入一个列表,python,pandas,Python,Pandas,我已经将输出保存到一个列表中,我想创建一个新列并在其中添加值 我可以用anaconda成功地做到这一点,但我现在正在尝试python,它不起作用 attendance = [] sleep(5) for x in df1['Login Time']: s1 = str(x) s2 = '12:00:00' s3 = '3:15:00' s4 = '02:00:00' FMT = '%H:%M:%S' tdelta = datetime.strpti

我已经将输出保存到一个列表中,我想创建一个新列并在其中添加值

我可以用anaconda成功地做到这一点,但我现在正在尝试python,它不起作用

attendance = []
sleep(5)
for x in df1['Login Time']:
    s1 = str(x)
    s2 = '12:00:00'
    s3 = '3:15:00'
    s4 = '02:00:00'
    FMT = '%H:%M:%S'
    tdelta = datetime.strptime(s2, FMT) - datetime.strptime(s1, FMT)
    if x <= s4:
        attendance.append('Aussi')
    elif (str(tdelta)) < s3:
        attendance.append('Present')
    else: 
        attendance.append('Check')
df1['Attendance'] = attendance
sleep(5)
df1.to_csv(datetime.strftime(datetime.now() - timedelta(1), '%Y-%m-%d')+".csv",index=False)
出勤率=[]
睡眠(5)
对于df1中的x[‘登录时间’]:
s1=str(x)
s2='12:00:00'
s3='3:15:00'
s4='02:00:00'
FMT='%H:%M:%S'
tdelta=datetime.strtime(s2,FMT)-datetime.strtime(s1,FMT)

如果x问题在于
df1['attention']=attention

试试这个:

attendance = []
sleep(5)
for x in df1['Login Time']:
    s1 = str(x)
    s2 = '12:00:00'
    s3 = '3:15:00'
    s4 = '02:00:00'
    FMT = '%H:%M:%S'
    tdelta = datetime.strptime(s2, FMT) - datetime.strptime(s1, FMT)
    if x <= s4:
        attendance.append('Aussi')
    elif (str(tdelta)) < s3:
        attendance.append('Present')
    else: 
        attendance.append('Check')
#From Here......................................................
attendance_df = pd.Dataframe(attendance,column=['Title']
df1.loc[attendance_df.index,'Login Time']=attendance_df['Title']
#...............................................................
sleep(5)
df1.to_csv(datetime.strftime(datetime.now() - timedelta(1), '%Y-%m-%d')+".csv",index=False)
出勤率=[]
睡眠(5)
对于df1中的x[‘登录时间’]:
s1=str(x)
s2='12:00:00'
s3='3:15:00'
s4='02:00:00'
FMT='%H:%M:%S'
tdelta=datetime.strtime(s2,FMT)-datetime.strtime(s1,FMT)

如果x我收到df1无效语法的错误['Attention']无效语法-df1请尝试上述更改。“登录时间”已更改我已上载了相同的屏幕截图最后一行似乎重复了?抱歉,请将“Attention.index”更改为“Attention_.index”,这不是错误。