Python 功能赢得'；t应用于熊猫数据帧，获取语法错误

我试图将此函数应用于pandas数据帧，以查看出租车上下车时间是否在我使用下面的arrivemin、ArriveMax变量创建的范围内

如果时间真的在这个范围内，我想保留这一行。如果超出范围，我想从数据帧中删除它

gametimestart = df['Start.Time'] 
gametimeend = df['End.Time'] 
arrivemin = gametimestart - datetime.timedelta(minutes=120) 
arrivemax = gametimeend - datetime.timedelta(minutes = 30) 
departmin = gametimeend - datetime.timedelta(minutes = 60) 
departmax = gametimeend + datetime.timedelta(minutes = 90)
df_filtered = df[(df['pickup_datetime'] > arrivemin) &
                 (df['pickup_datetime'] < arrivemax) &
                 (df['dropoff_datetime'] > departmin) & 
                 (df['dropoffdatetime'] < departmax)]

Start.Time、End.Time等都是datetime对象，因此时间功能应该可以正常工作

def time_function(df, row):
    gametimestart = df['Start.Time'] 
    gametimeend = df['End.Time'] 
    arrivemin = gametimestart - datetime.timedelta(minutes=120) 
    arrivemax = gametimeend - datetime.timedelta(minutes = 30) 
    departmin = gametimeend - datetime.timedelta(minutes = 60) 
    departmax = gametimeend + datetime.timedelta(minutes = 90)
    for not i in ((df['pickup_datetime'] > arrivemin) & (df['pickupdatetime'] < arrivemax) &(df['dropoff_datetime'] > departmin) & (df['dropoffdatetime'] < departmax)):
        df = df.drop[df[i.index]]
    return


for index, row in yankdf:
    time_function(yankdf, row)

def time_函数（df，行）：
gametimestart=df['Start.Time']
gametimeend=df['End.Time']
arrivemin=gametimestart-datetime.timedelta（分钟=120）
arrivemax=gametimeend-datetime.timedelta（分钟=30）
departmin=gametimeend-datetime.timedelta（分钟=60）
departmax=gametimeend+datetime.timedelta（分钟=90）
对于非i in（（df['picku datetime']>arrivemin）&（df['pickupdatetime']departmin）&（df['dropoffdatetime']

继续获取此语法错误：

 File "<ipython-input-25-bda6fb2db429>", line 17
    for not i in (((row['pickup_datetime'] > arrivemin) & (row['pickupdatetime'] < arrivemax)) | ((row['dropoff_datetime'] > departmin) & (row['dropoffdatetime'] < departmax)):
          ^
SyntaxError: invalid syntax

文件“”，第17行
对于不在（（第['Picku_datetime']>arrivemin行）和第['pickupdatetime']departmin行）和第['dropoffdatetime']

我认为您不需要该函数。只需执行一个基本子集，df_filtered应该是您的过滤数据帧

gametimestart = df['Start.Time'] 
gametimeend = df['End.Time'] 
arrivemin = gametimestart - datetime.timedelta(minutes=120) 
arrivemax = gametimeend - datetime.timedelta(minutes = 30) 
departmin = gametimeend - datetime.timedelta(minutes = 60) 
departmax = gametimeend + datetime.timedelta(minutes = 90)
df_filtered = df[(df['pickup_datetime'] > arrivemin) &
                 (df['pickup_datetime'] < arrivemax) &
                 (df['dropoff_datetime'] > departmin) & 
                 (df['dropoffdatetime'] < departmax)]

gametimestart=df['Start.Time']
gametimeend=df['End.Time']
arrivemin=gametimestart-datetime.timedelta（分钟=120）
arrivemax=gametimeend-datetime.timedelta（分钟=30）
departmin=gametimeend-datetime.timedelta（分钟=60）
departmax=gametimeend+datetime.timedelta（分钟=90）
df_filtered=df[（df['picku_datetime']>arrivemin）&
（df['picku datetime']departmin）和
（df['dropoffdatetime']

嗯，你的时间函数体缩进错误。可能是吗？哎呀……那只是一个复制粘贴错误。请在上面编辑你的问题，并包括完整的错误回溯。有趣的是……我的错误消息是重复的，似乎错误是“不是”而不是“离开”