Python 使用pandas从日志文件分析生成会话
我正在分析一个Apache日志文件,并将其导入到一个数据帧中 “65.55.52.118——[30/May/2013:06:58:52-0600]“GET/detaileddven.php?refId=7954&uId=2802 HTTP/1.1”200 4514“-”Mozilla/5.0(兼容;bingbot/2.0;+)” 我的数据帧:Python 使用pandas从日志文件分析生成会话,python,pandas,timedelta,dataframe,Python,Pandas,Timedelta,Dataframe,我正在分析一个Apache日志文件,并将其导入到一个数据帧中 “65.55.52.118——[30/May/2013:06:58:52-0600]“GET/detaileddven.php?refId=7954&uId=2802 HTTP/1.1”200 4514“-”Mozilla/5.0(兼容;bingbot/2.0;+)” 我的数据帧: 我想根据IP、代理和时差将其分组到会话中(如果持续时间大于30分钟,则应为新会话) 通过IP和代理对数据帧进行分组很容易,但是如何检查这个时差呢?希望问
我想根据IP、代理和时差将其分组到会话中(如果持续时间大于30分钟,则应为新会话) 通过IP和代理对数据帧进行分组很容易,但是如何检查这个时差呢?希望问题清楚
sessions = df.groupby(['IP', 'Agent']).size()
更新:df.index如下所示:
<class 'pandas.tseries.index.DatetimeIndex'>
[2013-05-30 06:00:41, ..., 2013-05-30 22:29:14]
Length: 31975, Freq: None, Timezone: None
[2013-05-30 06:00:41, ..., 2013-05-30 22:29:14]
长度:31975,频率:无,时区:无
我会使用a和a(这里是一个简单的例子,用数字代替时间-但它们的工作原理完全相同):
*对skipna=False的需求似乎是一个bug
然后,您可以将其用于:
现在,您可以按
'ip'
和'session\u number'
进行分组(并分析每个会话)。安迪·海登的回答既可爱又简洁,但如果您有大量用户/ip地址进行分组,则会变得非常缓慢。这里有另一种方法,更丑陋,但也更快
import pandas as pd
import numpy as np
sample = lambda x: np.random.choice(x, size=10000)
df = pd.DataFrame({'ip': sample(range(500)),
'time': sample([1., 1.1, 1.2, 2.7, 3.2, 3.8, 3.9])})
max_diff = 0.5 # Max time difference
def method_1(df):
df = df.sort_values('time')
g = df.groupby('ip')
df['session'] = g['time'].apply(
lambda s: (s - s.shift(1) > max_diff).fillna(0).cumsum(skipna=False)
)
return df['session']
def method_2(df):
# Sort by ip then time
df = df.sort_values(['ip', 'time'])
# Get locations where the ip changes
ip_change = df.ip != df.ip.shift()
time_or_ip_change = (df.time - df.time.shift() > max_diff) | ip_change
df['session'] = time_or_ip_change.cumsum()
# The cumsum operated over the whole series, so subtract out the first
# value for each IP
df['tmp'] = 0
df.loc[ip_change, 'tmp'] = df.loc[ip_change, 'session']
df['tmp'] = np.maximum.accumulate(df.tmp)
df['session'] = df.session - df.tmp
# Delete the temporary column
del df['tmp']
return df['session']
r1 = method_1(df)
r2 = method_2(df)
assert (r1.sort_index() == r2.sort_index()).all()
%timeit method_1(df)
%timeit method_2(df)
400 ms ± 195 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
11.6 ms ± 2.04 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)
谢谢你,安迪!过了很长一段时间,我得到了一个答案:),但为什么我会出现这个错误呢?AttributeError:“Timestamp”对象没有属性“shift”@NilaniAlgiriyage看起来您试图将shift应用于时间戳而不是列/序列(但不确定是如何实现的)。df['tval']=df.index df['delta']=(df['tval']-df['tval'].shift(1)>30)。fillna(0)。cumsum(skipna=False)上述代码正确吗?但它给出了另一个类型错误?上面的代码对我来说很有用。。。您的代码看起来不错,但我认为您应该使用
pd.offset.Minute(30).nanos,而不是30
。您能确认类型(df['tval'])的结果吗
?以及您的pandas版本(适用于0.11)。
In [21]: df = pd.DataFrame([[1.1, 1.7, 2.5, 2.6, 2.7, 3.4], list('AAABBB')]).T
In [22]: df.columns = ['time', 'ip']
In [23]: df
Out[23]:
time ip
0 1.1 A
1 1.7 A
2 2.5 A
3 2.6 B
4 2.7 B
5 3.4 B
In [24]: g = df.groupby('ip')
In [25]: df['session_number'] = g['time'].apply(lambda s: (s - s.shift(1) > 0.5).fillna(0).cumsum(skipna=False))
In [26]: df
Out[26]:
time ip session_number
0 1.1 A 0
1 1.7 A 1
2 2.5 A 2
3 2.6 B 0
4 2.7 B 0
5 3.4 B 1
import pandas as pd
import numpy as np
sample = lambda x: np.random.choice(x, size=10000)
df = pd.DataFrame({'ip': sample(range(500)),
'time': sample([1., 1.1, 1.2, 2.7, 3.2, 3.8, 3.9])})
max_diff = 0.5 # Max time difference
def method_1(df):
df = df.sort_values('time')
g = df.groupby('ip')
df['session'] = g['time'].apply(
lambda s: (s - s.shift(1) > max_diff).fillna(0).cumsum(skipna=False)
)
return df['session']
def method_2(df):
# Sort by ip then time
df = df.sort_values(['ip', 'time'])
# Get locations where the ip changes
ip_change = df.ip != df.ip.shift()
time_or_ip_change = (df.time - df.time.shift() > max_diff) | ip_change
df['session'] = time_or_ip_change.cumsum()
# The cumsum operated over the whole series, so subtract out the first
# value for each IP
df['tmp'] = 0
df.loc[ip_change, 'tmp'] = df.loc[ip_change, 'session']
df['tmp'] = np.maximum.accumulate(df.tmp)
df['session'] = df.session - df.tmp
# Delete the temporary column
del df['tmp']
return df['session']
r1 = method_1(df)
r2 = method_2(df)
assert (r1.sort_index() == r2.sort_index()).all()
%timeit method_1(df)
%timeit method_2(df)
400 ms ± 195 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
11.6 ms ± 2.04 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)