Adr*_*ian 3 datetime dataframe python-3.x pandas pandas-groupby
我有以下数据帧片段:
Full dataframe: ip time cik crawler
ts
2019-03-11 00:00:01 71.155.177.ide 00:00:01 1262327 0.0
2019-03-11 00:00:02 71.155.177.ide 00:00:02 1262329 0.0
2019-03-11 00:00:05 69.243.218.cah 00:00:05 751200 0.0
2019-03-11 00:00:08 172.173.121.efb 00:00:08 881890 0.0
2019-03-11 00:00:09 216.254.60.idd 00:00:09 1219169 0.0
2019-03-11 00:00:09 64.18.197.gjc 00:00:09 1261705 0.0
2019-03-11 00:00:09 64.18.197.gjc 00:00:09 1261734 0.0
2019-03-11 00:00:10 64.18.197.gjc 00:00:10 1263094 0.0
2019-03-11 00:00:10 64.18.197.gjc 00:00:10 1264242 0.0
2019-03-11 00:00:10 64.18.197.gjc 00:00:10 1264242 0.0
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我想按 IP 分组,然后使用一些函数来计数:
1) 1 分钟内每个 IP 有多少唯一 CIK
2) 1 分钟内每个 IP 有多少 CIK(总共)。
我已经尝试过 resample 功能,但我不知道如何按照我想要的方式计算它。我的代码如下:
dataframe = pd.read_csv(path + "log20060702.csv", usecols=['cik', 'ip', 'time', 'crawler'])
dataframe = dataframe[dataframe['crawler'] == 0]
dataframe['cik'] = pd.to_numeric(dataframe['cik'], downcast='integer')
dataframe['ts'] = pd.to_datetime((dataframe['time']))
dataframe = dataframe.set_index(['ts'])
print("Full dataframe: ", dataframe.head(10))
df_dict = dataframe.groupby("ip")
counter = 0
for key, df_values in df_dict:
counter += 1
print("df values: ", df_values)
# df_values = df_values.resample("5T").count()
if counter == 5:
break
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或者,如果有人可以告诉我如何按 IP 分组,每 1 分钟分组一次,其余时间我可以自己完成。我不一定要寻找完整的解决方案,非常感谢一些指导。
使用groupby与DataFrameGroupBy.resample和聚合SeriesGroupBy.nunique与计数DataFrameGroupBy.size:
df = dataframe.groupby("ip").resample('1Min')['cik'].agg(['nunique','size'])
print (df)
nunique size
ip ts
172.173.121.efb 2019-03-11 1 1
216.254.60.idd 2019-03-11 1 1
64.18.197.gjc 2019-03-11 4 5
69.243.218.cah 2019-03-11 1 1
71.155.177.ide 2019-03-11 2 2
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或使用Grouper:
df = dataframe.groupby(["ip", pd.Grouper(freq='1Min')])['cik'].agg(['nunique','size'])
print (df)
nunique size
ip ts
172.173.121.efb 2019-03-11 1 1
216.254.60.idd 2019-03-11 1 1
64.18.197.gjc 2019-03-11 4 5
69.243.218.cah 2019-03-11 1 1
71.155.177.ide 2019-03-11 2 2
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