Question

我是熊猫的新手，现在我不知道如何安排我的时间系列，看看它：

date & time of connection
19/06/2017 12:39
19/06/2017 12:40
19/06/2017 13:11
20/06/2017 12:02
20/06/2017 12:04
21/06/2017 09:32
21/06/2017 18:23
21/06/2017 18:51
21/06/2017 19:08
21/06/2017 19:50
22/06/2017 13:22
22/06/2017 13:41
22/06/2017 18:01
23/06/2017 16:18
23/06/2017 17:00
23/06/2017 19:25
23/06/2017 20:58
23/06/2017 21:03
23/06/2017 21:05

这是130 k raws数据集的示例，我尝试过： df.groupby('date & time of connection')['date & time of connection'].apply(list)

我猜不够

我想我应该：

创建一个索引从dd / mm / yyyy到dd / mm / yyyy
将“date＆amp;连接时间”类型dateTime更改为Date
分组和计数“连接日期和时间”的日期
把数字计算在字典里面？

您如何看待我的逻辑？你知道一些tutos吗？非常感谢你

Answer 1

您可以使用dt.floor转换为date，然后value_counts或groupby转换为size：

df = (pd.to_datetime(df['date & time of connection'])
       .dt.floor('d')
       .value_counts()
       .rename_axis('date')
       .reset_index(name='count'))
print (df)
        date  count
0 2017-06-23      6
1 2017-06-21      5
2 2017-06-19      3
3 2017-06-22      3
4 2017-06-20      2

或者：

s = pd.to_datetime(df['date & time of connection'])
df = s.groupby(s.dt.floor('d')).size().reset_index(name='count')
print (df)
  date & time of connection  count
0                2017-06-19      3
1                2017-06-20      2
2                2017-06-21      5
3                2017-06-22      3
4                2017-06-23      6

<强>计时：

np.random.seed(1542)

N = 220000
a = np.unique(np.random.randint(N, size=int(N/2)))
df = pd.DataFrame(pd.date_range('2000-01-01', freq='37T', periods=N)).drop(a)
df.columns = ['date & time of connection']
df['date & time of connection'] = df['date & time of connection'].dt.strftime('%d/%m/%Y %H:%M:%S')
print (df.head()) 

In [193]: %%timeit
     ...: df['date & time of connection']=pd.to_datetime(df['date & time of connection'])
     ...: df1 = df.groupby(by=df['date & time of connection'].dt.date).count()
     ...: 
539 ms ± 45.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

In [194]: %%timeit
     ...: df1 = (pd.to_datetime(df['date & time of connection'])
     ...:        .dt.floor('d')
     ...:        .value_counts()
     ...:        .rename_axis('date')
     ...:        .reset_index(name='count'))
     ...: 
12.4 ms ± 350 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [195]: %%timeit
     ...: s = pd.to_datetime(df['date & time of connection'])
     ...: df2 = s.groupby(s.dt.floor('d')).size().reset_index(name='count')
     ...: 
17.7 ms ± 140 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

确保您的列符合日期格式。

df['date & time of connection']=pd.to_datetime(df['date & time of connection'])

然后，您可以按日期对数据进行分组并进行计数：

df.groupby(by=df['date & time of connection'].dt.date).count()
Out[10]: 
                           date & time of connection
date & time of connection                           
2017-06-19                                         3
2017-06-20                                         2
2017-06-21                                         5
2017-06-22                                         3
2017-06-23                                         6

Answer 3

嘿，我发现重采样的简单方法。

# Set the date column as index column.
df = df.set_index('your_date_column')

# Make counts
df_counts = df.your_date_column.resample('D').count()

尽管您的列名很长并且包含空格，这使我有点胆怯。我会用破折号代替空格。

Python＆amp;熊猫 - 每天分组并计算每一天

3 个答案: