[英]Python & Pandas - Group by day and count for each day
我是熊貓的新手,現在我不知道如何安排我的時間系列,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
這是一個 130 k df.groupby('date & time of connection')['date & time of connection'].apply(list)
數據集的樣本,我試過: df.groupby('date & time of connection')['date & time of connection'].apply(list)
我猜還不夠
我想我應該:
你怎么看我的邏輯? 你知道一些教程嗎? 非常感謝你
您可以使用dt.floor
轉換為date
s,然后value_counts
或groupby
與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
要么:
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
時間:
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)
確保您的列采用日期格式。
df['date & time of connection']=pd.to_datetime(df['date & time of connection'])
然后您可以按日期對數據進行分組並進行計數:
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
嘿,我找到了使用重新采樣的簡單方法。
# 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()
雖然你的列名很長並且包含空格,這讓我有點害怕。 我會使用破折號而不是空格。
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