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Python & Pandas - 按天分组并计算每一天

转载 作者:太空狗 更新时间:2023-10-29 22:12:17 26 4
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我是 pandas 的新手,现在我不知道如何安排我的时间序列,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 原始数据集的样本,我试过:df.groupby('连接日期和时间')['连接日期和时间'].apply(list)

我想还不够

我想我应该:

  • 创建一个索引从 dd/mm/yyyy 到 dd/mm/yyyy 的字典
  • 将“连接日期和时间”类型的 dateTime 转换为 Date
  • 分组并统计“连接日期和时间”的日期
  • 将我数的数字放入字典中?

你觉得我的逻辑怎么样?你知道一些短裙吗?非常感谢

最佳答案

您可以使用 dt.floor用于转换为 date,然后是 value_countsgroupbysize :

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

时间:

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)

关于Python & Pandas - 按天分组并计算每一天,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48961892/

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