Question

我正在尝试计算在给定日期每小时登录系统的用户数。我的日期类似于：

df=

Name      Date
name_1    2012-07-12 22:20:00
name_1    2012-07-16 22:19:00
name_1    2013-12-16 17:50:00
...
name_2    2010-01-11 19:54:00 
name_2    2010-02-06 12:10:00
...
name_2    2012-07-18 22:12:00
...
name_5423 2013-11-23 10:21:00

因为我对用户名不感兴趣，所以我删除了该列。我设法使用以下命令

创建分组数据结构和新数据帧df2

grp = df.groupby(by=[df.Date.map(lambda x: (x.year, x.month, x.day, x.hour))])
df2 = pd.DataFrame({'Count' : grp.size()}).reset_index()

grp将datetime类型转换为(year, month, day, hour)的元组。

我可以使用datetime循环将其转换回for类型

for i in range(len(df2)):
    proper_date = datetime.datetime(*df2['Date'][i])
    df2.set_value(i, 'Date', proper_date)

我想知道的是，是否有更好/更有效的方法来解决这个问题？

Answer 1

您可以groupby列Date转换为h并汇总size：

print (df.Date.values.astype('datetime64[h]'))
['2012-07-12T22+0200' '2012-07-16T22+0200' '2013-12-16T17+0100'
 '2010-01-11T19+0100' '2010-02-06T12+0100' '2012-07-18T22+0200'
 '2013-11-23T10+0100']

print (df.Name.groupby([df.Date.values.astype('datetime64[h]')]).size())
2010-01-11 19:00:00    1
2010-02-06 12:00:00    1
2012-07-12 22:00:00    1
2012-07-16 22:00:00    1
2012-07-18 22:00:00    1
2013-11-23 10:00:00    1
2013-12-16 17:00:00    1
dtype: int64

另一种解决方案：

print (df.Date.values.astype('<M8[h]'))
['2012-07-12T22+0200' '2012-07-16T22+0200' '2013-12-16T17+0100'
 '2010-01-11T19+0100' '2010-02-06T12+0100' '2012-07-18T22+0200'
 '2013-11-23T10+0100']

print (df.Name.groupby([df.Date.values.astype('<M8[h]')]).size())
2010-01-11 19:00:00    1
2010-02-06 12:00:00    1
2012-07-12 22:00:00    1
2012-07-16 22:00:00    1
2012-07-18 22:00:00    1
2013-11-23 10:00:00    1
2013-12-16 17:00:00    1
dtype: int64

Answer 2

使用重新取样的另一个答案。我认为不是很有效，但有趣。

# Test data
d = {'Date': ['2012-07-12 22:20:00', '2012-07-12 22:19:00', '2013-12-16 17:50:00', '2010-01-11 19:54:00', '2010-02-06 12:10:00', '2012-07-18 22:12:00'],
     'Name': ['name_1', 'name_1', 'name_1', 'name_2', 'name_2', 'name_5']}

df = pd.DataFrame(d)
df['Date'] = pd.to_datetime(df['Date'])
result = df.set_index('Date')

# Resampling data for each hour
result = result.resample('H').count()
# Filtering to keep only hours with at least one row
result[result['Name'] > 0]

                     Name
Date                     
2010-01-11 19:00:00     1
2010-02-06 12:00:00     1
2012-07-12 22:00:00     2
2012-07-18 22:00:00     1
2013-12-16 17:00:00     1

熊猫按日期时间分组

2 个答案: