Python 数据帧中连续天数的平均值
我有一个熊猫数据帧Python 数据帧中连续天数的平均值,python,python-3.x,pandas,Python,Python 3.x,Pandas,我有一个熊猫数据帧dfas: Date Val WD 1/3/2019 2.65 Thursday 1/4/2019 2.51 Friday 1/5/2019 2.95 Saturday 1/6/2019 3.39 Sunday 1/7/2019 3.39 Monday 1/12/2019 2.23 Saturday 1/13/2019 2.50 Sunday 1/14/2019 3.62
df
as:
Date Val WD
1/3/2019 2.65 Thursday
1/4/2019 2.51 Friday
1/5/2019 2.95 Saturday
1/6/2019 3.39 Sunday
1/7/2019 3.39 Monday
1/12/2019 2.23 Saturday
1/13/2019 2.50 Sunday
1/14/2019 3.62 Monday
1/15/2019 3.81 Tuesday
1/16/2019 3.75 Wednesday
1/17/2019 3.69 Thursday
1/18/2019 3.47 Friday
我需要从上面获取以下df2
:
Date Val WD
1/3/2019 2.65 Thursday
1/4/2019 2.51 Friday
1/5/2019 3.24 Saturday
1/6/2019 3.24 Sunday
1/7/2019 3.24 Monday
1/12/2019 2.78 Saturday
1/13/2019 2.78 Sunday
1/14/2019 2.78 Monday
1/15/2019 3.81 Tuesday
1/16/2019 3.75 Wednesday
1/17/2019 3.69 Thursday
1/18/2019 3.47 Friday
其中,df2值更新为具有连续Sat、Sun和Mon值的平均值
i、 e.日期2019年5月1日、2019年6月1日、2019年7月1日的df中2.95、3.39、3.39的平均值为3.24,因此在df2中,我将2019年5月1日、2019年6月1日、2019年7月1日的值替换为3.24
诀窍在于找到连续的周六、周日和周一。不确定如何处理此问题。此逻辑创建一个系列
,该系列为数据帧中的连续Sat/Sun/Mon行组分配唯一ID。然后确保其中有3个(不仅仅是Sat/Sun或Sun/Mon),并转换这些值的平均值:
import pandas as pd
#df['Date'] = pd.to_datetime(df.Date)
s = (~(df.Date.dt.dayofweek.isin([0,6])
& (df.Date - df.Date.shift(1)).dt.days.eq(1))).cumsum()
to_trans = s[s.groupby(s).transform('size').eq(3)]
df.loc[to_trans.index, 'Val'] = df.loc[to_trans.index].groupby(to_trans).Val.transform('mean')
输出:
扩展输入数据
您可以使用CustomBusinessDay
和pd.grouper
创建组列:
# if you want to only find the mean if all three days are found
from pandas.tseries.offsets import CustomBusinessDay
days = CustomBusinessDay(weekmask='Tue Wed Thu Fri Sat')
df['group_col'] = df.groupby(pd.Grouper(key='Date', freq=days)).ngroup()
df.update(df[df.groupby('group_col')['Val'].transform('size').eq(3)].groupby('group_col').transform('mean'))
Date Val WD group_col
0 2019-01-03 2.650000 Thursday 0
1 2019-01-04 2.510000 Friday 1
2 2019-01-05 3.243333 Saturday 2
3 2019-01-06 3.243333 Sunday 2
4 2019-01-07 3.243333 Monday 2
5 2019-01-12 2.783333 Saturday 7
6 2019-01-13 2.783333 Sunday 7
7 2019-01-14 2.783333 Monday 7
8 2019-01-15 3.810000 Tuesday 8
9 2019-01-16 3.750000 Wednesday 9
10 2019-01-17 3.690000 Thursday 10
11 2019-01-18 3.470000 Friday 11
或者,如果你想找到在同一周的星期六和星期一的任何组合的平均值
days = CustomBusinessDay(weekmask='Tue Wed Thu Fri Sat')
df['group_col'] = df.groupby(pd.Grouper(key='Date', freq=days)).ngroup()
df['Val'] = df.groupby('group_col')['Val'].transform('mean')
一种方法是计算周数,然后使用groupby
计算特定天数的平均值,并将其映射回原始数据帧
df['Date'] = pd.to_datetime(df['Date'])
# consider Monday to belong to previous week
week, weekday = df['Date'].dt.week, df['Date'].dt.weekday
df['Week'] = np.where(weekday.eq(0), week - 1, week)
# take means of Fri, Sat, Sun, then map back
mask = weekday.isin([5, 6, 0])
week_val_map = df[mask].groupby('Week')['Val'].mean()
df.loc[mask, 'Val'] = df['Week'].map(week_val_map)
print(df)
Date Val WD Week
0 2019-01-03 2.650000 Thursday 1
1 2019-01-04 2.510000 Friday 1
2 2019-01-05 3.243333 Saturday 1
3 2019-01-06 3.243333 Sunday 1
4 2019-01-07 3.243333 Monday 1
5 2019-01-12 2.783333 Saturday 2
6 2019-01-13 2.783333 Sunday 2
7 2019-01-14 2.783333 Monday 2
8 2019-01-15 3.810000 Tuesday 3
9 2019-01-16 3.750000 Wednesday 3
10 2019-01-17 3.690000 Thursday 3
11 2019-01-18 3.470000 Friday 3
请注意,'WD'
列是完全不必要的,因为您可以使用Series.dt.dayofweek
访问该信息,即使没有周六数据,这也将是周日/周一的平均值。@Alolz您完全正确,感谢您指出这一点。它已被更正。如果您只有连续的星期六和星期天(没有星期一),甚至只有星期六和星期一,会发生什么?在这些情况下,您还是要取平均值,还是保持数据不变?
days = CustomBusinessDay(weekmask='Tue Wed Thu Fri Sat')
df['group_col'] = df.groupby(pd.Grouper(key='Date', freq=days)).ngroup()
df['Val'] = df.groupby('group_col')['Val'].transform('mean')
df['Date'] = pd.to_datetime(df['Date'])
# consider Monday to belong to previous week
week, weekday = df['Date'].dt.week, df['Date'].dt.weekday
df['Week'] = np.where(weekday.eq(0), week - 1, week)
# take means of Fri, Sat, Sun, then map back
mask = weekday.isin([5, 6, 0])
week_val_map = df[mask].groupby('Week')['Val'].mean()
df.loc[mask, 'Val'] = df['Week'].map(week_val_map)
print(df)
Date Val WD Week
0 2019-01-03 2.650000 Thursday 1
1 2019-01-04 2.510000 Friday 1
2 2019-01-05 3.243333 Saturday 1
3 2019-01-06 3.243333 Sunday 1
4 2019-01-07 3.243333 Monday 1
5 2019-01-12 2.783333 Saturday 2
6 2019-01-13 2.783333 Sunday 2
7 2019-01-14 2.783333 Monday 2
8 2019-01-15 3.810000 Tuesday 3
9 2019-01-16 3.750000 Wednesday 3
10 2019-01-17 3.690000 Thursday 3
11 2019-01-18 3.470000 Friday 3