问题
I have the following dataset,
Day Element Data_Value
6786 01-01 TMAX 112
9333 01-01 TMAX 101
9330 01-01 TMIN 60
11049 01-01 TMIN 0
6834 01-01 TMIN 25
11862 01-01 TMAX 113
1781 01-01 TMAX 115
11042 01-01 TMAX 105
1110 01-01 TMAX 111
651 01-01 TMIN 44
11350 01-01 TMIN 83
1798 01-02 TMAX 70
4975 01-02 TMAX 79
12774 01-02 TMIN 0
3977 01-02 TMIN 60
2485 01-02 TMAX 73
4888 01-02 TMIN 31
11836 01-02 TMIN 26
11368 01-02 TMAX 71
2483 01-02 TMIN 26
I want to group by the Day and then find the overall min of TMIN an the max of TMAX and put these in to a data frame, so I get an output like...
Day DayMin DayMax
01-01 0 115
01-02 0 79
I know I need to do,
df.groupby(by='Day')
but I am a stuck with the next step - should create columns to store the TMAX and TMIN values?
回答1:
You can use a assign
+ abs
, followed by groupby
+ agg
:
df = (df.assign(Data_Value=df['Data_Value'].abs())
.groupby(['Day'])['Data_Value'].agg([('Min' , 'min'), ('Max', 'max')])
.add_prefix('Day'))
df
DayMin DayMax
Day
01-01 0 115
01-02 0 79
回答2:
Use
In [5265]: def maxmin(x):
...: mx = x[x.Element == 'TMAX'].Data_Value.max()
...: mn = x[x.Element == 'TMIN'].Data_Value.min()
...: return pd.Series({'DayMin': mn, 'DayMax': mx})
...:
In [5266]: df.groupby('Day').apply(maxmin)
Out[5266]:
DayMax DayMin
Day
01-01 115 0
01-02 79 0
Also,
In [5268]: df.groupby('Day').apply(maxmin).reset_index()
Out[5268]:
Day DayMax DayMin
0 01-01 115 0
1 01-02 79 0
Or, use query
instead of x[x.Element == 'TMAX']
as x.query("Element == 'TMAX'")
回答3:
Create duplicate columns and find min and max using agg i.e
ndf = df.assign(DayMin = df['Data_Value'].abs(),DayMax=df['Data_Value'].abs()).groupby('Day')\
.agg({'DayMin':'min','DayMax':'max'})
DayMax DayMin Day 01-01 115 0 01-02 79 0
Incase you want both TMIN and TMAX then groupby(['Day','Element'])
来源:https://stackoverflow.com/questions/46501703/groupby-column-and-find-min-and-max-of-each-group