问题
How do I find all rows in a pandas dataframe which have the max value for count column, after grouping by [\'Sp\',\'Mt\'] columns?
Example 1: the following dataFrame, which I group by [\'Sp\',\'Mt\']:
Sp Mt Value count
0 MM1 S1 a **3**
1 MM1 S1 n 2
2 MM1 S3 cb 5
3 MM2 S3 mk **8**
4 MM2 S4 bg **10**
5 MM2 S4 dgd 1
6 MM4 S2 rd 2
7 MM4 S2 cb 2
8 MM4 S2 uyi **7**
Expected output: get the result rows whose count is max between the groups, like:
0 MM1 S1 a **3**
1 3 MM2 S3 mk **8**
4 MM2 S4 bg **10**
8 MM4 S2 uyi **7**
Example 2: this dataframe, which I group by [\'Sp\',\'Mt\']:
Sp Mt Value count
4 MM2 S4 bg 10
5 MM2 S4 dgd 1
6 MM4 S2 rd 2
7 MM4 S2 cb 8
8 MM4 S2 uyi 8
For the above example, I want to get all the rows where count equals max, in each group e.g :
MM2 S4 bg 10
MM4 S2 cb 8
MM4 S2 uyi 8
回答1:
In [1]: df
Out[1]:
Sp Mt Value count
0 MM1 S1 a 3
1 MM1 S1 n 2
2 MM1 S3 cb 5
3 MM2 S3 mk 8
4 MM2 S4 bg 10
5 MM2 S4 dgd 1
6 MM4 S2 rd 2
7 MM4 S2 cb 2
8 MM4 S2 uyi 7
In [2]: df.groupby(['Mt'], sort=False)['count'].max()
Out[2]:
Mt
S1 3
S3 8
S4 10
S2 7
Name: count
To get the indices of the original DF you can do:
In [3]: idx = df.groupby(['Mt'])['count'].transform(max) == df['count']
In [4]: df[idx]
Out[4]:
Sp Mt Value count
0 MM1 S1 a 3
3 MM2 S3 mk 8
4 MM2 S4 bg 10
8 MM4 S2 uyi 7
Note that if you have multiple max values per group, all will be returned.
Update
On a hail mary chance that this is what the OP is requesting:
In [5]: df['count_max'] = df.groupby(['Mt'])['count'].transform(max)
In [6]: df
Out[6]:
Sp Mt Value count count_max
0 MM1 S1 a 3 3
1 MM1 S1 n 2 3
2 MM1 S3 cb 5 8
3 MM2 S3 mk 8 8
4 MM2 S4 bg 10 10
5 MM2 S4 dgd 1 10
6 MM4 S2 rd 2 7
7 MM4 S2 cb 2 7
8 MM4 S2 uyi 7 7
回答2:
You can sort the dataFrame by count and then remove duplicates. I think it's easier:
df.sort_values('count', ascending=False).drop_duplicates(['Sp','Mt'])
回答3:
Easy solution would be to apply : idxmax() function to get indices of rows with max values. This would filter out all the rows with max value in the group.
In [365]: import pandas as pd
In [366]: df = pd.DataFrame({
'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4','MM4'],
'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
'count' : [3,2,5,8,10,1,2,2,7]
})
In [367]: df
Out[367]:
count mt sp val
0 3 S1 MM1 a
1 2 S1 MM1 n
2 5 S3 MM1 cb
3 8 S3 MM2 mk
4 10 S4 MM2 bg
5 1 S4 MM2 dgb
6 2 S2 MM4 rd
7 2 S2 MM4 cb
8 7 S2 MM4 uyi
### Apply idxmax() and use .loc() on dataframe to filter the rows with max values:
In [368]: df.loc[df.groupby(["sp", "mt"])["count"].idxmax()]
Out[368]:
count mt sp val
0 3 S1 MM1 a
2 5 S3 MM1 cb
3 8 S3 MM2 mk
4 10 S4 MM2 bg
8 7 S2 MM4 uyi
### Just to show what values are returned by .idxmax() above:
In [369]: df.groupby(["sp", "mt"])["count"].idxmax().values
Out[369]: array([0, 2, 3, 4, 8])
回答4:
Having tried the solution suggested by Zelazny on a relatively large DataFrame (~400k rows) I found it to be very slow. Here is an alternative that I found to run orders of magnitude faster on my data set.
df = pd.DataFrame({
'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4', 'MM4'],
'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
'count' : [3,2,5,8,10,1,2,2,7]
})
df_grouped = df.groupby(['sp', 'mt']).agg({'count':'max'})
df_grouped = df_grouped.reset_index()
df_grouped = df_grouped.rename(columns={'count':'count_max'})
df = pd.merge(df, df_grouped, how='left', on=['sp', 'mt'])
df = df[df['count'] == df['count_max']]
回答5:
You may not need to do with group by , using sort_values+ drop_duplicates
df.sort_values('count').drop_duplicates(['Sp','Mt'],keep='last')
Out[190]:
Sp Mt Value count
0 MM1 S1 a 3
2 MM1 S3 cb 5
8 MM4 S2 uyi 7
3 MM2 S3 mk 8
4 MM2 S4 bg 10
Also almost same logic by using tail
df.sort_values('count').groupby(['Sp', 'Mt']).tail(1)
Out[52]:
Sp Mt Value count
0 MM1 S1 a 3
2 MM1 S3 cb 5
8 MM4 S2 uyi 7
3 MM2 S3 mk 8
4 MM2 S4 bg 10
回答6:
For me, the easiest solution would be keep value when count is equal to the maximum. Therefore, the following one line command is enough :
df[df['count'] == df.groupby(['Mt'])['count'].transform(max)]
回答7:
Use groupby and idxmax methods:
transfer col
datetodatetime:df['date']=pd.to_datetime(df['date'])get the index of
maxof columndate, aftergroupyby ad_id:idx=df.groupby(by='ad_id')['date'].idxmax()get the wanted data:
df_max=df.loc[idx,]
Out[54]:
ad_id price date
7 22 2 2018-06-11
6 23 2 2018-06-22
2 24 2 2018-06-30
3 28 5 2018-06-22
回答8:
df = pd.DataFrame({
'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4','MM4'],
'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
'count' : [3,2,5,8,10,1,2,2,7]
})
df.groupby(['sp', 'mt']).apply(lambda grp: grp.nlargest(1, 'count'))
回答9:
I've been using this functional style for many group operations:
df = pd.DataFrame({
'Sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4', 'MM4'],
'Mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
'Val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
'Count' : [3,2,5,8,10,1,2,2,7]
})
df.groupby('Mt')\
.apply(lambda group: group[group.Count == group.Count.max()])\
.reset_index(drop=True)
sp mt val count
0 MM1 S1 a 3
1 MM4 S2 uyi 7
2 MM2 S3 mk 8
3 MM2 S4 bg 10
.reset_index(drop=True) gets you back to the original index by dropping the group-index.
回答10:
Realizing that "applying" "nlargest" to groupby object works just as fine:
Additional advantage - also can fetch top n values if required:
In [85]: import pandas as pd
In [86]: df = pd.DataFrame({
...: 'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4','MM4'],
...: 'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
...: 'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
...: 'count' : [3,2,5,8,10,1,2,2,7]
...: })
## Apply nlargest(1) to find the max val df, and nlargest(n) gives top n values for df:
In [87]: df.groupby(["sp", "mt"]).apply(lambda x: x.nlargest(1, "count")).reset_index(drop=True)
Out[87]:
count mt sp val
0 3 S1 MM1 a
1 5 S3 MM1 cb
2 8 S3 MM2 mk
3 10 S4 MM2 bg
4 7 S2 MM4 uyi
回答11:
Try using "nlargest" on the groupby object. The advantage of using nlargest is that it returns the index of the rows where "the nlargest item(s)" were fetched from. Note: we slice the second(1) element of our index since our index in this case consist of tuples(eg.(s1, 0)).
df = pd.DataFrame({
'sp' : ['MM1', 'MM1', 'MM1', 'MM2', 'MM2', 'MM2', 'MM4', 'MM4','MM4'],
'mt' : ['S1', 'S1', 'S3', 'S3', 'S4', 'S4', 'S2', 'S2', 'S2'],
'val' : ['a', 'n', 'cb', 'mk', 'bg', 'dgb', 'rd', 'cb', 'uyi'],
'count' : [3,2,5,8,10,1,2,2,7]
})
d = df.groupby('mt')['count'].nlargest(1) # pass 1 since we want the max
df.iloc[[i[1] for i in d.index], :] # pass the index of d as list comprehension
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来源：https://stackoverflow.com/questions/15705630/get-the-rows-which-have-the-max-value-in-groups-using-groupby