'Get the row(s) which have the max value in groups using groupby
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 in each group, like:
0 MM1 S1 a **3**
2 MM1 S3 cb **5**
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
Solution 1:[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
Solution 2:[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'])
Solution 3:[3]
Easy solution would be to apply the 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])
Solution 4:[4]
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
Solution 5:[5]
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']]
Solution 6:[6]
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
Solution 7:[7]
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)]
Solution 8:[8]
Summarizing, there are many ways, but which one is faster?
import pandas as pd
import numpy as np
import time
df = pd.DataFrame(np.random.randint(1,10,size=(1000000, 2)), columns=list('AB'))
start_time = time.time()
df1idx = df.groupby(['A'])['B'].transform(max) == df['B']
df1 = df[df1idx]
print("---1 ) %s seconds ---" % (time.time() - start_time))
start_time = time.time()
df2 = df.sort_values('B').groupby(['A']).tail(1)
print("---2 ) %s seconds ---" % (time.time() - start_time))
start_time = time.time()
df3 = df.sort_values('B').drop_duplicates(['A'],keep='last')
print("---3 ) %s seconds ---" % (time.time() - start_time))
start_time = time.time()
df3b = df.sort_values('B', ascending=False).drop_duplicates(['A'])
print("---3b) %s seconds ---" % (time.time() - start_time))
start_time = time.time()
df4 = df[df['B'] == df.groupby(['A'])['B'].transform(max)]
print("---4 ) %s seconds ---" % (time.time() - start_time))
start_time = time.time()
d = df.groupby('A')['B'].nlargest(1)
df5 = df.iloc[[i[1] for i in d.index], :]
print("---5 ) %s seconds ---" % (time.time() - start_time))
And the winner is...
- --1 ) 0.03337574005126953 seconds ---
- --2 ) 0.1346898078918457 seconds ---
- --3 ) 0.10243558883666992 seconds ---
- --3b) 0.1004343032836914 seconds ---
- --4 ) 0.028397560119628906 seconds ---
- --5 ) 0.07552886009216309 seconds ---
Solution 9:[9]
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
Solution 10:[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
Solution 11:[11]
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'))
Solution 12:[12]
If you sort your DataFrame that ordering will be preserved in the groupby. You can then just grab the first or last element and reset the index.
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.sort_values("count", ascending=False).groupby(["sp", "mt"]).first().reset_index()
Solution 13:[13]
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.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow