python - Get the row(s) which have the max value in groups using groupby

312

votes

How do I find all rows in a pandas data frame 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**
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

pythonpandasmaxpandas-groupby

What format is your data frame in? – David Robinson

I don't get it. What exactly is a group? Why does the second line in the result begin with 1 3? – Jo So

stackoverflow.com/questions/18879782/… Could be useful – J_Arthur

This answer is the fastest solution I could find: stackoverflow.com/a/21007047/778533 – tommy.carstensen

Similar to this question, could anybody please answer this: stackoverflow.com/questions/62069465/… Thanks. – ds_Abc

428

votes

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

207

votes

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'])

75

votes

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])

37

votes

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']]

32

votes

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

9

votes

Use groupby and idxmax methods:

transfer col date to datetime:
```
df['date']=pd.to_datetime(df['date'])
```
get the index of max of column date, after groupyby 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

7

votes

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)]

5

votes

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

enter image description here

4

votes

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

2

votes

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'))

1

votes

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.

1

votes

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 ---

0

votes

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()

python - Get the row(s) which have the max value in groups using groupby

13 Answers