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

Question

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

I don't get it. What exactly is a group? Why does the second line in the result begin with `1 3`? — Jo So, Mar 29 '13 at 14:53
http://stackoverflow.com/questions/18879782/python-pandas-groupby-forloop-idxmax Could be useful — J_Arthur, Sep 24 '13 at 22:30
This answer is the fastest solution I could find: http://stackoverflow.com/a/21007047/778533 — tommy.carstensen, Mar 26 '17 at 02:17
Similar to this question, could anybody please answer this: https://stackoverflow.com/questions/62069465/match-lists-and-get-value-of-one-column-based-on-values-of-other-columns-from-da Thanks. — ds_Abc, May 29 '20 at 05:42

Zelazny7 · Accepted Answer · 2013-03-29T15:42:14.357

512

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

edited Mar 29 '13 at 15:42

answered Mar 29 '13 at 15:09

Zelazny7

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@Zelazny7, is there a way to adopt this answer to apply to grouping by a column and then looking at 2 columns and doing a max of them to get a greater of the two? I can't get that to work. What I currently have is:def Greater(Merge, maximumA, maximumB): a = Merge[maximumA] b = Merge[maximumB] return max(a, b) Merger.groupby("Search_Term").apply(Greater,"Ratio_x","Ratio_y") – bernando_vialli Nov 15 '17 at 20:35
4

@Zelazny7 I'm using the second, `idx` approach. But, I can only afford to a single maximum for each group (and my data has a few duplicate-max's). is there a way to get around this with your solution? – 3pitt Jan 03 '18 at 20:36
actually, that does not work for me. I can not track the problem, because dataframe if quit big, but the solution by @Rani works good – Ladenkov Vladislav Feb 18 '18 at 18:09
Hi Zealzny, If I want to take top 3 maximum row instead of one max value, How can I tweak your code? – Zephyr Nov 13 '18 at 15:51
`transform` method may have pool performance when the data set is large enough, get the max value first then merge the dataframes will be better. – Woods Chen Apr 10 '19 at 02:54
You are my hero! :-D I had the almost same situation which I couldn't fix. In my case I was working with a Django queryset. – Peter Dec 23 '21 at 11:55

score 250 · Answer 2 · answered Nov 16 '16 at 10:14

250

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'])

answered Nov 16 '16 at 10:14

Rani

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7

Very nice! Fast with largish frames (25k rows) – Nolan Conaway Sep 27 '17 at 18:23
3

For those who are somewhat new with Python, you will need to assign this to a new variable, it doesn't change the current df variable. – Tyler Dec 27 '18 at 17:14
3

@Samir or use `inplace = True` as an argument to `drop_duplicates` – TMrtSmith Feb 04 '19 at 13:11
10

This is a great answer when need only one of rows with the same max values, however it wont work as expected if I need all the rows with max values. – Woods Chen Apr 10 '19 at 02:50
@WoodsChen it should work. It keeps the max value per key of [sp,mt] – Rani Apr 10 '19 at 04:28
2

I mean if the dataframe is pd.DataFrame({'sp':[1, 1, 2], 'mt':[1, 1, 2], 'value':[2, 2, 3]}, then there will be 2 rows with the same max value 2 in the group where sp==1 and mt==2. @Rani – Woods Chen Apr 11 '19 at 09:37
1

@WoodsChen, it drops duplicates of [sp, mt], therefore in your example, the output should be only one row. – Rani Apr 11 '19 at 15:54

score 96 · Answer 3 · edited Feb 19 '22 at 00:42

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])

The questioner here specified `"I want to get ALL the rows where count equals max in each group"`, while `idxmax` `Return[s] index of first occurrence of maximum over requested axis"` according to the docs (0.21). — Max Power, Dec 19 '17 at 11:55

BENY · Answer 4 · 2019-02-27T16:24:06.533

53

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

edited Feb 27 '19 at 16:24

answered Jan 04 '19 at 14:55

BENY

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Not only is this an order of magnitude faster than the other solutions (at least for my use case), it has the added benefit of simply chaining as part of the construction of the original dataframe. – Clay Aug 09 '19 at 13:49
When you see this answer, you realize that all the others are wrong. This is clearly the way to do it. Thanks. – Hunaphu Feb 25 '21 at 15:32
One should add `na_position="first"` to `sort_values` in order to ignore `NaN`s. – Antoine Aug 20 '21 at 10:57
I found this to be fast for my DF of several million rows. – John Stud Feb 10 '22 at 02:07

landewednack · Answer 5 · 2014-02-11T18:06:24.633

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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']]

edited Feb 11 '14 at 18:06

answered Feb 11 '14 at 17:54

landewednack

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indeed this is much faster. transform seems to be slow for large dataset. – goh Jul 11 '14 at 06:30
1

Can you add comments to explain what each line does? – tommy.carstensen Mar 26 '17 at 00:47
1

fwiw: I found the more elegant-looking solution from @Zelazny7 took a long time to execute for my set of ~100K rows, but this one ran pretty quickly. (I'm running a now way-obsolete 0.13.0, which might account for slowness). – Roland May 04 '17 at 21:25
3

But doing this `df[df['count'] == df['count_max']]` will lose NaN rows, as well as the answers above. – Qy Zuo Jul 20 '17 at 07:38
I highly suggest to use this approach, for bigger data frames it is much faster to use .appy() or .agg(). – Gerard Sep 18 '18 at 05:37
I am using pandas version 0.25.1 and timed everything in jupyter (extended the lists with `*1000000`). I found the @Zelazny7 solution to be faster. Did I miss something in the test or is it due to the new pandas version? – Eulenfuchswiesel Sep 30 '19 at 11:38

score 15 · Answer 6 · edited Jul 24 '18 at 14:38

15

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

edited Jul 24 '18 at 14:38

Tim Diekmann

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answered Jul 24 '18 at 10:45

blueear

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score 12 · Answer 7 · answered Jul 02 '15 at 12:52

12

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)]

answered Jul 02 '15 at 12:52

PAC

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score 7 · Answer 8 · edited Apr 16 '22 at 09:30

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

score 7 · Answer 9 · answered Mar 02 '21 at 11:42

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 ---

score 5 · Answer 10 · answered Apr 10 '19 at 02:38

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

score 2 · Answer 11 · answered Aug 08 '18 at 18:25

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'))

score 2 · Answer 12 · answered Apr 20 '21 at 21:51

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()

score 1 · Answer 13 · answered Jan 14 '19 at 10:03

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.

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

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