By jojo12

2013-03-29 14:48:13 8 Comments

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


@escha 2019-07-08 16:37:59

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

@Surya 2019-04-10 02:38:11

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)
   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

@YOBEN_S 2019-01-04 14:55:23

You may not need to do with group by , using sort_values+ drop_duplicates

    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)
    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

@Clay 2019-08-09 13:49:45

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.

@Datanovice 2019-11-26 12:50:09

I was scratching my head thinking surely this is simple, thanks for your brilliant answer as always mr Wen.

@joh-mue 2019-01-14 10:03:03

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]

  .apply(lambda group: group[group.Count == group.Count.max()])\

    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.

@George Liu 2018-08-08 18:25:07

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'))

@blueear 2018-07-24 10:45:21

Use groupby and idxmax methods:

  1. transfer col date to datetime:

  2. get the index of max of column date, after groupyby ad_id:

  3. get the wanted data:



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

@Surya 2017-07-07 00:40:35

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                                                                                                       
   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()]                                                       
   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])

@Max Power 2017-12-19 11:55:23

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).

@Carlos Souza 2019-10-27 18:40:12

This is a great solution, but for a different problem

@Rani 2016-11-16 10:14:22

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'])

@Nolan Conaway 2017-09-27 18:23:53

Very nice! Fast with largish frames (25k rows)

@Tyler 2018-12-27 17:14:52

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.

@TMrtSmith 2019-02-04 13:11:47

@Samir or use inplace = True as an argument to drop_duplicates

@Woods Chen 2019-04-10 02:50:40

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.

@Rani 2019-04-10 04:28:11

@WoodsChen it should work. It keeps the max value per key of [sp,mt]

@Woods Chen 2019-04-11 09:37:07

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

@Rani 2019-04-11 15:54:37

@WoodsChen, it drops duplicates of [sp, mt], therefore in your example, the output should be only one row.

@PAC 2015-07-02 12:52:33

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)]

@landewednack 2014-02-11 17:54:50

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']]

@goh 2014-07-11 06:30:52

indeed this is much faster. transform seems to be slow for large dataset.

@tommy.carstensen 2017-03-26 00:47:26

Can you add comments to explain what each line does?

@Roland 2017-05-04 21:25:45

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).

@Qy Zuo 2017-07-20 07:38:28

But doing this df[df['count'] == df['count_max']] will lose NaN rows, as well as the answers above.

@Touya D. Serdan 2018-09-18 05:37:42

I highly suggest to use this approach, for bigger data frames it is much faster to use .appy() or .agg().

@Eulenfuchswiesel 2019-09-30 11:38:30

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?

@Zelazny7 2013-03-29 15:09:40

In [1]: df
    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()
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]
    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.


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
    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

@mathlover 2017-11-15 20:35:01

@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")

@Bjorks number one fan 2018-01-03 20:36:10

@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?

@Ladenkov Vladislav 2018-02-18 18:09:48

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

@Zephyr 2018-11-13 15:51:32

Hi Zealzny, If I want to take top 3 maximum row instead of one max value, How can I tweak your code?

@Woods Chen 2019-04-10 02:54:17

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.

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