Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Showing content from https://stackoverflow.com/questions/14941366/pandas-sort-by-group-aggregate-and-column below:

python - Pandas sort by group aggregate and column

Given the following dataframe

In [31]: rand = np.random.RandomState(1)
         df = pd.DataFrame({'A': ['foo', 'bar', 'baz'] * 2,
                            'B': rand.randn(6),
                            'C': rand.rand(6) > .5})

In [32]: df
Out[32]:      A         B      C
         0  foo  1.624345  False
         1  bar -0.611756   True
         2  baz -0.528172  False
         3  foo -1.072969   True
         4  bar  0.865408  False
         5  baz -2.301539   True 

I would like to sort it in groups (A) by the aggregated sum of B, and then by the value in C (not aggregated). So basically get the order of the A groups with

In [28]: df.groupby('A').sum().sort('B')
Out[28]:             B  C
         A               
         baz -2.829710  1
         bar  0.253651  1
         foo  0.551377  1

And then by True/False, so that it ultimately looks like this:

In [30]: df.ix[[5, 2, 1, 4, 3, 0]]
Out[30]: A         B      C
    5  baz -2.301539   True
    2  baz -0.528172  False
    1  bar -0.611756   True
    4  bar  0.865408  False
    3  foo -1.072969   True
    0  foo  1.624345  False

How can this be done?

asked Feb 18, 2013 at 16:55

21.2k1919 gold badges8080 silver badges9797 bronze badges

Groupby A:

In [0]: grp = df.groupby('A')

Within each group, sum over B and broadcast the values using transform. Then sort by B:

In [1]: grp[['B']].transform(sum).sort('B')
Out[1]:
          B
2 -2.829710
5 -2.829710
1  0.253651
4  0.253651
0  0.551377
3  0.551377

Index the original df by passing the index from above. This will re-order the A values by the aggregate sum of the B values:

In [2]: sort1 = df.ix[grp[['B']].transform(sum).sort('B').index]

In [3]: sort1
Out[3]:
     A         B      C
2  baz -0.528172  False
5  baz -2.301539   True
1  bar -0.611756   True
4  bar  0.865408  False
0  foo  1.624345  False
3  foo -1.072969   True

Finally, sort the 'C' values within groups of 'A' using the sort=False option to preserve the A sort order from step 1:

In [4]: f = lambda x: x.sort('C', ascending=False)

In [5]: sort2 = sort1.groupby('A', sort=False).apply(f)

In [6]: sort2
Out[6]:
         A         B      C
A
baz 5  baz -2.301539   True
    2  baz -0.528172  False
bar 1  bar -0.611756   True
    4  bar  0.865408  False
foo 3  foo -1.072969   True
    0  foo  1.624345  False

Clean up the df index by using reset_index with drop=True:

In [7]: sort2.reset_index(0, drop=True)
Out[7]:
     A         B      C
5  baz -2.301539   True
2  baz -0.528172  False
1  bar -0.611756   True
4  bar  0.865408  False
3  foo -1.072969   True
0  foo  1.624345  False

answered Feb 18, 2013 at 22:11

40.7k1818 gold badges7171 silver badges8585 bronze badges

Here's a more concise approach...

df['a_bsum'] = df.groupby('A')['B'].transform(sum)
df.sort(['a_bsum','C'], ascending=[True, False]).drop('a_bsum', axis=1)

The first line adds a column to the data frame with the groupwise sum. The second line performs the sort and then removes the extra column.

Result:

    A       B           C
5   baz     -2.301539   True
2   baz     -0.528172   False
1   bar     -0.611756   True
4   bar      0.865408   False
3   foo     -1.072969   True
0   foo      1.624345   False

NOTE: sort is deprecated, use sort_values instead

15.6k55 gold badges4242 silver badges5050 bronze badges

answered May 14, 2013 at 14:03

842k202202 gold badges1.6k1.6k silver badges1.5k1.5k bronze badges

One way to do this is to insert a dummy column with the sums in order to sort:

In [10]: sum_B_over_A = df.groupby('A').sum().B

In [11]: sum_B_over_A
Out[11]: 
A
bar    0.253652
baz   -2.829711
foo    0.551376
Name: B

in [12]: df['sum_B_over_A'] = df.A.apply(sum_B_over_A.get_value)

In [13]: df
Out[13]: 
     A         B      C  sum_B_over_A
0  foo  1.624345  False      0.551376
1  bar -0.611756   True      0.253652
2  baz -0.528172  False     -2.829711
3  foo -1.072969   True      0.551376
4  bar  0.865408  False      0.253652
5  baz -2.301539   True     -2.829711

In [14]: df.sort(['sum_B_over_A', 'A', 'B'])
Out[14]: 
     A         B      C   sum_B_over_A
5  baz -2.301539   True      -2.829711
2  baz -0.528172  False      -2.829711
1  bar -0.611756   True       0.253652
4  bar  0.865408  False       0.253652
3  foo -1.072969   True       0.551376
0  foo  1.624345  False       0.551376

and maybe you would drop the dummy row:

In [15]: df.sort(['sum_B_over_A', 'A', 'B']).drop('sum_B_over_A', axis=1)
Out[15]: 
     A         B      C
5  baz -2.301539   True
2  baz -0.528172  False
1  bar -0.611756   True
4  bar  0.865408  False
3  foo -1.072969   True
0  foo  1.624345  False

answered Feb 18, 2013 at 18:06

377k110110 gold badges639639 silver badges543543 bronze badges

The question is difficult to understand. However, group by A and sum by B then sort values descending. The column A sort order depends on B. You can then use filtering to create a new dataframe filter by A values order the dataframe.

rand = np.random.RandomState(1)
df = pd.DataFrame({'A': ['foo', 'bar', 'baz'] * 2,
                        'B': rand.randn(6),
                        'C': rand.rand(6) > .5})
grouped=df.groupby('A')['B'].sum().sort_values(ascending=False)
print(grouped)
print(grouped.index.get_level_values(0))

Output:

A
foo    0.551377
bar    0.253651
baz   -2.829710

answered Jul 12, 2021 at 14:58

Start asking to get answers

Find the answer to your question by asking.

Explore related questions

See similar questions with these tags.

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4