Sorting the grouped data as per group size in Pandas

Question:

I have two columns in my dataset, col1 and col2. I want group the data as per col1 and then sort the data as per the size of each group. That is, I want to display groups in ascending order of their size.

I have written the code for grouping and displaying the data as follows:

grouped_data = df.groupby('col1')
"""code for sorting comes here"""
for name,group in grouped_data:
          print (name)
          print (group)

Before displaying the data, I need to sort it as per group size, which I am not able to do.

Asked By: krackoder

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Answers:

You can use python’s sorted:

In [11]: df = pd.DataFrame([[1, 2], [1, 4], [5, 6]], index=['a', 'b', 'c'], columns=['A', 'B'])

In [12]: g = df.groupby('A')

In [13]: sorted(g,  # iterates pairs of (key, corresponding subDataFrame)
                key=lambda x: len(x[1]),  # sort by number of rows (len of subDataFrame)
                reverse=True)  # reverse the sort i.e. largest first
Out[13]: 
[(1,    A  B
     a  1  2
     b  1  4),
 (5,    A  B
     c  5  6)]

Note: as an iterator g, iterates over pairs of the key and the corresponding subframe:

In [14]: list(g)  # happens to be the same as the above...
Out[14]:
[(1,    A  B
     a  1  2
     b  1  4,
 (5,    A  B
     c  5  6)]
Answered By: Andy Hayden

For Pandas 0.17+, use sort_values:

df.groupby('col1').size().sort_values(ascending=False)

For pre-0.17, you can use size().order():

df.groupby('col1').size().order(ascending=False)
Answered By: Victor Yan
df = pandas.DataFrame([[5, 5], [9, 7], [1, 8], [1, 7], [7, 8],
                       [9, 5], [5, 6], [1, 2], [1, 4], [5, 6]],
                      columns=['A', 'B'])

  A   B
0   5   5
1   9   7
2   1   8
3   1   7
4   7   8
5   9   5
6   5   6
7   1   2
8   1   4
9   5   6

group = df.groupby('A')

count = group.size()

count  
A  

1   4
5   3
7   1
9   2
dtype: int64

grp_len = count[count.index.isin(count.nlargest(2).index)]

grp_len   
A  

1   4
5   3
dtype: int64

Answered By: srishti k