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.
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)]
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)
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
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.
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)]
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)
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