'Get statistics for each group (such as count, mean, etc) using pandas GroupBy?
I have a data frame df and I use several columns from it to groupby:
df['col1','col2','col3','col4'].groupby(['col1','col2']).mean()
In the above way I almost get the table (data frame) that I need. What is missing is an additional column that contains number of rows in each group. In other words, I have mean but I also would like to know how many number were used to get these means. For example in the first group there are 8 values and in the second one 10 and so on.
In short: How do I get group-wise statistics for a dataframe?
Solution 1:[1]
On groupby object, the agg function can take a list to apply several aggregation methods at once. This should give you the result you need:
df[['col1', 'col2', 'col3', 'col4']].groupby(['col1', 'col2']).agg(['mean', 'count'])
Solution 2:[2]
Quick Answer:
The simplest way to get row counts per group is by calling .size(), which returns a Series:
df.groupby(['col1','col2']).size()
Usually you want this result as a DataFrame (instead of a Series) so you can do:
df.groupby(['col1', 'col2']).size().reset_index(name='counts')
If you want to find out how to calculate the row counts and other statistics for each group continue reading below.
Detailed example:
Consider the following example dataframe:
In [2]: df
Out[2]:
col1 col2 col3 col4 col5 col6
0 A B 0.20 -0.61 -0.49 1.49
1 A B -1.53 -1.01 -0.39 1.82
2 A B -0.44 0.27 0.72 0.11
3 A B 0.28 -1.32 0.38 0.18
4 C D 0.12 0.59 0.81 0.66
5 C D -0.13 -1.65 -1.64 0.50
6 C D -1.42 -0.11 -0.18 -0.44
7 E F -0.00 1.42 -0.26 1.17
8 E F 0.91 -0.47 1.35 -0.34
9 G H 1.48 -0.63 -1.14 0.17
First let's use .size() to get the row counts:
In [3]: df.groupby(['col1', 'col2']).size()
Out[3]:
col1 col2
A B 4
C D 3
E F 2
G H 1
dtype: int64
Then let's use .size().reset_index(name='counts') to get the row counts:
In [4]: df.groupby(['col1', 'col2']).size().reset_index(name='counts')
Out[4]:
col1 col2 counts
0 A B 4
1 C D 3
2 E F 2
3 G H 1
Including results for more statistics
When you want to calculate statistics on grouped data, it usually looks like this:
In [5]: (df
...: .groupby(['col1', 'col2'])
...: .agg({
...: 'col3': ['mean', 'count'],
...: 'col4': ['median', 'min', 'count']
...: }))
Out[5]:
col4 col3
median min count mean count
col1 col2
A B -0.810 -1.32 4 -0.372500 4
C D -0.110 -1.65 3 -0.476667 3
E F 0.475 -0.47 2 0.455000 2
G H -0.630 -0.63 1 1.480000 1
The result above is a little annoying to deal with because of the nested column labels, and also because row counts are on a per column basis.
To gain more control over the output I usually split the statistics into individual aggregations that I then combine using join. It looks like this:
In [6]: gb = df.groupby(['col1', 'col2'])
...: counts = gb.size().to_frame(name='counts')
...: (counts
...: .join(gb.agg({'col3': 'mean'}).rename(columns={'col3': 'col3_mean'}))
...: .join(gb.agg({'col4': 'median'}).rename(columns={'col4': 'col4_median'}))
...: .join(gb.agg({'col4': 'min'}).rename(columns={'col4': 'col4_min'}))
...: .reset_index()
...: )
...:
Out[6]:
col1 col2 counts col3_mean col4_median col4_min
0 A B 4 -0.372500 -0.810 -1.32
1 C D 3 -0.476667 -0.110 -1.65
2 E F 2 0.455000 0.475 -0.47
3 G H 1 1.480000 -0.630 -0.63
Footnotes
The code used to generate the test data is shown below:
In [1]: import numpy as np
...: import pandas as pd
...:
...: keys = np.array([
...: ['A', 'B'],
...: ['A', 'B'],
...: ['A', 'B'],
...: ['A', 'B'],
...: ['C', 'D'],
...: ['C', 'D'],
...: ['C', 'D'],
...: ['E', 'F'],
...: ['E', 'F'],
...: ['G', 'H']
...: ])
...:
...: df = pd.DataFrame(
...: np.hstack([keys,np.random.randn(10,4).round(2)]),
...: columns = ['col1', 'col2', 'col3', 'col4', 'col5', 'col6']
...: )
...:
...: df[['col3', 'col4', 'col5', 'col6']] = \
...: df[['col3', 'col4', 'col5', 'col6']].astype(float)
...:
Disclaimer:
If some of the columns that you are aggregating have null values, then you really want to be looking at the group row counts as an independent aggregation for each column. Otherwise you may be misled as to how many records are actually being used to calculate things like the mean because pandas will drop NaN entries in the mean calculation without telling you about it.
Solution 3:[3]
Swiss Army Knife: GroupBy.describe
Returns count, mean, std, and other useful statistics per-group.
df.groupby(['A', 'B'])['C'].describe()
count mean std min 25% 50% 75% max
A B
bar one 1.0 0.40 NaN 0.40 0.40 0.40 0.40 0.40
three 1.0 2.24 NaN 2.24 2.24 2.24 2.24 2.24
two 1.0 -0.98 NaN -0.98 -0.98 -0.98 -0.98 -0.98
foo one 2.0 1.36 0.58 0.95 1.15 1.36 1.56 1.76
three 1.0 -0.15 NaN -0.15 -0.15 -0.15 -0.15 -0.15
two 2.0 1.42 0.63 0.98 1.20 1.42 1.65 1.87
To get specific statistics, just select them,
df.groupby(['A', 'B'])['C'].describe()[['count', 'mean']]
count mean
A B
bar one 1.0 0.400157
three 1.0 2.240893
two 1.0 -0.977278
foo one 2.0 1.357070
three 1.0 -0.151357
two 2.0 1.423148
Note: if you only need to compute 1 or 2 stats then it might be faster to use
groupby.aggand just compute those columns otherwise you are performing wasteful computation.
describe works for multiple columns (change ['C'] to ['C', 'D']—or remove it altogether—and see what happens, the result is a MultiIndexed columned dataframe).
You also get different statistics for string data. Here's an example,
df2 = df.assign(D=list('aaabbccc')).sample(n=100, replace=True)
with pd.option_context('precision', 2):
display(df2.groupby(['A', 'B'])
.describe(include='all')
.dropna(how='all', axis=1))
C D
count mean std min 25% 50% 75% max count unique top freq
A B
bar one 14.0 0.40 5.76e-17 0.40 0.40 0.40 0.40 0.40 14 1 a 14
three 14.0 2.24 4.61e-16 2.24 2.24 2.24 2.24 2.24 14 1 b 14
two 9.0 -0.98 0.00e+00 -0.98 -0.98 -0.98 -0.98 -0.98 9 1 c 9
foo one 22.0 1.43 4.10e-01 0.95 0.95 1.76 1.76 1.76 22 2 a 13
three 15.0 -0.15 0.00e+00 -0.15 -0.15 -0.15 -0.15 -0.15 15 1 c 15
two 26.0 1.49 4.48e-01 0.98 0.98 1.87 1.87 1.87 26 2 b 15
For more information, see the documentation.
pandas >= 1.1: DataFrame.value_counts
This is available from pandas 1.1 if you just want to capture the size of every group, this cuts out the GroupBy and is faster.
df.value_counts(subset=['col1', 'col2'])
Minimal Example
# Setup
np.random.seed(0)
df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar',
'foo', 'bar', 'foo', 'foo'],
'B' : ['one', 'one', 'two', 'three',
'two', 'two', 'one', 'three'],
'C' : np.random.randn(8),
'D' : np.random.randn(8)})
df.value_counts(['A', 'B'])
A B
foo two 2
one 2
three 1
bar two 1
three 1
one 1
dtype: int64
Other Statistical Analysis Tools
If you didn't find what you were looking for above, the User Guide has a comprehensive listing of supported statical analysis, correlation, and regression tools.
Solution 4:[4]
To get multiple stats, collapse the index, and retain column names:
df = df.groupby(['col1','col2']).agg(['mean', 'count'])
df.columns = [ ' '.join(str(i) for i in col) for col in df.columns]
df.reset_index(inplace=True)
df
Produces:
Solution 5:[5]
We can easily do it by using groupby and count. But, we should remember to use reset_index().
df[['col1','col2','col3','col4']].groupby(['col1','col2']).count().\
reset_index()
Solution 6:[6]
Please try this code
new_column=df[['col1', 'col2', 'col3', 'col4']].groupby(['col1', 'col2']).count()
df['count_it']=new_column
df
I think that code will add a column called 'count it' which count of each group
Solution 7:[7]
Create a group object and call methods like below example:
grp = df.groupby(['col1', 'col2', 'col3'])
grp.max()
grp.mean()
grp.describe()
Solution 8:[8]
If you are familiar with tidyverse R packages, here is a way to do it in python:
from datar.all import tibble, rnorm, f, group_by, summarise, mean, n, rep
df = tibble(
col1=rep(['A', 'B'], 5),
col2=rep(['C', 'D'], each=5),
col3=rnorm(10),
col4=rnorm(10)
)
df >> group_by(f.col1, f.col2) >> summarise(
count=n(),
col3_mean=mean(f.col3),
col4_mean=mean(f.col4)
)
col1 col2 n mean_col3 mean_col4
0 A C 3 -0.516402 0.468454
1 A D 2 -0.248848 0.979655
2 B C 2 0.545518 -0.966536
3 B D 3 -0.349836 -0.915293
[Groups: ['col1'] (n=2)]
I am the author of the datar package. Please feel free to submit issues if you have any questions about using it.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | |
| Solution 3 | |
| Solution 4 | Jake Drew |
| Solution 5 | |
| Solution 6 | Ichsan |
| Solution 7 | Mahendra |
| Solution 8 | Panwen Wang |