Asked  7 Months ago    Answers:  5   Viewed   48 times
  1. How to perform aggregation with pandas?
  2. No DataFrame after aggregation! What happened?
  3. How to aggregate mainly strings columns (to lists, tuples, strings with separator)?
  4. How to aggregate counts?
  5. How to create new column filled by aggregated values?

I've seen these recurring questions asking about various faces of the pandas aggregate functionality. Most of the information regarding aggregation and its various use cases today is fragmented across dozens of badly worded, unsearchable posts. The aim here is to collate some of the more important points for posterity.

This Q/A is meant to be the next instalment in a series of helpful user-guides:

  • How to pivot a dataframe,
  • Pandas concat
  • How do I operate on a DataFrame with a Series for every column
  • Pandas Merging 101

Please note that this post is not meant to be a replacement for the documentation about aggregation and about groupby, so please read that as well!

 Answers

28

Question 1

How to perform aggregation with pandas ?

Expanded aggregation documentation.

Aggregating functions are the ones that reduce the dimension of the returned objects. It means output Series/DataFrame have less or same rows like original. Some common aggregating functions are tabulated below:

Function    Description
mean()      Compute mean of groups
sum()       Compute sum of group values
size()      Compute group sizes
count()     Compute count of group
std()       Standard deviation of groups
var()       Compute variance of groups
sem()       Standard error of the mean of groups
describe()  Generates descriptive statistics
first()     Compute first of group values
last()      Compute last of group values
nth()       Take nth value, or a subset if n is a list
min()       Compute min of group values
max()       Compute max of group values
np.random.seed(123)

df = pd.DataFrame({'A' : ['foo', 'foo', 'bar', 'foo', 'bar', 'foo'],
                   'B' : ['one', 'two', 'three','two', 'two', 'one'],
                   'C' : np.random.randint(5, size=6),
                   'D' : np.random.randint(5, size=6),
                   'E' : np.random.randint(5, size=6)})
print (df)
     A      B  C  D  E
0  foo    one  2  3  0
1  foo    two  4  1  0
2  bar  three  2  1  1
3  foo    two  1  0  3
4  bar    two  3  1  4
5  foo    one  2  1  0

Aggregation by filtered columns and cython implemented functions:

df1 = df.groupby(['A', 'B'], as_index=False)['C'].sum()
print (df1)
     A      B  C
0  bar  three  2
1  bar    two  3
2  foo    one  4
3  foo    two  5

Aggregate function is using for all columns without specified in groupby function, here A, B columns:

df2 = df.groupby(['A', 'B'], as_index=False).sum()
print (df2)
     A      B  C  D  E
0  bar  three  2  1  1
1  bar    two  3  1  4
2  foo    one  4  4  0
3  foo    two  5  1  3

You can also specify only some columns used for aggregation in a list after groupby function:

df3 = df.groupby(['A', 'B'], as_index=False)['C','D'].sum()
print (df3)
     A      B  C  D
0  bar  three  2  1
1  bar    two  3  1
2  foo    one  4  4
3  foo    two  5  1

Same results by using function DataFrameGroupBy.agg:

df1 = df.groupby(['A', 'B'], as_index=False)['C'].agg('sum')
print (df1)
     A      B  C
0  bar  three  2
1  bar    two  3
2  foo    one  4
3  foo    two  5

df2 = df.groupby(['A', 'B'], as_index=False).agg('sum')
print (df2)
     A      B  C  D  E
0  bar  three  2  1  1
1  bar    two  3  1  4
2  foo    one  4  4  0
3  foo    two  5  1  3

For multiple functions applied for one column use a list of tuples - names of new columns and aggregated functions:

df4 = (df.groupby(['A', 'B'])['C']
         .agg([('average','mean'),('total','sum')])
         .reset_index())
print (df4)
     A      B  average  total
0  bar  three      2.0      2
1  bar    two      3.0      3
2  foo    one      2.0      4
3  foo    two      2.5      5

If want to pass multiple functions is possible pass list of tuples:

df5 = (df.groupby(['A', 'B'])
         .agg([('average','mean'),('total','sum')]))

print (df5)
                C             D             E      
          average total average total average total
A   B                                              
bar three     2.0     2     1.0     1     1.0     1
    two       3.0     3     1.0     1     4.0     4
foo one       2.0     4     2.0     4     0.0     0
    two       2.5     5     0.5     1     1.5     3

    

Then get MultiIndex in columns:

print (df5.columns)
MultiIndex(levels=[['C', 'D', 'E'], ['average', 'total']],
           labels=[[0, 0, 1, 1, 2, 2], [0, 1, 0, 1, 0, 1]])
           

And for converting to columns, flattening MultiIndex use map with join:

df5.columns = df5.columns.map('_'.join)
df5 = df5.reset_index()
print (df5)
     A      B  C_average  C_total  D_average  D_total  E_average  E_total
0  bar  three        2.0        2        1.0        1        1.0        1
1  bar    two        3.0        3        1.0        1        4.0        4
2  foo    one        2.0        4        2.0        4        0.0        0
3  foo    two        2.5        5        0.5        1        1.5        3

Another solution is pass list of aggregate functions, then flatten MultiIndex and for another columns names use str.replace:

df5 = df.groupby(['A', 'B']).agg(['mean','sum'])
    
df5.columns = (df5.columns.map('_'.join)
                  .str.replace('sum','total')
                  .str.replace('mean','average'))
df5 = df5.reset_index()
print (df5)
     A      B  C_average  C_total  D_average  D_total  E_average  E_total
0  bar  three        2.0        2        1.0        1        1.0        1
1  bar    two        3.0        3        1.0        1        4.0        4
2  foo    one        2.0        4        2.0        4        0.0        0
3  foo    two        2.5        5        0.5        1        1.5        3

If want specified each column with aggregated function separately pass dictionary:

df6 = (df.groupby(['A', 'B'], as_index=False)
         .agg({'C':'sum','D':'mean'})
         .rename(columns={'C':'C_total', 'D':'D_average'}))
print (df6)
     A      B  C_total  D_average
0  bar  three        2        1.0
1  bar    two        3        1.0
2  foo    one        4        2.0
3  foo    two        5        0.5

You can pass custom function too:

def func(x):
    return x.iat[0] + x.iat[-1]

df7 = (df.groupby(['A', 'B'], as_index=False)
         .agg({'C':'sum','D': func})
         .rename(columns={'C':'C_total', 'D':'D_sum_first_and_last'}))
print (df7)
     A      B  C_total  D_sum_first_and_last
0  bar  three        2                     2
1  bar    two        3                     2
2  foo    one        4                     4
3  foo    two        5                     1

Question 2

No DataFrame after aggregation! What happened?

Aggregation by 2 or more columns:

df1 = df.groupby(['A', 'B'])['C'].sum()
print (df1)
A    B    
bar  three    2
     two      3
foo  one      4
     two      5
Name: C, dtype: int32

First check Index and type of pandas object:

print (df1.index)
MultiIndex(levels=[['bar', 'foo'], ['one', 'three', 'two']],
           labels=[[0, 0, 1, 1], [1, 2, 0, 2]],
           names=['A', 'B'])

print (type(df1))
<class 'pandas.core.series.Series'>

There are 2 solutions how get MultiIndex Series to columns:

  • add parameter as_index=False
df1 = df.groupby(['A', 'B'], as_index=False)['C'].sum()
print (df1)
     A      B  C
0  bar  three  2
1  bar    two  3
2  foo    one  4
3  foo    two  5
  • use Series.reset_index:
df1 = df.groupby(['A', 'B'])['C'].sum().reset_index()
print (df1)
     A      B  C
0  bar  three  2
1  bar    two  3
2  foo    one  4
3  foo    two  5

If group by one column:

df2 = df.groupby('A')['C'].sum()
print (df2)
A
bar    5
foo    9
Name: C, dtype: int32

... get Series with Index:

print (df2.index)
Index(['bar', 'foo'], dtype='object', name='A')

print (type(df2))
<class 'pandas.core.series.Series'>

And solution is same like in MultiIndex Series:

df2 = df.groupby('A', as_index=False)['C'].sum()
print (df2)
     A  C
0  bar  5
1  foo  9

df2 = df.groupby('A')['C'].sum().reset_index()
print (df2)
     A  C
0  bar  5
1  foo  9

Question 3

How to aggregate mainly strings columns (to lists, tuples, strings with separator)?

df = pd.DataFrame({'A' : ['a', 'c', 'b', 'b', 'a', 'c', 'b'],
                   'B' : ['one', 'two', 'three','two', 'two', 'one', 'three'],
                   'C' : ['three', 'one', 'two', 'two', 'three','two', 'one'],
                   'D' : [1,2,3,2,3,1,2]})
print (df)
   A      B      C  D
0  a    one  three  1
1  c    two    one  2
2  b  three    two  3
3  b    two    two  2
4  a    two  three  3
5  c    one    two  1
6  b  three    one  2

Instead of an aggregetion function it is possible to pass list, tuple, set for converting column:

df1 = df.groupby('A')['B'].agg(list).reset_index()
print (df1)
   A                    B
0  a           [one, two]
1  b  [three, two, three]
2  c           [two, one]

Alternative is use GroupBy.apply:

df1 = df.groupby('A')['B'].apply(list).reset_index()
print (df1)
   A                    B
0  a           [one, two]
1  b  [three, two, three]
2  c           [two, one]

For converting to strings with separator use .join only if string column:

df2 = df.groupby('A')['B'].agg(','.join).reset_index()
print (df2)
   A                B
0  a          one,two
1  b  three,two,three
2  c          two,one

If numeric column use lambda function with astype for converting to strings:

df3 = (df.groupby('A')['D']
         .agg(lambda x: ','.join(x.astype(str)))
         .reset_index())
print (df3)
   A      D
0  a    1,3
1  b  3,2,2
2  c    2,1

Another solution is converting to strings before groupby:

df3 = (df.assign(D = df['D'].astype(str))
         .groupby('A')['D']
         .agg(','.join).reset_index())
print (df3)
   A      D
0  a    1,3
1  b  3,2,2
2  c    2,1

For converting all columns pass no list of column(s) after groupby. There is no column D because automatic exclusion of 'nuisance' columns, it means all numeric columns are excluded.

df4 = df.groupby('A').agg(','.join).reset_index()
print (df4)
   A                B            C
0  a          one,two  three,three
1  b  three,two,three  two,two,one
2  c          two,one      one,two

So it's necessary to convert all columns into strings, then get all columns:

df5 = (df.groupby('A')
         .agg(lambda x: ','.join(x.astype(str)))
         .reset_index())
print (df5)
   A                B            C      D
0  a          one,two  three,three    1,3
1  b  three,two,three  two,two,one  3,2,2
2  c          two,one      one,two    2,1

Question 4

How to aggregate counts?

df = pd.DataFrame({'A' : ['a', 'c', 'b', 'b', 'a', 'c', 'b'],
                   'B' : ['one', 'two', 'three','two', 'two', 'one', 'three'],
                   'C' : ['three', np.nan, np.nan, 'two', 'three','two', 'one'],
                   'D' : [np.nan,2,3,2,3,np.nan,2]})
print (df)
   A      B      C    D
0  a    one  three  NaN
1  c    two    NaN  2.0
2  b  three    NaN  3.0
3  b    two    two  2.0
4  a    two  three  3.0
5  c    one    two  NaN
6  b  three    one  2.0

Function GroupBy.size for size of each group:

df1 = df.groupby('A').size().reset_index(name='COUNT')
print (df1)
   A  COUNT
0  a      2
1  b      3
2  c      2

Function GroupBy.count exclude missing values:

df2 = df.groupby('A')['C'].count().reset_index(name='COUNT')
print (df2)
   A  COUNT
0  a      2
1  b      2
2  c      1

Function should be used fo multiple columns for count non missing values:

df3 = df.groupby('A').count().add_suffix('_COUNT').reset_index()
print (df3)
   A  B_COUNT  C_COUNT  D_COUNT
0  a        2        2        1
1  b        3        2        3
2  c        2        1        1

Related function Series.value_counts return size object containing counts of unique values in descending order so that the first element is the most frequently-occurring element. Excludes NaNs values by default.

df4 = (df['A'].value_counts()
              .rename_axis('A')
              .reset_index(name='COUNT'))
print (df4)
   A  COUNT
0  b      3
1  a      2
2  c      2

If you want same output like using function groupby + size add Series.sort_index:

df5 = (df['A'].value_counts()
              .sort_index()
              .rename_axis('A')
              .reset_index(name='COUNT'))
print (df5)
   A  COUNT
0  a      2
1  b      3
2  c      2

Question 5

How to create new column filled by aggregated values?

Method GroupBy.transform returns an object that is indexed the same (same size) as the one being grouped

Pandas documentation for more information.

np.random.seed(123)

df = pd.DataFrame({'A' : ['foo', 'foo', 'bar', 'foo', 'bar', 'foo'],
                    'B' : ['one', 'two', 'three','two', 'two', 'one'],
                    'C' : np.random.randint(5, size=6),
                    'D' : np.random.randint(5, size=6)})
print (df)
     A      B  C  D
0  foo    one  2  3
1  foo    two  4  1
2  bar  three  2  1
3  foo    two  1  0
4  bar    two  3  1
5  foo    one  2  1


df['C1'] = df.groupby('A')['C'].transform('sum')
df['C2'] = df.groupby(['A','B'])['C'].transform('sum')


df[['C3','D3']] = df.groupby('A')['C','D'].transform('sum')
df[['C4','D4']] = df.groupby(['A','B'])['C','D'].transform('sum')

print (df)

     A      B  C  D  C1  C2  C3  D3  C4  D4
0  foo    one  2  3   9   4   9   5   4   4
1  foo    two  4  1   9   5   9   5   5   1
2  bar  three  2  1   5   2   5   2   2   1
3  foo    two  1  0   9   5   9   5   5   1
4  bar    two  3  1   5   3   5   2   3   1
5  foo    one  2  1   9   4   9   5   4   4
Tuesday, June 1, 2021
 
jsuissa
answered 7 Months ago
43

Look at this variants. The first is Andrews' curves and the second is a multiline plot which are grouped by one column Month. The dataframe data includes three columns Temperature, Day, and Month:

import pandas as pd
import statsmodels.api as sm
import matplotlib.pylab as plt
from pandas.tools.plotting import andrews_curves

data = sm.datasets.get_rdataset('airquality').data
fig, (ax1, ax2) = plt.subplots(nrows = 2, ncols = 1)
data = data[data.columns.tolist()[3:]] # use only Temp, Month, Day

# Andrews' curves
andrews_curves(data, 'Month', ax=ax1)

# multiline plot with group by
for key, grp in data.groupby(['Month']): 
    ax2.plot(grp['Day'], grp['Temp'], label = "Temp in {0:02d}".format(key))
plt.legend(loc='best')    
plt.show()

When you plot Andrews' curve your data salvaged to one function. It means that Andrews' curves that are represented by functions close together suggest that the corresponding data points will also be close together.

enter image description here

Wednesday, August 4, 2021
 
peixotorms
answered 5 Months ago
37

You can use iloc (to get by position):

df.iloc[3,4]

I recommend reading the indexing section of the docs.

Monday, August 9, 2021
 
Sunny Shah
answered 4 Months ago
39

You could do something like the following:

target_value = 15
df['max_duration'] = df.groupby('Date')['Duration'].transform('max')
df.query('max_duration == Duration')
  .assign(dist=lambda df: np.abs(df['Value'] - target_value))
  .assign(min_dist=lambda df: df.groupby('Date')['dist'].transform('min'))
  .query('min_dist == dist')
  .loc[:, ['Date', 'ID']

Results:

        Date ID
4   1/1/2018  e
11  1/2/2018  e
Saturday, August 28, 2021
 
Carlo Pellegrini
answered 4 Months ago
52

Use isin for compare columns and groupby by columns with aggregate sum, last cast to int and reset_index for columns from MultiIndex:

a = (df['material1'].isin(df['material2']))
df = a.groupby([df['claim'], df['event']]).sum().astype(int).reset_index(name='matches')

Solution with assign to new column:

df['matches'] = df['material1'].isin(df['material2']).astype(int)
df = df.groupby(['claim', 'event'])['matches'].sum().reset_index()

Solutions by @Wen, thank you:

df['matches'] = df['material1'].isin(df['material2']).astype(int)
df = df.groupby(['claim', 'event'], as_index=False)['matches'].sum()

I think it should be slowier in larger DataFrames:

df = (df.groupby(['claim', 'event'])
                  .apply(lambda x : x['material1'].isin(x['material2']).astype(int).sum())
                  .reset_index(name='matches'))

print (df)
  claim event  matches
0     A     X        3
1     A     Y        1
2     B     Z        0
Monday, October 18, 2021
 
pirtle
answered 2 Months ago
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