CODEWITHSUNDEEP
pythonlisttuplesaverage
gilgameshbk
gilgameshbk

Reputation: 121

How to group tuples by common items and find average per each group

I have a list of tuples named data:

data = [('A', 2), 
        ('B', 2), ('B', 4), ('B', 6), ('B', 8), ('B', 6), ('B', 4), ('B', 3),
        ('C', 10), ('C', 10), ('C', 10),
        ('D', 12),
        ('E', 12),
        ('F', 10), ('F', 8), ('F', 6)]
average = []

I would like an average of the values for each same letter:

Expected Output:

average = [('A', 2), ('B', 5), ('C', 10), ('D', 12), ('E', 12), ('F', 8)]

Upvotes: 2

Views: 770

Answers (3)

Georgy
Georgy

Reputation: 13747

An alternative solution which you might consider, especially when dealing with large data sets, is to use pandas. Here, groupby and mean will do the job:

import pandas as pd

data = [('A', 2), 
        ('B', 2), ('B', 4), ('B', 6), ('B', 8), ('B', 6), ('B', 4), ('B', 3),
        ('C', 10), ('C', 10), ('C', 10),
        ('D', 12),
        ('E', 12),
        ('F', 10), ('F', 8), ('F', 6)]

df = pd.DataFrame(data, columns=['letter', 'number'])
print(df)
#    letter  number
# 0       A       2
# 1       B       2
# 2       B       4
# 3       B       6
# 4       B       8
# 5       B       6
# 6       B       4
# 7       B       3
# 8       C      10
# 9       C      10
# 10      C      10
# 11      D      12
# 12      E      12
# 13      F      10
# 14      F       8
# 15      F       6

print(df.groupby('letter').mean())
#            number
# letter           
# A        2.000000
# B        4.714286
# C       10.000000
# D       12.000000
# E       12.000000
# F        8.000000

print(df.groupby('letter').mean().round().astype(int))
#         number
# letter        
# A            2
# B            5
# C           10
# D           12
# E           12
# F            8

You can get back your list of tuples as follows:

averages = df.groupby('letter').mean().round().astype(int)
result = list(result.to_records())
print(result)
# [('A', 2), ('B', 5), ('C', 10), ('D', 12), ('E', 12), ('F', 8)]

Upvotes: -2

Rahul K P
Rahul K P

Reputation: 16081

Try with groupby

from itertools import groupby
data_ = [(n,[i[1] for i in g]) for n,g in groupby(data, key = lambda x:x[0])]   
result = [(i,float(sum(j))/float(len(j))) for i,j in data_]

Result 

[('A', 2.0),
 ('B', 4.714285714285714),
 ('C', 10.0),
 ('D', 12.0),
 ('E', 12.0),
 ('F', 8.0)]

Upvotes: 2

akuiper
akuiper

Reputation: 215107

Here is an option with a defaultdict:

from collections import defaultdict
avg = defaultdict(lambda :{'count': 0, 'sum': 0})
​
# calculate the sum and count for each key
for k, v in data:
    avg[k]['count'] += 1
    avg[k]['sum'] += v

# calculate the average
[(k, v['sum']/v['count']) for k, v in avg.items()]

#[('A', 2.0),
# ('D', 12.0),
# ('F', 8.0),
# ('E', 12.0),
# ('B', 4.714285714285714),
# ('C', 10.0)]

Upvotes: 4

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