List Of Lists Of Tuples, Group By First Element And Add Second Elements

Let's say I have the following list of lists of tuples: tuples = [ [ ('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'),

Solution 1:

The solution using itertools.chain.from_iterable, itertools.groupby and sum functions:

import itertools, operator

tuples = [
         [('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
         [('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
         ]

result = [(k, "%.2f" % sum(float(t[1]) for t in g)) 
          for k,g in itertools.groupby(sorted(itertools.chain.from_iterable(tuples)), operator.itemgetter(0))]

print(result)

The output:

[('2017-04-11', '2000000.00'), ('2017-04-12', '1472943.00'), ('2017-04-13', '4000000.00')]

Solution 2:

First, flat a list of tuples out of a list of lists of tuples, and then use itertools.groupby,

import itertools 
import operator

lists = [
         [('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
         [('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
         ]

# Step 1: Flat a list of tuples outof a list of lists of tuples
list_tuples = [t for sublist in lists for t in sublist]
'''
[('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00'), ('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
'''

# Step 2: Groupby
results = list()

for key, groupin itertools.groupby(sorted(list_tuples), operator.itemgetter(0)):
    s =sum(float(t[1]) for t ingroup)
    results.append((key, s))

print(results)
#[('2017-04-11', 2000000.0), ('2017-04-12', 1472943.0), ('2017-04-13', 4000000.0)]

Solution 3:

My way is to convert the nested list to flat list and iterate it:

t = [
         [('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
         [('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
         ]
a={}

for i,j insum(t,[]):
    a[i]=a[i]+float(j) if i in a elsefloat(j)

print(a)

Output:

{'2017-04-11': 2000000.0, '2017-04-13': 4000000.0, '2017-04-12': 1472943.0}

If you want a list, you can use [(k,v) for k,v in a.items()])

Solution 4:

Flattening the list then using defaultdict:

from collections import defaultdict

flattened_tuples = [item for sublist in tuples for item in sublist]

result= defaultdict(float)
fordate, valuein flattened_tuples:
    result[date] +=float(value)
print(result)

returns {'2017-04-11': 2000000.0, '2017-04-12': 1472943.0, '2017-04-13': 4000000.0}