Python : Adding the same elements of a list iteratively

Question

I have a list of elements as given below-

[(a,1),(b,2),(c,3),(d,4),(e,5),(f,6),(a,7),(b,8),(c,9),(d,10),(e,11),(f,12)]

I am trying to add the value given next to a letter with a different value given next to the same letter.

For example- "a" has a value of 1, the program should compare "a" to all the terms in the list until it finds a match. Once it finds another term "a" having a value of 7, it should add the two values so that we get a=8.

Expected output is-

a=8, b=10, c=12, d=14, e=16, f=18

Citing the rules "Questions asking for homework help must include a summary of the work you've done so far to solve the problem, and a description of the difficulty you are having solving it." You have not done that. — Nearoo
– Nearoo, Commented Oct 11, 2018 at 11:41
@Nearoo Well after seeing the answers below i can see that my code was completely wrong, should i still post it? — SalazarSid
– SalazarSid, Commented Oct 11, 2018 at 12:30

DeepSpace · Accepted Answer · 2018-10-11 11:55:30Z

5

The solution you suggest is going to have a complexity of O(n²).

You can do it in O(n) by using defaultdict (since it requires a single pass over the entire list):

from collections import defaultdict

li = [('a', 1), ('b', 2), ('c', 3), ('d', 4), ('e', 5), ('f', 6), ('a', 7), ('b', 8), 
      ('c', 9), ('d', 10), ('e', 11), ('f', 12)]

output = defaultdict(int)

for letter, number in li:
    output[letter] += number

print(output)
# defaultdict(<class 'int'>, {'a': 8, 'b': 10, 'c': 12, 'd': 14, 'e': 16, 'f': 18})

This solution of course requires the elements to be hashable, but this can be remedied by using the string representation of the elements if they are not.

edited Oct 11, 2018 at 11:55

answered Oct 11, 2018 at 11:39

DeepSpace

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Comments

Melih Taşdizen · Accepted Answer · 2018-10-11 11:48:41Z

1

An answer that doesn't utilize the defaultdict.

_list = [("a",1),("b",2),("c",3),("d",4),("e",5),("f",6),("a",7),("b",8),("c",9),("d",10),("e",11),("f",12)]

tmp = dict()

for k in _list:
    try:
        tmp[k[0]] += k[1]
    except KeyError:
        tmp[k[0]] = k[1]

print(tmp)

Result:

{'a': 8, 'b': 10, 'c': 12, 'd': 14, 'e': 16, 'f': 18}

answered Oct 11, 2018 at 11:48

Melih Taşdizen

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Comments

vash_the_stampede · Accepted Answer · 2018-10-11 16:50:33Z

1

Using itertools.groupby and operator.itemgetter, can be substituted with lambda

from itertools import groupby
from operator import itemgetter

lst = sorted(lst, key=itemgetter(0))
d = {k: sum(i[1] for i in g) for k, g in groupby(lst, key=itemgetter(0))}
# {'a': 8, 'b': 10, 'c': 12, 'd': 14, 'e': 16, 'f': 18}

Dictionary comprehension expanded:

d = {}
for k, g in groupby(lst, key=itemgetter(0)):
    d[k] = sum(i[1] for i in g)

edited Oct 11, 2018 at 16:50

answered Oct 11, 2018 at 15:10

vash_the_stampede

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Comments

Daniel Hepper · Accepted Answer · 2018-10-11 11:51:16Z

0

from collections import defaultdict

l = [('a',1),('b',2),('c',3),('d',4),('e',5),('f',6),('a',7),('b',8),('c',9),('d',10),('e',11),('f',12)]

d = defaultdict(int)
for k, v in l:
    d[k] += v

Result:

defaultdict(<class 'int'>, {'a': 8, 'b': 10, 'c': 12, 'd': 14, 'e': 16, 'f': 18})

edited Oct 11, 2018 at 11:51

answered Oct 11, 2018 at 11:40

Daniel Hepper

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1 Comment

Jay Patel Over a year ago

I think you forgot to initialize d with defaultdict

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