Python: adding value from list of lists

Question

below is my list of lists;

db_rows = [('a','b','c',4),
           ('a','s','f',6),
           ('a','c','d',6),
           ('a','b','f',2),
           ('a','b','c',6),
           ('a','b','f',8),
           ('a','s','f',6),
           ('a','b','f',7),
           ('a','s','f',5),
           ('a','b','f',2)]

if first three values are same in the inner list then I need to add 4th value to create new list

I need result list like this:

final_list = [('a','b','c',10),
              ('a','s','f',17),
              ('a','c','d',6),
              ('a','b','f',19)]

I have tried the below script (not working):

final_list = []
for row in db_rows:
    temp_flag=False
    temp_list = []
    val = 0
    for ref_row in db_rows:
        if row != ref_row:
            if row[0]==ref_row[0] and row[1]==ref_row[1] and row[2]==ref_row[2]:
                val = val + ref_row[3]
                temp_flag=True
    temp_list=(row[0],row[1],row[2],val)
    if temp_flag==False:
        temp_list=row
    final_list.append(temp_list)

please advice me.

Try using a dictionary with the first three values as the key. — Dov Grobgeld
– Dov Grobgeld, Commented Jun 7, 2015 at 10:43
Please correct the desired output. It should be ('a','c','d',6) instead of ('a','b','d',6). — Rahul Gupta
– Rahul Gupta, Commented Jun 7, 2015 at 11:21

falsetru · Accepted Answer · 2015-06-07 10:46:20Z

5

Use a dictionary as Dov Grobgeld commented, then convert the dictionary back to the list.

from collections import defaultdict

db_rows = [('a','b','c',4),
           ('a','s','f',6),
           ('a','c','d',6),
           ('a','b','f',2),
           ('a','b','c',6),
           ('a','b','f',8),
           ('a','s','f',6),
           ('a','b','f',7),
           ('a','s','f',5),
           ('a','b','f',2)]

sums = defaultdict(int)
for row in db_rows:
    sums[row[:3]] += row[3]

final_list = [key + (value,) for key, value in sums.iteritems()]

Printing final_list outputs:

[('a', 'b', 'c', 10), ('a', 's', 'f', 17), ('a', 'b', 'f', 19), ('a', 'c', 'd', 6)]

See collections.defaultdict.

answered Jun 7, 2015 at 10:46

falsetru

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Padraic Cunningham · Accepted Answer · 2015-06-07 10:47:40Z

1

from collections import defaultdict
db_rows = [('a','b','c',4),
           ('a','s','f',6),
           ('a','c','d',6),
           ('a','b','f',2),
           ('a','b','c',6),
           ('a','b','f',8),
           ('a','s','f',6),
           ('a','b','f',7),
           ('a','s','f',5),
           ('a','b','f',2)]

d = defaultdict(int)


for ele in db_rows:
    d[ele[:-1]] += ele[-1]

print([(k +(v,)) for k,v in d.items()])

answered Jun 7, 2015 at 10:47

Padraic Cunningham

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Comments

Rahul Gupta · Accepted Answer · 2015-06-07 11:06:03Z

Here, I am using OrderedDict to preserve the order in which first three values are found.

from collections import OrderedDict

db_rows = [('a','b','c',4),
           ('a','s','f',6),
           ('a','c','d',6),
           ('a','b','f',2),
           ('a','b','c',6),
           ('a','b','f',8),
           ('a','s','f',6),
           ('a','b','f',7),
           ('a','s','f',5),
           ('a','b','f',2)]

temp_dict = OrderedDict()

for x in db_rows:
    temp_dict[x[:3]] = temp_dict.get(x[:3], 0) + x[3]

final_list =  [k +(v,) for k,v in temp_dict.iteritems()]

We can now check value of final_list which gives the desired result in the order as mentioned above.

final_list
[('a', 'b', 'c', 10),
 ('a', 's', 'f', 17),
 ('a', 'c', 'd', 6),
 ('a', 'b', 'f', 19)]

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