If you are using python 3.5 or greater you can merge your dictionaries using the following syntax:

appointment1 = { 'soccer' : {
                        'day' : 20,
                       'month' : 'april'
                       }

            }
appointment2 = { 'soccer' : {
                        'day' : 20,
                        'month' : 'april'
                       },
              'gym' : {
                        'day' : 5,
                        'month' : 'may'
                       }
            }
appointment = {**appointment1,**appointment2}

In python, append is only for lists, not dictionaries.

This should do what you want:

d['A']['b'] = 3

Explanation: When you write d['A'] you are getting another dictionary (the one whose key is A), and you can then use another set of brackets to add or access entries in the second dictionary.

You told your code to assign newly created dict to key e[0]. It's always replaced blindly and it does not look at previously stored value.

Instead you need something like:

for e in keyList:
    if e[0] not in nested_dict:
        nested_dict[e[0]] = {}
    nested_dict[e[0]].update({e[2] : e[3:]})

If conditional is required to handle 'first key' case. Alternatively defaultdict can be used.

from collections import defaultdict
nested_dict = defaultdict(dict)
for e in keyList:
    nested_dict[e[0]].update({e[2] : e[3:]})

.update is designed to be used with iterable feed. One of possible ways is providing dict, then value of keys in dict_things would be updated to corresponding values in provided dict. That is you might do:

dict_things = {}
for iteration, i_parameter in enumerate(list_of_things):
   dict_things.update({iteration: Function_a(i_parameter, matrix)})

Above example is just for showing how dict.update can be used - this task might be done using dict-comprehension resulting in more readable code.

You can use recursion with collections.defaultdict:

from collections import defaultdict
import re
def merge(*d):
   v = defaultdict(list)
   for i in d:
      for a, b in i.items():
         v[re.sub('^\s+', '', a)].append(b)
   return {a:merge(*b) if all(isinstance(j, dict) for j in b) 
            else [i for j in b for i in (j if isinstance(j, list) else [j])] 
              for a, b in v.items()}

print(merge(grouped1, grouped2))

Output:

{'LabelStat': {'Carrier': ['1', '1', '2', '2'], 'FormID': ['0201', '0430', '10201', '10430']}, 'McAfee': {'DatDate': ['Not Available', 'Available'], 'DatVersion': ['Not Available', 'Available']}}

Expanding on idjaw's comment and keksnicoh's answer, I think you can make your life a little bit easier by using a defaultdict.

>>> from collections import defaultdict
>>> d = defaultdict(lambda: defaultdict(list))
>>> d['s1']['port1'].append([0, 0, 0])
>>> d['s1']['port1'].append([1, 1, 1])
>>> d['s1']['port2'].append([2, 2, 2])
>>> d
defaultdict(<function <lambda> at 0x7f5d217e2b90>, {'s1': defaultdict(<type 'list'>, {'port2': [[2, 2, 2]], 'port1': [[0, 0, 0], [1, 1, 1]]})})

You can use it just like a regular dictionary:

>>> d['s1']
defaultdict(<type 'list'>, {'port2': [[2, 2, 2]], 'port1': [[0, 0, 0], [1, 1, 1]]})
>>> d['s1']['port1']
[[0, 0, 0], [1, 1, 1]]

Provided I've understood your requirements correctly:

In [25]: from collections import defaultdict

In [26]: d = defaultdict(lambda: defaultdict(list))

In [30]: for group, name, obj in [('g1','n1','o1'),('g1','n2','o2'),('g1','n1','o3'),('g2','n1','o4')]:
   ....:     d[group][name].append(obj)

I think what you're trying to do is make a nested dictionary. If I understand your input correctly, your code could be fixed just by setting report_dict[row[0]] to an empty dict, by changing

report_dict[row[0]] = row[3]

to

report_dict[row[0]] = {}

This would make the next line,

report_dict[row[0]][row[3]] = row[1]

add the key row[3] with value row[1] to the dictionary report_dict[row[0]]. This would give us:

report_dict = {row[0]: {row[3]: row[1]}, ...}

which I think is your expected output.

here's solution without defaultdict:

d = {'Spring Savings 0413' : {1 : 3000, 2: 2000, 4:1000}, 
 'Back to School 0812' : {1: 4000, 3:3000, 4:2000}}

r = {}

for s, l in d.items():
    for i in range(1, 5):
        if i not in r: r[i] = {}
        r[i][s] = l.get(i, 0) + r.get(i - 1, {}).get(s, 0)

{1: {'Back to School 0812': 4000, 'Spring Savings 0413': 3000},
 2: {'Back to School 0812': 4000, 'Spring Savings 0413': 5000},
 3: {'Back to School 0812': 7000, 'Spring Savings 0413': 5000},
 4: {'Back to School 0812': 9000, 'Spring Savings 0413': 6000}}