根据Python 3.4上键的名称，将字典分为嵌套词典

Question

I have the following dictionary (short version, real data is much larger):

dict = {'C-STD-B&M-SUM:-1': 0, 'C-STD-B&M-SUM:-10': 4.520475, 'H-NSW-BAC-ART:-9': 0.33784000000000003, 'H-NSW-BAC-ART:0': 0, 'H-NSW-BAC-ENG:-59': 0.020309999999999998, 'H-NSW-BAC-ENG:-6': 0,}

I want to divide it into smaller nested dictionaries, depending on a part of the key name.

Expected output would be:

# fixed closing brackets
dict1 = {'C-STD-B&M-SUM: {'-1': 0, '-10': 4.520475}}
dict2 = {'H-NSW-BAC-ART: {'-9': 0.33784000000000003, '0': 0}}
dict3 = {'H-NSW-BAC-ENG: {'-59': 0.020309999999999998, '-6': 0}}

Logic behind is:

dict1: if the part of the key name is 'C-STD-B&M-SUM', add to dict1.
dict2: if the part of the key name is 'H-NSW-BAC-ART', add to dict2.
dict3: if the part of the key name is 'H-NSW-BAC-ENG', add to dict3.

Partial code so far:

def divide_dictionaries(dict):
    c_std_bem_sum = {}
    for k, v in dict.items():
        if k[0:13] == 'C-STD-B&M-SUM':
            c_std_bem_sum = k[14:17], v

What I'm trying to do is to create the nested dictionaries that I need and then I'll create the dictionary and add the nested one to it, but I'm not sure if it's a good way to do it.

When I run the code above, the variable c_std_bem_sum becomes a tuple, with only two values that are changed at each iteration. How can I make it be a dictionary, so I can later create another dictionary, and use this one as the value for one of the keys?

Answer 1

One way to approach it would be to do something like

d = {'C-STD-B&M-SUM:-1': 0, 'C-STD-B&M-SUM:-10': 4.520475, 'H-NSW-BAC-ART:-9': 0.33784000000000003, 'H-NSW-BAC-ART:0': 0, 'H-NSW-BAC-ENG:-59': 0.020309999999999998, 'H-NSW-BAC-ENG:-6': 0,}

def divide_dictionaries(somedict):
    out = {}
    for k,v in somedict.items():
        head, tail = k.split(":")
        subdict = out.setdefault(head, {})
        subdict[tail] = v
    return out

which gives

>>> dnew = divide_dictionaries(d)
>>> import pprint
>>> pprint.pprint(dnew)
{'C-STD-B&M-SUM': {'-1': 0, '-10': 4.520475},
 'H-NSW-BAC-ART': {'-9': 0.33784000000000003, '0': 0},
 'H-NSW-BAC-ENG': {'-59': 0.020309999999999998, '-6': 0}}

A few notes:

(1) We're using nested dictionaries instead of creating separate named dictionaries, which aren't convenient.

(2) We used setdefault , which is a handy way to say "give me the value in the dictionary, but if there isn't one, add this to the dictionary and return it instead.". Saves an if .

(3) We can use .split(":") instead of hardcoding the width, which isn't very robust -- at least assuming that's the delimiter, anyway!

(4) It's a bad idea to use dict , the name of a builtin type, as a variable name.

Answer 2

That's because you're setting your dictionary and overriding it with a tuple:

>>> a = 1, 2
>>> print a
>>> (1,2)

Now for your example:

 >>> def divide_dictionaries(dict):
 >>>      c_std_bem_sum = {}
 >>>      for k, v in dict.items():
 >>>          if k[0:13] == 'C-STD-B&M-SUM':
 >>>              new_key = k[14:17]         # sure you don't want [14:], open ended?
 >>>              c_std_bem_sum[new_key] = v

Basically, this grabs the rest of the key (or 3 characters, as you have it, the [14:None] or [14:] would get the rest of the string) and then uses that as the new key for the dict.