Applying dictionary comprehension to a defaultdict

Question

I have the following, which parses three columns of tabular data (an openpyxl worksheet) into a defaultdict.

def campaigns_and_adsets_and_pageviews_from_ga(ourTab):
    d = defaultdict(lambda: defaultdict(int))
    for row in ourTab.rows[1:-1]:
        if ('Facebook' in row[0].value) and ('(not set)' not in row[2].value):
            d[row[1].value][row[2].value] += row[4].value
    return d

The output of which looks like the following:

In [790]: campaigns_and_adsets_and_pageviews_from_ga(ourTab)
Out[790]: 
defaultdict(<function __main__.<lambda>>,
            {u'XXX 20160314': defaultdict(int,
                         {u'Carnival desktopfeed': 2.0,
                          u'Carnival mobilefeed': 588.0,
                          u'PYS Broad desktopfeed': 371.0,
                          u'PYS Broad mobilefeed': 1192.0}),
             u'YYY Intl 20150903': defaultdict(int,
                         {u'CA desktopfeed': 2.0}),

What I want to do is multiply the final value in each element (i.e. 2.0, 588.0 etc.) by a constant, resulting in another defaultdict (or even a regular nested dict would be fine).

Can the defaultdict be deconstructed somehow back into a nested dict in order to allow the transformation to be possible? Or what other approaches are possible?

You need `.items()` so you're actually working through the entries. — g.d.d.c, May 19 '16 at 22:42
In view of two bugs being spotted immediately, I suggest you first bodge/mock your `campaigns_and_adsets_and_pageviews_from_ga` function to return just a dictionary literal defining the nested dictionaries (never mind `defaultdict` yet): `return {u'XXX 20160314' : {u'Carnival desktopfeed': 2.0, ... }}`. Then write code that correctly copies it and doubles the numbers. Then switch back to returning the `defaultdict` with the computed results. That way you won't be distracted by assuming that everything which goes wrong must be to do with `defaultdict` :-) — Steve Jessop, May 19 '16 at 22:43

score 4 · Accepted Answer · answered May 19 '16 at 23:33

You could use a simple recursive function that multiplies the numbers with given value and for every dict instance constructs a new dictionary:

from numbers import Number

def multiply(o, mul):
    if isinstance(o, dict):
        return {k: multiply(v, mul) for k, v in o.items()}
    elif isinstance(o, Number):
        return o * mul
    else:
        return o

Given your example defaultdict and multiplier 2 as input the output looks following:

{
    u'YYY Intl 20150903': {u'CA desktopfeed': 4.0}, 
    u'XXX 20160314': {
        u'Carnival desktopfeed': 4.0, 
        u'PYS Broad desktopfeed': 742.0, 
        u'PYS Broad mobilefeed': 2384.0, 
        u'Carnival mobilefeed': 1176.0
    }
}

Note that the example doesn't work with lists, for those you'd need to add some more code.

Now *that's* thinking outside of the box. Was not aware of the `numbers` module. Thank you on both fronts. — Pyderman, May 20 '16 at 00:01
@Pyderman: Had to search it too after I figured that there needs to be an easier way than handling `float` and `int` separately. — niemmi, May 20 '16 at 00:12

Applying dictionary comprehension to a defaultdict

1 Answers1