How to map values from a list to nested dictionaries in Python?

Question

I have a list of values, and a dictionary of dictionaries. They look something like this:

d = {1 : {1 : '', 2 : '', 3 : ''},2 : {1 : '', 2 : '', 3 : ''}}
l = ['Erich Martin', 'Zia Michael', 'Olga Williams', 'Uma Gates']

I am trying to map the values from the list into the dictionary, filling each dictionary before moving on to the next nested dictionary. The last dictionary would have some empty slots, and that's fine. I can't seem to wrap my head around what I need to do; I run into the end of the list and get a keyerror, as there are no more values. Here's the gist of what I've got so far:

for g,s in d.items():
        for i in s:
                s[i] = l.pop()

Using Python 3.4.

Thanks!

What does it meap to map the values from the list into the dict? can you provide an example of what you would like to get? — bgusach
– bgusach, Commented Nov 6, 2014 at 20:16
It looks like you really want to chop up your list of names. You can try this: [l[i:i+3] for i in range(0, len(l), 3)]. This will create a list with items grouped 3 at a time. — vikramls
– vikramls, Commented Nov 6, 2014 at 20:19

Steven Mercatante · Accepted Answer · 2014-11-06 21:11:35Z

1

Try this:

edited based on ikaros45's comment

for g,s in d.items():
    for i in s:
        if not l:
            break
        s[i] = l.pop()

This will yield:

{1: {1: 'Uma Gates', 2: 'Olga Williams', 3: 'Zia Michael'}, 2: {1: 'Erich Martin', 2: '', 3: ''}}

edited Nov 6, 2014 at 21:11

answered Nov 6, 2014 at 20:17

Steven Mercatante

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3 Comments

tsz Over a year ago

That did it; I had tried some while len(l) > 0: without success. this makes more sense and is cleaner.

bgusach Over a year ago

In python there is no need to check for length of a list. Just do if not l: break

Steven Mercatante Over a year ago

You're right, @ikaros45. I've updated my answer with your suggestion.

Kevin · Accepted Answer · 2014-11-06 20:16:46Z

0

You are modifying the dictionary as you iterate over it. Fix that first:

for g,s in d.items():
    for i in list(s):
        s[i] = l.pop()

You also need to stop when the list is empty:

try:
    for g,s in d.items():
        for i in list(s):
            s[i] = l.pop()
except IndexError:
    pass
else:
    if l:
        # There weren't enough slots, handle it or raise an exception

answered Nov 6, 2014 at 20:16

Kevin

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Comments

bgusach · Accepted Answer · 2014-11-06 20:36:23Z

I assume you want to put the names into the dict values, replacing the empty string. In that case, I would dump your initial dictionaries and do it this way:

from itertools import count

def generate(lst):
    target = {}
    for index in count(1):
        target[index] = {}
        for subindex in xrange(1, 4):
            target[index][subindex] = lst.pop()
            if not lst:
                return target

generate(['Erich Martin', 'Zia Michael', 'Olga Williams', 'Uma Gates'])

or more elegantly

from itertools import izip_longest

def generate(lst):
    groups = izip_longest(fillvalue='', *([iter(lst)] * 3))
    dictgroups = [dict(enumerate(group, 1)) for group in groups]
    return dict(enumerate(dictgroups, 1))

generate(['Erich Martin', 'Zia Michael', 'Olga Williams', 'Uma Gates'])

Both solutions would work with any input list, no restriction on length as you would have with the approach of mutating an existing dict.

neil · Accepted Answer · 2014-11-06 20:42:44Z

0

A different approach so as to deal with the inherent unsorted nature of a dict. Also, popping from the list seems to give the values in reverse order and destroys the original list (which may be ok if you have no further use for it), so I have used iter to iterate through the list as required.

d = {1 : {1 : '', 2 : '', 3 : ''},2 : {1 : '', 2 : '', 3 : ''}}
l = ['Erich Martin', 'Zia Michael', 'Olga Williams', 'Uma Gates']

i = iter(l)

for outer in sorted(d.keys()):
    for inner in sorted(d[outer].keys()):
        try:
            d[outer][inner]=next(i)
        except StopIteration:
            break

answered Nov 6, 2014 at 20:42

neil

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