Extract results from a dictionary

Question

I have a dictionary as follows:

  D = { "America": { "Washington": { "Seattle": ('park', 'museum'), "Kent": ("market",) }, 'Colorado': { "Boulder": ("hiking",) } } }

how can I make the following results using that dictionary.

America Wahington Seattle park
America Wahington Seattle museum
America Wahington Kent market
America Colorado Boulder hiking

I tried as follows:

for x in D.iteritems(): print x

Could not figure out how to extract each elements after that. Alos wan to to how is the good way of getting the above result.

If it is always nested three levels deep as in your example, you have to use a three level nested `for` loop too. Or use recursion. — Selcuk, Oct 10 '16 at 03:23
You can work into the direction of flattening the dictionary to solve it in a generic fashion: http://stackoverflow.com/questions/6027558/flatten-nested-python-dictionaries-compressing-keys. — alecxe, Oct 10 '16 at 03:24

furas · Accepted Answer · 2016-10-10T03:40:38.530

1

This versions should be more readable for you. But they are not so universal like other versions.

for country in D:
    for state in D[country]:
        for city in D[country][state]:
            for place in D[country][state][city]:
                print(country, state, city, place)


for country, A in D.items():
    for state, B in A.items():
        for city, C in B.items():
            for place in C:
                print(country, state, city, place)

edited Oct 10 '16 at 03:40

answered Oct 10 '16 at 03:36

furas

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+1 for lols... I don't even think to write specific solutions 9/10 times, and honestly this didn't even occur to me :p – kpie Oct 10 '16 at 03:39

score 0 · Answer 2 · edited May 23 '17 at 12:00

Here is a solution almost entirely based on the "recursive"-approach answer for the "flattening the dictionary" problem:

import collections


def flatten(d, parent_key='', sep=' '):
    items = []
    for k, v in d.items():
        new_key = parent_key + sep + k if parent_key else k
        if isinstance(v, collections.MutableMapping):
            items.extend(flatten(v, new_key, sep=sep).items())
        else:
            items.append((new_key, v))
    return dict(items)

Usage:

$ ipython -i test.py
In [1]: D = {"America": {"Washington": {"Seattle": ('park', 'museum'), "Kent": ("market",)},
   ...:                  'Colorado': {"Boulder": ("hiking",)}}}

In [2]: for key, values in flatten(D, sep=" ").items():
   ...:     for value in values:
   ...:         print(key + " " + value)
   ...:         
America Washington Seattle park
America Washington Seattle museum
America Colorado Boulder hiking
America Washington Kent market

One of the biggest advantages of this approach is that it is scalable - it would work for different depths of the dictionary D.

kpie · Answer 3 · 2016-10-10T03:37:48.750

Recursion is your friend...

D = { "America": { "Washington": { "Seattle": ('park', 'museum'), "Kent": ("market",) }, 'Colorado': { "Boulder": ("hiking",) } } }

def printOut(d,s):
    if(type(d)==dict):
        for k in d.keys():
            printOut(d[k],s+" "+str(k)) # add the key to a string and pass the contained dictionary and the new string to the same function. (this is the recursive part...)
    else:
        try:                      # this try catch allows the method to handle iterable and non-iterable leaf nodes in your dictionary.
            for k in d:
                print s+" "+str(k)
        except TypeError:
                print s+" "+str(d)

printOut(D,"")

prints::

 America Washington Seattle park
 America Washington Seattle museum
 America Washington Kent market
 America Colorado Boulder hiking

notice that there is a leading space so this code might fail a test if it looking for a specific output, to eliminate the leading space we would simply add the line s = s[1:] if len(s)>0 else s immediately after the else.

score 0 · Answer 4 · answered Oct 10 '16 at 03:38

I would suggest recursive solution to work with your different kind of data:

def printX(data, prefix=''):
    if isinstance(data, dict):
        for itemKey, itemValue in data.iteritems():
            newPrefix = '{} {}'.format(prefix, itemKey)
            printX(itemValue, newPrefix)
    elif isinstance(data, tuple):
        for itemValue in data:
            printX(itemValue, prefix)
    else:
        print('{} {}'.format(prefix, data))

Did a test as below:

D = {
    'America': { 
        'Washington': {
            'Seattle': ('park', 'museum'),
            'Kent': ('market',) 
        },
        'Colorado': {
            'Boulder': ('hiking',)
        } 
    } 
}
printX(D)

Output is as below

America Washington Seattle park
America Washington Seattle museum
America Washington Kent market
America Colorado Boulder hiking

Extract results from a dictionary

4 Answers4