Python: Tuples/dictionaries as keys, select, sort

Question

Suppose I have quantities of fruits of different colors, e.g., 24 blue bananas, 12 green apples, 0 blue strawberries and so on. I\'d like to organize them in a data structur

Answer 1

Database, dict of dicts, dictionary of list of dictionaries, named tuple (it's a subclass), sqlite, redundancy... I didn't believe my eyes. What else ?

"It might well be that dictionaries with tuples as keys are not the proper way to handle this situation."

"my gut feeling is that a database is overkill for the OP's needs; "

Yeah! I thought

So, in my opinion, a list of tuples is plenty enough :

from operator import itemgetter

li = [  ('banana',     'blue'   , 24) ,
        ('apple',      'green'  , 12) ,
        ('strawberry', 'blue'   , 16 ) ,
        ('banana',     'yellow' , 13) ,
        ('apple',      'gold'   , 3 ) ,
        ('pear',       'yellow' , 10) ,
        ('strawberry', 'orange' , 27) ,
        ('apple',      'blue'   , 21) ,
        ('apple',      'silver' , 0 ) ,
        ('strawberry', 'green'  , 4 ) ,
        ('banana',     'brown'  , 14) ,
        ('strawberry', 'yellow' , 31) ,
        ('apple',      'pink'   , 9 ) ,
        ('strawberry', 'gold'   , 0 ) ,
        ('pear',       'gold'   , 66) ,
        ('apple',      'yellow' , 9 ) ,
        ('pear',       'brown'  , 5 ) ,
        ('strawberry', 'pink'   , 8 ) ,
        ('apple',      'purple' , 7 ) ,
        ('pear',       'blue'   , 51) ,
        ('chesnut',    'yellow',  0 )   ]


print set( u[1] for u in li ),': all potential colors'
print set( c for f,c,n in li if n!=0),': all effective colors'
print [ c for f,c,n in li if f=='banana' ],': all potential colors of bananas'
print [ c for f,c,n in li if f=='banana' and n!=0],': all effective colors of bananas'
print

print set( u[0] for u in li ),': all potential fruits'
print set( f for f,c,n in li if n!=0),': all effective fruits'
print [ f for f,c,n in li if c=='yellow' ],': all potential fruits being yellow'
print [ f for f,c,n in li if c=='yellow' and n!=0],': all effective fruits being yellow'
print

print len(set( u[1] for u in li )),': number of all potential colors'
print len(set(c for f,c,n in li if n!=0)),': number of all effective colors'
print len( [c for f,c,n in li if f=='strawberry']),': number of potential colors of strawberry'
print len( [c for f,c,n in li if f=='strawberry' and n!=0]),': number of effective colors of strawberry'
print

# sorting li by name of fruit
print sorted(li),'  sorted li by name of fruit'
print

# sorting li by number 
print sorted(li, key = itemgetter(2)),'  sorted li by number'
print

# sorting li first by name of color and secondly by name of fruit
print sorted(li, key = itemgetter(1,0)),'  sorted li first by name of color and secondly by name of fruit'
print

result

set(['blue', 'brown', 'gold', 'purple', 'yellow', 'pink', 'green', 'orange', 'silver']) : all potential colors
set(['blue', 'brown', 'gold', 'purple', 'yellow', 'pink', 'green', 'orange']) : all effective colors
['blue', 'yellow', 'brown'] : all potential colors of bananas
['blue', 'yellow', 'brown'] : all effective colors of bananas

set(['strawberry', 'chesnut', 'pear', 'banana', 'apple']) : all potential fruits
set(['strawberry', 'pear', 'banana', 'apple']) : all effective fruits
['banana', 'pear', 'strawberry', 'apple', 'chesnut'] : all potential fruits being yellow
['banana', 'pear', 'strawberry', 'apple'] : all effective fruits being yellow

9 : number of all potential colors
8 : number of all effective colors
6 : number of potential colors of strawberry
5 : number of effective colors of strawberry

[('apple', 'blue', 21), ('apple', 'gold', 3), ('apple', 'green', 12), ('apple', 'pink', 9), ('apple', 'purple', 7), ('apple', 'silver', 0), ('apple', 'yellow', 9), ('banana', 'blue', 24), ('banana', 'brown', 14), ('banana', 'yellow', 13), ('chesnut', 'yellow', 0), ('pear', 'blue', 51), ('pear', 'brown', 5), ('pear', 'gold', 66), ('pear', 'yellow', 10), ('strawberry', 'blue', 16), ('strawberry', 'gold', 0), ('strawberry', 'green', 4), ('strawberry', 'orange', 27), ('strawberry', 'pink', 8), ('strawberry', 'yellow', 31)]   sorted li by name of fruit

[('apple', 'silver', 0), ('strawberry', 'gold', 0), ('chesnut', 'yellow', 0), ('apple', 'gold', 3), ('strawberry', 'green', 4), ('pear', 'brown', 5), ('apple', 'purple', 7), ('strawberry', 'pink', 8), ('apple', 'pink', 9), ('apple', 'yellow', 9), ('pear', 'yellow', 10), ('apple', 'green', 12), ('banana', 'yellow', 13), ('banana', 'brown', 14), ('strawberry', 'blue', 16), ('apple', 'blue', 21), ('banana', 'blue', 24), ('strawberry', 'orange', 27), ('strawberry', 'yellow', 31), ('pear', 'blue', 51), ('pear', 'gold', 66)]   sorted li by number

[('apple', 'blue', 21), ('banana', 'blue', 24), ('pear', 'blue', 51), ('strawberry', 'blue', 16), ('banana', 'brown', 14), ('pear', 'brown', 5), ('apple', 'gold', 3), ('pear', 'gold', 66), ('strawberry', 'gold', 0), ('apple', 'green', 12), ('strawberry', 'green', 4), ('strawberry', 'orange', 27), ('apple', 'pink', 9), ('strawberry', 'pink', 8), ('apple', 'purple', 7), ('apple', 'silver', 0), ('apple', 'yellow', 9), ('banana', 'yellow', 13), ('chesnut', 'yellow', 0), ('pear', 'yellow', 10), ('strawberry', 'yellow', 31)]   sorted li first by name of color and secondly by name of fruit

Answer 2

Personally, one of the things I love about python is the tuple-dict combination. What you have here is effectively a 2d array (where x = fruit name and y = color), and I am generally a supporter of the dict of tuples for implementing 2d arrays, at least when something like numpy or a database isn't more appropriate. So in short, I think you've got a good approach.

Note that you can't use dicts as keys in a dict without doing some extra work, so that's not a very good solution.

That said, you should also consider namedtuple(). That way you could do this:

>>> from collections import namedtuple
>>> Fruit = namedtuple("Fruit", ["name", "color"])
>>> f = Fruit(name="banana", color="red")
>>> print f
Fruit(name='banana', color='red')
>>> f.name
'banana'
>>> f.color
'red'

Now you can use your fruitcount dict:

>>> fruitcount = {Fruit("banana", "red"):5}
>>> fruitcount[f]
5

Other tricks:

>>> fruits = fruitcount.keys()
>>> fruits.sort()
>>> print fruits
[Fruit(name='apple', color='green'), 
 Fruit(name='apple', color='red'), 
 Fruit(name='banana', color='blue'), 
 Fruit(name='strawberry', color='blue')]
>>> fruits.sort(key=lambda x:x.color)
>>> print fruits
[Fruit(name='banana', color='blue'), 
 Fruit(name='strawberry', color='blue'), 
 Fruit(name='apple', color='green'), 
 Fruit(name='apple', color='red')]

Echoing chmullig, to get a list of all colors of one fruit, you would have to filter the keys, i.e.

bananas = [fruit for fruit in fruits if fruit.name=='banana']

Answer 3

You want to use two keys independently, so you have two choices:

1. Store the data redundantly with two dicts as {'banana' : {'blue' : 4, ...}, .... } and {'blue': {'banana':4, ...} ...}. Then, searching and sorting is easy but you have to make sure you modify the dicts together.

2. Store it just one dict, and then write functions that iterate over them eg.:

   d = {'banana' : {'blue' : 4, 'yellow':6}, 'apple':{'red':1} }
   
   blueFruit = [(fruit,d[fruit]['blue']) if d[fruit].has_key('blue') for fruit in d.keys()]
   

Answer 4

Your best option will be to create a simple data structure to model what you have. Then you can store these objects in a simple list and sort/retrieve them any way you wish.

For this case, I'd use the following class:

class Fruit:
    def __init__(self, name, color, quantity): 
        self.name = name
        self.color = color
        self.quantity = quantity

    def __str__(self):
        return "Name: %s, Color: %s, Quantity: %s" % \
     (self.name, self.color, self.quantity)

Then you can simply construct "Fruit" instances and add them to a list, as shown in the following manner:

fruit1 = Fruit("apple", "red", 12)
fruit2 = Fruit("pear", "green", 22)
fruit3 = Fruit("banana", "yellow", 32)
fruits = [fruit3, fruit2, fruit1] 

The simple list fruits will be much easier, less confusing, and better-maintained.

Some examples of use:

All outputs below is the result after running the given code snippet followed by:

for fruit in fruits:
    print fruit

Unsorted list:

Displays:

Name: banana, Color: yellow, Quantity: 32
Name: pear, Color: green, Quantity: 22
Name: apple, Color: red, Quantity: 12

Sorted alphabetically by name:

fruits.sort(key=lambda x: x.name.lower())

Displays:

Name: apple, Color: red, Quantity: 12
Name: banana, Color: yellow, Quantity: 32
Name: pear, Color: green, Quantity: 22

Sorted by quantity:

fruits.sort(key=lambda x: x.quantity)

Displays:

Name: apple, Color: red, Quantity: 12
Name: pear, Color: green, Quantity: 22
Name: banana, Color: yellow, Quantity: 32

Where color == red:

red_fruit = filter(lambda f: f.color == "red", fruits)

Displays:

Name: apple, Color: red, Quantity: 12

Answer 5

A dictionary probably isn't what you should be using in this case. A more full featured library would be a better alternative. Probably a real database. The easiest would be sqlite. You can keep the whole thing in memory by passing in the string ':memory:' instead of a filename.

If you do want to continue down this path, you can do it with the extra attributes in the key or the value. However a dictionary can't be the key to a another dictionary, but a tuple can. The docs explain what's allowable. It must be an immutable object, which includes strings, numbers and tuples that contain only strings and numbers (and more tuples containing only those types recursively...).

You could do your first example with d = {('apple', 'red') : 4}, but it'll be very hard to query for what you want. You'd need to do something like this:

#find all apples
apples = [d[key] for key in d.keys() if key[0] == 'apple']

#find all red items
red = [d[key] for key in d.keys() if key[1] == 'red']

#the red apple
redapples = d[('apple', 'red')]

Answer 6

This type of data is efficiently pulled from a Trie-like data structure. It also allows for fast sorting. The memory efficiency might not be that great though.

A traditional trie stores each letter of a word as a node in the tree. But in your case your "alphabet" is different. You are storing strings instead of characters.

it might look something like this:

root:                Root
                     /|\
                    / | \
                   /  |  \     
fruit:       Banana Apple Strawberry
              / |      |     \
             /  |      |      \
color:     Blue Yellow Green  Blue
            /   |       |       \
           /    |       |        \
end:      24   100      12        0

see this link: trie in python