views:

369

answers:

6

I'm trying to convert a Python dictionary into a Python list, in order to perform some calculations.

#My dictionary
dict = {}
dict['Capital']="London"
dict['Food']="Fish&Chips"
dict['2012']="Olympics"

#lists
temp = []
dictList = []

#My attempt:
for key, value in dict.iteritems():
    aKey = key
    aValue = value
    temp.append(aKey)
    temp.append(aValue)
    dictList.append(temp) 
    aKey = ""
    aValue = ""

That's my attempt at it... but I can't work out what's wrong?

+15  A: 
dict.items()

Does the trick.

Björn
I knew there'd be some ridiculous Pythonic way of doing this, thank you very much indeed. I just find the documentation for Python rather hard to read!
day_trader
ridiculous? it's right there where it should be: http://docs.python.org/library/stdtypes.html#dict.items. And you are saying "pythonic" as if it means something bad and possibly perverted
shylent
+5  A: 

Your problem is that you have key and value in quotes making them strings, i.e. you're setting aKey to contain the string "key" and not the value of the variable key. Also, you're not clearing out the temp list, so you're adding to it each time, instead of just having two items in it.

To fix your code, try something like:

for key, value in dict.iteritems():
    temp = [key,value]
    dictlist.append(temp)

You don't need to copy the loop variables key and value into another variable before using them so I dropped them out. Similarly, you don't need to use append to build up a list, you can just specify it between square brackets as shown above. And we could have done dictlist.append([key,value]) if we wanted to be as brief as possible.

Or just use dict.items() as has been suggested.

Dave Webb
Correct answer due to explaining what I was doing wrong. As you can tell, I'm fairly knew to Python and any explanations are greatly appreciated. Cheers!
day_trader
A: 
 >>> a = {'foo': 'bar', 'baz': 'quux', 'hello': 'world'}
 >>> list(reduce(lambda x, y: x + y, a.items()))
 ['foo', 'bar', 'baz', 'quux', 'hello', 'world']
Martin DeMello
+3  A: 

Probably you just want this:

dictList = dict.items()

Your approach has two problems. For one you use key and value in quotes, which are strings with the letters "key" and "value", not related to the variables of that names. Also you keep adding elements to the "temporary" list and never get rid of old elements that are already in it from previous iterations. Make sure you have a new and empty temp list in each iteration and use the key and value variables:

for key, value in dict.iteritems():
    temp = []
    aKey = key
    aValue = value
    temp.append(aKey)
    temp.append(aValue)
    dictList.append(temp)

Also note that this could be written shorter without the temporary variables:

for key, value in dict.iteritems():
    dictList.append([key, value])
sth
+1  A: 

you should use dict.items().

here is a one liner solution for your problem

[(k,v) for k,v in dict.items()]

for the result: [('Food', 'Fish&Chips'), ('2012', 'Olympics'), ('Capital', 'London')] or you can do

l=[]
[l.extend([k,v]) for k,v in dict.items()]

for:

['Food', 'Fish&Chips', '2012', 'Olympics', 'Capital', 'London']

Shay
In python 2.6, at least, you do not need dict.items(): `[(k,v) for k,v in dict]`.
hughdbrown
A: 

If you're making a dictionary only to make a list of tuples, as creating dicts like you are may be a pain, you might look into using zip()

Its especialy useful if you've got one heading, and multiple rows. For instance if I assume that you want Olympics stats for countries:

headers = ['Capital', 'Food', 'Year']
countries = [
    ['London', 'Fish & Chips', '2012'],
    ['Beijing', 'Noodles', '2008'],
]

for olympics in countries:
    print zip(headers, olympics)

gives

[('Capital', 'London'), ('Food', 'Fish & Chips'), ('Year', '2012')]
[('Capital', 'Beijing'), ('Food', 'Noodles'), ('Year', '2008')]

Don't know if thats the end goal, and my be off topic, but it could be something to keep in mind.

Morgan