CODEWITHSUNDEEP
pythonstack
Shilpa
Shilpa

Reputation: 1925

dictionary and stack in python problem

I have made a dictionary and I put the keys of the dict in a list. My list contains elements like this:

s =  [((5, 4), 'South', 1), ((4, 5), 'West', 1)]

I made a dict from this:

child = dict((t[0], t[1]) for t in s)

keys = child.keys()
print keys

The output is : [(4, 5), (5, 4)]

Now I need to put (4,5) and (5,4) into stack. What should I do?

I tried, but when I do pop from the stack it is giving me the 2 elements together. like stack.pop() - output is : [(4, 5), (5, 4)]. I want to pop one by one... (4,5) and then (5,4)

Upvotes: 1

Views: 329

Answers (3)

Mark Byers
Mark Byers

Reputation: 838336

You can use a list as a stack:

stack = list(child.keys())
print stack.pop()
print stack.pop()

Result:

(5, 4)
(4, 5)

Important note: the keys of a dictionary are not ordered so if you want the items in a specific order you need to handle that yourself. For example if you want them in normal sorted order you could use sorted. If you want them to pop off in reverse order from the order they were in s you could skip the conversion to a dictionary and just go straight from s to your stack:

stack = [x[0] for x in s]
print stack.pop()
print stack.pop()

Result:

(4, 5)
(5, 4)

Upvotes: 2

eruciform
eruciform

Reputation: 7736

I think user wants this:

# python
> stack = [(4, 5), (5, 4)]
> stack.pop(0)
(4,5)
> stack.pop(0)
(5,4)

Just a reminder, though, this is not a proper stack. This is a proper stack:

# python
> stack=[]
> stack.append( (4,5) )    
> stack.append( (5,4) )
> stack.pop()
(5,4)
> stack.pop()
(4,5)

Upvotes: 0

Quonux
Quonux

Reputation: 2991

use

element = stack[0]
if len(stack) > 0:
   stack = stack[1:]

print element

but it is not that kind of a stack :/

Upvotes: 0

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