List of references to items in a dictionary












0















Is there a way, in python, for me to create a list where each element is a reference to an element in a dictionary? This way, I can maintain a sorted list, while also being able to change the value by indexing into the dictionary.



d = {

  'price_one': 12,

  'price_two': 13,

  'price_three': 5

}



sorted_list = [ptr_to_price_three, ptr_to_price_one, ptr_to_price_two]

print(sorted_list) # [5, 12, 13]



d['price_one'] = 1



sorted_list.sort() # [ptr_to_price_one, ptr_to_price_three, ptr_to_price_two]

print(sorted_list) # [1, 5, 13]





python python-3.x list dictionary pointers







share|improve this question


















  • 1





    Not sure what the actual use case is here, but would it be solved by OrderedDict? docs.python.org/3/library/…

    – JacobIRR
    Jan 3 at 0:11













  • @JacobIRR Interesting. It looks like this could work for me. If I were to change a value in the dictionary, would the order also change?

    – bpgeck
    Jan 3 at 0:15








  • 1





    OrderedDict (and normal dict on python 3.7+ / ipython 3.6+) are insert ordered - the only way to keep them ordered is to create them a-new: d =dict(d.items()) after changing something - which is not really good. Why do you need this and whatfor - how big is the dict? Why not create a def get_ordered_items(d): return sorted(d.values()) funciton?

    – Patrick Artner
    Jan 3 at 0:21













  • @PatrickArtner It seems likes I need to resort after every modification anyway so this may be the best option

    – bpgeck
    Jan 3 at 0:29






  • 1





    Are you trying to efficiently maintain a sorted container? Because then you should just use sorted containers

    – juanpa.arrivillaga
    Jan 3 at 2:33
















0















Is there a way, in python, for me to create a list where each element is a reference to an element in a dictionary? This way, I can maintain a sorted list, while also being able to change the value by indexing into the dictionary.



d = {

  'price_one': 12,

  'price_two': 13,

  'price_three': 5

}



sorted_list = [ptr_to_price_three, ptr_to_price_one, ptr_to_price_two]

print(sorted_list) # [5, 12, 13]



d['price_one'] = 1



sorted_list.sort() # [ptr_to_price_one, ptr_to_price_three, ptr_to_price_two]

print(sorted_list) # [1, 5, 13]





python python-3.x list dictionary pointers







share|improve this question


















  • 1





    Not sure what the actual use case is here, but would it be solved by OrderedDict? docs.python.org/3/library/…

    – JacobIRR
    Jan 3 at 0:11













  • @JacobIRR Interesting. It looks like this could work for me. If I were to change a value in the dictionary, would the order also change?

    – bpgeck
    Jan 3 at 0:15








  • 1





    OrderedDict (and normal dict on python 3.7+ / ipython 3.6+) are insert ordered - the only way to keep them ordered is to create them a-new: d =dict(d.items()) after changing something - which is not really good. Why do you need this and whatfor - how big is the dict? Why not create a def get_ordered_items(d): return sorted(d.values()) funciton?

    – Patrick Artner
    Jan 3 at 0:21













  • @PatrickArtner It seems likes I need to resort after every modification anyway so this may be the best option

    – bpgeck
    Jan 3 at 0:29






  • 1





    Are you trying to efficiently maintain a sorted container? Because then you should just use sorted containers

    – juanpa.arrivillaga
    Jan 3 at 2:33














0












0








0








Is there a way, in python, for me to create a list where each element is a reference to an element in a dictionary? This way, I can maintain a sorted list, while also being able to change the value by indexing into the dictionary.



d = {

  'price_one': 12,

  'price_two': 13,

  'price_three': 5

}



sorted_list = [ptr_to_price_three, ptr_to_price_one, ptr_to_price_two]

print(sorted_list) # [5, 12, 13]



d['price_one'] = 1



sorted_list.sort() # [ptr_to_price_one, ptr_to_price_three, ptr_to_price_two]

print(sorted_list) # [1, 5, 13]





python python-3.x list dictionary pointers







share|improve this question














Is there a way, in python, for me to create a list where each element is a reference to an element in a dictionary? This way, I can maintain a sorted list, while also being able to change the value by indexing into the dictionary.



d = {

  'price_one': 12,

  'price_two': 13,

  'price_three': 5

}



sorted_list = [ptr_to_price_three, ptr_to_price_one, ptr_to_price_two]

print(sorted_list) # [5, 12, 13]



d['price_one'] = 1



sorted_list.sort() # [ptr_to_price_one, ptr_to_price_three, ptr_to_price_two]

print(sorted_list) # [1, 5, 13]





python python-3.x list dictionary pointers




python python-3.x list dictionary pointers






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 3 at 0:08









bpgeckbpgeck

1,160517




1,160517








  • 1





    Not sure what the actual use case is here, but would it be solved by OrderedDict? docs.python.org/3/library/…

    – JacobIRR
    Jan 3 at 0:11













  • @JacobIRR Interesting. It looks like this could work for me. If I were to change a value in the dictionary, would the order also change?

    – bpgeck
    Jan 3 at 0:15








  • 1





    OrderedDict (and normal dict on python 3.7+ / ipython 3.6+) are insert ordered - the only way to keep them ordered is to create them a-new: d =dict(d.items()) after changing something - which is not really good. Why do you need this and whatfor - how big is the dict? Why not create a def get_ordered_items(d): return sorted(d.values()) funciton?

    – Patrick Artner
    Jan 3 at 0:21













  • @PatrickArtner It seems likes I need to resort after every modification anyway so this may be the best option

    – bpgeck
    Jan 3 at 0:29






  • 1





    Are you trying to efficiently maintain a sorted container? Because then you should just use sorted containers

    – juanpa.arrivillaga
    Jan 3 at 2:33














  • 1





    Not sure what the actual use case is here, but would it be solved by OrderedDict? docs.python.org/3/library/…

    – JacobIRR
    Jan 3 at 0:11













  • @JacobIRR Interesting. It looks like this could work for me. If I were to change a value in the dictionary, would the order also change?

    – bpgeck
    Jan 3 at 0:15








  • 1





    OrderedDict (and normal dict on python 3.7+ / ipython 3.6+) are insert ordered - the only way to keep them ordered is to create them a-new: d =dict(d.items()) after changing something - which is not really good. Why do you need this and whatfor - how big is the dict? Why not create a def get_ordered_items(d): return sorted(d.values()) funciton?

    – Patrick Artner
    Jan 3 at 0:21













  • @PatrickArtner It seems likes I need to resort after every modification anyway so this may be the best option

    – bpgeck
    Jan 3 at 0:29






  • 1





    Are you trying to efficiently maintain a sorted container? Because then you should just use sorted containers

    – juanpa.arrivillaga
    Jan 3 at 2:33








1




1





Not sure what the actual use case is here, but would it be solved by OrderedDict? docs.python.org/3/library/…

– JacobIRR
Jan 3 at 0:11







Not sure what the actual use case is here, but would it be solved by OrderedDict? docs.python.org/3/library/…

– JacobIRR
Jan 3 at 0:11















@JacobIRR Interesting. It looks like this could work for me. If I were to change a value in the dictionary, would the order also change?

– bpgeck
Jan 3 at 0:15







@JacobIRR Interesting. It looks like this could work for me. If I were to change a value in the dictionary, would the order also change?

– bpgeck
Jan 3 at 0:15






1




1





OrderedDict (and normal dict on python 3.7+ / ipython 3.6+) are insert ordered - the only way to keep them ordered is to create them a-new: d =dict(d.items()) after changing something - which is not really good. Why do you need this and whatfor - how big is the dict? Why not create a def get_ordered_items(d): return sorted(d.values()) funciton?

– Patrick Artner
Jan 3 at 0:21







OrderedDict (and normal dict on python 3.7+ / ipython 3.6+) are insert ordered - the only way to keep them ordered is to create them a-new: d =dict(d.items()) after changing something - which is not really good. Why do you need this and whatfor - how big is the dict? Why not create a def get_ordered_items(d): return sorted(d.values()) funciton?

– Patrick Artner
Jan 3 at 0:21















@PatrickArtner It seems likes I need to resort after every modification anyway so this may be the best option

– bpgeck
Jan 3 at 0:29





@PatrickArtner It seems likes I need to resort after every modification anyway so this may be the best option

– bpgeck
Jan 3 at 0:29




1




1





Are you trying to efficiently maintain a sorted container? Because then you should just use sorted containers

– juanpa.arrivillaga
Jan 3 at 2:33





Are you trying to efficiently maintain a sorted container? Because then you should just use sorted containers

– juanpa.arrivillaga
Jan 3 at 2:33












3 Answers
3






active

oldest

votes


















1














You mentioned Python in general. You could use a dataframe:



import pandas as pd



d = {

  'price_one': 12,

  'price_two': 13,

  'price_three': 5

}



df = pd.DataFrame(list(d.values()), columns=['val'], index=d.keys())

df.loc['price_one'] = 1

df.sort_values(['val'])


Outputs:



enter image description here






share|improve this answer



















  • 1





    Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

    – ShadowRanger
    Jan 3 at 1:55



















0














Try this:



d = {

    'price_one': 12,

    'price_two': 13,

    'price_three': 5

}



def get_sorted(d):

    sorted_list = sorted(list(d.values()))

    return tuple(zip(sorted_list, d[key] for key in sorted_list))

# end get_sorted



print(get_sorted(d))



d['price_one'] = 1



print(get_sorted(d))


Note that you would have to call get_sorted(d) each time...



Also, it returns a tuple of pairs (key, value) sorted with their keys. If you wanted to access say the 3rd value, do get_sorted(d)[2][1]. The 2 for the third pair, the 1 for the value.






share|improve this answer


























  • I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

    – bpgeck
    Jan 3 at 0:21











  • I've changed it so it's a function instead. It might not be pretty but it works.

    – GeeTransit
    Jan 3 at 0:22











  • Gotcha thanks for the changes

    – bpgeck
    Jan 3 at 0:29











  • No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

    – GeeTransit
    Jan 3 at 0:32



















0














The dictionary keys are references to particular entries in a dictionary.



A function to generate a list of dictionary keys in the order of the dictionary entry values is:



def sorted_keys(d):

    return sorted(d.keys(), key=lambda k: d[k])


so to access the smallest price in your example:



ks = sorted_keys(d)

smallest = d[ ks[0] ]


The problem is that you have to recalculate your ordering on any insert, change or delete of the dictionary. These would be hard to intercept so better to keep control and recalculate the ordering just when you want to use it.



You could bury all this in a class SortedDict(dict) that overrides the get function of a dictionary.






share|improve this answer























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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    You mentioned Python in general. You could use a dataframe:



    import pandas as pd
    

    
d = {
    
  'price_one': 12,
    
  'price_two': 13,
    
  'price_three': 5
    
}
    

    
df = pd.DataFrame(list(d.values()), columns=['val'], index=d.keys())
    
df.loc['price_one'] = 1
    
df.sort_values(['val'])


    Outputs:



    enter image description here






    share|improve this answer



















    • 1





      Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

      – ShadowRanger
      Jan 3 at 1:55
















    1














    You mentioned Python in general. You could use a dataframe:



    import pandas as pd
    

    
d = {
    
  'price_one': 12,
    
  'price_two': 13,
    
  'price_three': 5
    
}
    

    
df = pd.DataFrame(list(d.values()), columns=['val'], index=d.keys())
    
df.loc['price_one'] = 1
    
df.sort_values(['val'])


    Outputs:



    enter image description here






    share|improve this answer



















    • 1





      Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

      – ShadowRanger
      Jan 3 at 1:55














    1












    1








    1







    You mentioned Python in general. You could use a dataframe:



    import pandas as pd
    

    
d = {
    
  'price_one': 12,
    
  'price_two': 13,
    
  'price_three': 5
    
}
    

    
df = pd.DataFrame(list(d.values()), columns=['val'], index=d.keys())
    
df.loc['price_one'] = 1
    
df.sort_values(['val'])


    Outputs:



    enter image description here






    share|improve this answer













    You mentioned Python in general. You could use a dataframe:



    import pandas as pd
    

    
d = {
    
  'price_one': 12,
    
  'price_two': 13,
    
  'price_three': 5
    
}
    

    
df = pd.DataFrame(list(d.values()), columns=['val'], index=d.keys())
    
df.loc['price_one'] = 1
    
df.sort_values(['val'])


    Outputs:



    enter image description here







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Jan 3 at 0:45









    deckarddeckard

    31527




    31527








    • 1





      Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

      – ShadowRanger
      Jan 3 at 1:55














    • 1





      Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

      – ShadowRanger
      Jan 3 at 1:55








    1




    1





    Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

    – ShadowRanger
    Jan 3 at 1:55





    Pulling in pandas (and incredibly heavyweight framework) for this seems a bit much. It's fine if you're using pandas anyway, but I wouldn't use it as a go-to solution in general.

    – ShadowRanger
    Jan 3 at 1:55













    0














    Try this:



    d = {
    
    'price_one': 12,
    
    'price_two': 13,
    
    'price_three': 5
    
}
    

    
def get_sorted(d):
    
    sorted_list = sorted(list(d.values()))
    
    return tuple(zip(sorted_list, d[key] for key in sorted_list))
    
# end get_sorted
    

    
print(get_sorted(d))
    

    
d['price_one'] = 1
    

    
print(get_sorted(d))


    Note that you would have to call get_sorted(d) each time...



    Also, it returns a tuple of pairs (key, value) sorted with their keys. If you wanted to access say the 3rd value, do get_sorted(d)[2][1]. The 2 for the third pair, the 1 for the value.






    share|improve this answer


























    • I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

      – bpgeck
      Jan 3 at 0:21











    • I've changed it so it's a function instead. It might not be pretty but it works.

      – GeeTransit
      Jan 3 at 0:22











    • Gotcha thanks for the changes

      – bpgeck
      Jan 3 at 0:29











    • No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

      – GeeTransit
      Jan 3 at 0:32
















    0














    Try this:



    d = {
    
    'price_one': 12,
    
    'price_two': 13,
    
    'price_three': 5
    
}
    

    
def get_sorted(d):
    
    sorted_list = sorted(list(d.values()))
    
    return tuple(zip(sorted_list, d[key] for key in sorted_list))
    
# end get_sorted
    

    
print(get_sorted(d))
    

    
d['price_one'] = 1
    

    
print(get_sorted(d))


    Note that you would have to call get_sorted(d) each time...



    Also, it returns a tuple of pairs (key, value) sorted with their keys. If you wanted to access say the 3rd value, do get_sorted(d)[2][1]. The 2 for the third pair, the 1 for the value.






    share|improve this answer


























    • I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

      – bpgeck
      Jan 3 at 0:21











    • I've changed it so it's a function instead. It might not be pretty but it works.

      – GeeTransit
      Jan 3 at 0:22











    • Gotcha thanks for the changes

      – bpgeck
      Jan 3 at 0:29











    • No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

      – GeeTransit
      Jan 3 at 0:32














    0












    0








    0







    Try this:



    d = {
    
    'price_one': 12,
    
    'price_two': 13,
    
    'price_three': 5
    
}
    

    
def get_sorted(d):
    
    sorted_list = sorted(list(d.values()))
    
    return tuple(zip(sorted_list, d[key] for key in sorted_list))
    
# end get_sorted
    

    
print(get_sorted(d))
    

    
d['price_one'] = 1
    

    
print(get_sorted(d))


    Note that you would have to call get_sorted(d) each time...



    Also, it returns a tuple of pairs (key, value) sorted with their keys. If you wanted to access say the 3rd value, do get_sorted(d)[2][1]. The 2 for the third pair, the 1 for the value.






    share|improve this answer















    Try this:



    d = {
    
    'price_one': 12,
    
    'price_two': 13,
    
    'price_three': 5
    
}
    

    
def get_sorted(d):
    
    sorted_list = sorted(list(d.values()))
    
    return tuple(zip(sorted_list, d[key] for key in sorted_list))
    
# end get_sorted
    

    
print(get_sorted(d))
    

    
d['price_one'] = 1
    

    
print(get_sorted(d))


    Note that you would have to call get_sorted(d) each time...



    Also, it returns a tuple of pairs (key, value) sorted with their keys. If you wanted to access say the 3rd value, do get_sorted(d)[2][1]. The 2 for the third pair, the 1 for the value.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Jan 3 at 0:21

























    answered Jan 3 at 0:18









    GeeTransitGeeTransit

    694316




    694316













    • I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

      – bpgeck
      Jan 3 at 0:21











    • I've changed it so it's a function instead. It might not be pretty but it works.

      – GeeTransit
      Jan 3 at 0:22











    • Gotcha thanks for the changes

      – bpgeck
      Jan 3 at 0:29











    • No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

      – GeeTransit
      Jan 3 at 0:32



















    • I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

      – bpgeck
      Jan 3 at 0:21











    • I've changed it so it's a function instead. It might not be pretty but it works.

      – GeeTransit
      Jan 3 at 0:22











    • Gotcha thanks for the changes

      – bpgeck
      Jan 3 at 0:29











    • No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

      – GeeTransit
      Jan 3 at 0:32

















    I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

    – bpgeck
    Jan 3 at 0:21





    I'm pretty sure this won't work. When this line is executed d['price_one'] = 1, the value will not change in the list

    – bpgeck
    Jan 3 at 0:21













    I've changed it so it's a function instead. It might not be pretty but it works.

    – GeeTransit
    Jan 3 at 0:22





    I've changed it so it's a function instead. It might not be pretty but it works.

    – GeeTransit
    Jan 3 at 0:22













    Gotcha thanks for the changes

    – bpgeck
    Jan 3 at 0:29





    Gotcha thanks for the changes

    – bpgeck
    Jan 3 at 0:29













    No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

    – GeeTransit
    Jan 3 at 0:32





    No problem. If it works, you can help other programmers like you get this answer when they see this question by pressing the checkmark to mark it as a correct answer. Keep up the great work :D

    – GeeTransit
    Jan 3 at 0:32











    0














    The dictionary keys are references to particular entries in a dictionary.



    A function to generate a list of dictionary keys in the order of the dictionary entry values is:



    def sorted_keys(d):
    
    return sorted(d.keys(), key=lambda k: d[k])


    so to access the smallest price in your example:



    ks = sorted_keys(d)
    
smallest = d[ ks[0] ]


    The problem is that you have to recalculate your ordering on any insert, change or delete of the dictionary. These would be hard to intercept so better to keep control and recalculate the ordering just when you want to use it.



    You could bury all this in a class SortedDict(dict) that overrides the get function of a dictionary.






    share|improve this answer




























      0














      The dictionary keys are references to particular entries in a dictionary.



      A function to generate a list of dictionary keys in the order of the dictionary entry values is:



      def sorted_keys(d):
      
    return sorted(d.keys(), key=lambda k: d[k])


      so to access the smallest price in your example:



      ks = sorted_keys(d)
      
smallest = d[ ks[0] ]


      The problem is that you have to recalculate your ordering on any insert, change or delete of the dictionary. These would be hard to intercept so better to keep control and recalculate the ordering just when you want to use it.



      You could bury all this in a class SortedDict(dict) that overrides the get function of a dictionary.






      share|improve this answer


























        0












        0








        0







        The dictionary keys are references to particular entries in a dictionary.



        A function to generate a list of dictionary keys in the order of the dictionary entry values is:



        def sorted_keys(d):
        
    return sorted(d.keys(), key=lambda k: d[k])


        so to access the smallest price in your example:



        ks = sorted_keys(d)
        
smallest = d[ ks[0] ]


        The problem is that you have to recalculate your ordering on any insert, change or delete of the dictionary. These would be hard to intercept so better to keep control and recalculate the ordering just when you want to use it.



        You could bury all this in a class SortedDict(dict) that overrides the get function of a dictionary.






        share|improve this answer













        The dictionary keys are references to particular entries in a dictionary.



        A function to generate a list of dictionary keys in the order of the dictionary entry values is:



        def sorted_keys(d):
        
    return sorted(d.keys(), key=lambda k: d[k])


        so to access the smallest price in your example:



        ks = sorted_keys(d)
        
smallest = d[ ks[0] ]


        The problem is that you have to recalculate your ordering on any insert, change or delete of the dictionary. These would be hard to intercept so better to keep control and recalculate the ordering just when you want to use it.



        You could bury all this in a class SortedDict(dict) that overrides the get function of a dictionary.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Jan 3 at 1:23









        Mike RobinsMike Robins

        1,315511




        1,315511






























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