Reverse dictionary where values are listsConvert dictionary of lists of tuples to list of lists/tuples in...

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Reverse dictionary where values are lists


Convert dictionary of lists of tuples to list of lists/tuples in pythonMulti-dictionary API using Python Flask-RESTfulUpdating a dict with dict valuesMatching two lists of dicts, strictly and more looselyCreate dictionary with default immutable keysDictionary of dictionary - Get 2nd keysList all possible permutations from a python dictionary of listsFlatten Dictionary Python challengeRemove the second number from every group of three numbersTransforming a dict of lists into a list of dicts













2












$begingroup$


I'm working on the following problem: I've got a dictionary like this one:



dic={0:[0,1,2],1:[3,4,5]}


And I want to reverse it so it looks like this:



dic2={0:0,1:0,2:0,3:1,4:1,5:1}


I managed to make it, but doing this:



dic2={}
for i in dic:
for j in dic[i]:
dic2[j]=i


I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?










share|improve this question









New contributor




Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    2












    $begingroup$


    I'm working on the following problem: I've got a dictionary like this one:



    dic={0:[0,1,2],1:[3,4,5]}


    And I want to reverse it so it looks like this:



    dic2={0:0,1:0,2:0,3:1,4:1,5:1}


    I managed to make it, but doing this:



    dic2={}
    for i in dic:
    for j in dic[i]:
    dic2[j]=i


    I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?










    share|improve this question









    New contributor




    Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      I'm working on the following problem: I've got a dictionary like this one:



      dic={0:[0,1,2],1:[3,4,5]}


      And I want to reverse it so it looks like this:



      dic2={0:0,1:0,2:0,3:1,4:1,5:1}


      I managed to make it, but doing this:



      dic2={}
      for i in dic:
      for j in dic[i]:
      dic2[j]=i


      I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?










      share|improve this question









      New contributor




      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I'm working on the following problem: I've got a dictionary like this one:



      dic={0:[0,1,2],1:[3,4,5]}


      And I want to reverse it so it looks like this:



      dic2={0:0,1:0,2:0,3:1,4:1,5:1}


      I managed to make it, but doing this:



      dic2={}
      for i in dic:
      for j in dic[i]:
      dic2[j]=i


      I know about list and dict comprehensions and this code reeks of it, but I'm not good at them when there are nested for and dicts. How would you make it more efficiently?







      python dictionary






      share|improve this question









      New contributor




      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 4 hours ago









      Sᴀᴍ Onᴇᴌᴀ

      10.1k62167




      10.1k62167






      New contributor




      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 5 hours ago









      Juan CJuan C

      1185




      1185




      New contributor




      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Juan C is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          There's really not much to say about this code, it is straightforward.



          Style



          These are only nitpicks.




          • Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

          • Collections should be named by their purpose in the application, not their type.

          • By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.


          Code improvements



          When iterating over the keys and values of a dictionary at the same time, you can use



          for k, v in dic.items():
          # ... use k, v ...


          instead of



          for k in dic:
          v = dic[k]
          # ...


          The nested loop can be transformed to a dictionary comprehension like this:



          dic2 = {v: k for k, values in dic.items() for v in values}


          You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.



          Potential pitfalls



          You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:



          >>> dic = {0: [1, 2], 1: [2, 3]}
          >>> {v: k for k, values in dic.items() for v in values}
          {1: 0, 2: 1, 3: 1} # missing 2: 0


          To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.



          If the input represents a directed graph (mapping nodes to lists of neighbours), this corresponds to computing the transposed or reversed graph. It can be done by using a collections.defaultdict. I don't see an easy way to write it as a comprehension in this case.



          from collections import defaultdict

          graph = {0: [1, 2], 1: [2, 3]}

          transposed_graph = defaultdict(list)
          for node, neighbours in graph.items():
          for neighbour in neighbours:
          transposed_graph[neighbour].append(node)

          # {1: [0], 2: [0, 1], 3: [1]}





          share|improve this answer









          $endgroup$














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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            There's really not much to say about this code, it is straightforward.



            Style



            These are only nitpicks.




            • Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

            • Collections should be named by their purpose in the application, not their type.

            • By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.


            Code improvements



            When iterating over the keys and values of a dictionary at the same time, you can use



            for k, v in dic.items():
            # ... use k, v ...


            instead of



            for k in dic:
            v = dic[k]
            # ...


            The nested loop can be transformed to a dictionary comprehension like this:



            dic2 = {v: k for k, values in dic.items() for v in values}


            You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.



            Potential pitfalls



            You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:



            >>> dic = {0: [1, 2], 1: [2, 3]}
            >>> {v: k for k, values in dic.items() for v in values}
            {1: 0, 2: 1, 3: 1} # missing 2: 0


            To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.



            If the input represents a directed graph (mapping nodes to lists of neighbours), this corresponds to computing the transposed or reversed graph. It can be done by using a collections.defaultdict. I don't see an easy way to write it as a comprehension in this case.



            from collections import defaultdict

            graph = {0: [1, 2], 1: [2, 3]}

            transposed_graph = defaultdict(list)
            for node, neighbours in graph.items():
            for neighbour in neighbours:
            transposed_graph[neighbour].append(node)

            # {1: [0], 2: [0, 1], 3: [1]}





            share|improve this answer









            $endgroup$


















              3












              $begingroup$

              There's really not much to say about this code, it is straightforward.



              Style



              These are only nitpicks.




              • Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

              • Collections should be named by their purpose in the application, not their type.

              • By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.


              Code improvements



              When iterating over the keys and values of a dictionary at the same time, you can use



              for k, v in dic.items():
              # ... use k, v ...


              instead of



              for k in dic:
              v = dic[k]
              # ...


              The nested loop can be transformed to a dictionary comprehension like this:



              dic2 = {v: k for k, values in dic.items() for v in values}


              You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.



              Potential pitfalls



              You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:



              >>> dic = {0: [1, 2], 1: [2, 3]}
              >>> {v: k for k, values in dic.items() for v in values}
              {1: 0, 2: 1, 3: 1} # missing 2: 0


              To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.



              If the input represents a directed graph (mapping nodes to lists of neighbours), this corresponds to computing the transposed or reversed graph. It can be done by using a collections.defaultdict. I don't see an easy way to write it as a comprehension in this case.



              from collections import defaultdict

              graph = {0: [1, 2], 1: [2, 3]}

              transposed_graph = defaultdict(list)
              for node, neighbours in graph.items():
              for neighbour in neighbours:
              transposed_graph[neighbour].append(node)

              # {1: [0], 2: [0, 1], 3: [1]}





              share|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                There's really not much to say about this code, it is straightforward.



                Style



                These are only nitpicks.




                • Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

                • Collections should be named by their purpose in the application, not their type.

                • By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.


                Code improvements



                When iterating over the keys and values of a dictionary at the same time, you can use



                for k, v in dic.items():
                # ... use k, v ...


                instead of



                for k in dic:
                v = dic[k]
                # ...


                The nested loop can be transformed to a dictionary comprehension like this:



                dic2 = {v: k for k, values in dic.items() for v in values}


                You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.



                Potential pitfalls



                You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:



                >>> dic = {0: [1, 2], 1: [2, 3]}
                >>> {v: k for k, values in dic.items() for v in values}
                {1: 0, 2: 1, 3: 1} # missing 2: 0


                To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.



                If the input represents a directed graph (mapping nodes to lists of neighbours), this corresponds to computing the transposed or reversed graph. It can be done by using a collections.defaultdict. I don't see an easy way to write it as a comprehension in this case.



                from collections import defaultdict

                graph = {0: [1, 2], 1: [2, 3]}

                transposed_graph = defaultdict(list)
                for node, neighbours in graph.items():
                for neighbour in neighbours:
                transposed_graph[neighbour].append(node)

                # {1: [0], 2: [0, 1], 3: [1]}





                share|improve this answer









                $endgroup$



                There's really not much to say about this code, it is straightforward.



                Style



                These are only nitpicks.




                • Generic dictionary keys are typically named k instead of i or j, but in a specific application a more descriptive name would be even better.

                • Collections should be named by their purpose in the application, not their type.

                • By convention, assignments and other binary operators should be surrounded by spaces: a = b, not a=b. See PEP 8 for reference, if you're not already aware of it. Following this style guide makes your code easier to read for others.


                Code improvements



                When iterating over the keys and values of a dictionary at the same time, you can use



                for k, v in dic.items():
                # ... use k, v ...


                instead of



                for k in dic:
                v = dic[k]
                # ...


                The nested loop can be transformed to a dictionary comprehension like this:



                dic2 = {v: k for k, values in dic.items() for v in values}


                You can remember that the order of for clauses in the comprehension is the same as the order of corresponding nested for loops.



                Potential pitfalls



                You should be aware that this transformation from a dictionary of lists to a reverse dictionary only works if all items in the original lists are unique. Counterexample:



                >>> dic = {0: [1, 2], 1: [2, 3]}
                >>> {v: k for k, values in dic.items() for v in values}
                {1: 0, 2: 1, 3: 1} # missing 2: 0


                To correct this, the output dictionary should have the same format as the input, namely mapping each of the items in the input lists to a list of corresponding keys.



                If the input represents a directed graph (mapping nodes to lists of neighbours), this corresponds to computing the transposed or reversed graph. It can be done by using a collections.defaultdict. I don't see an easy way to write it as a comprehension in this case.



                from collections import defaultdict

                graph = {0: [1, 2], 1: [2, 3]}

                transposed_graph = defaultdict(list)
                for node, neighbours in graph.items():
                for neighbour in neighbours:
                transposed_graph[neighbour].append(node)

                # {1: [0], 2: [0, 1], 3: [1]}






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 4 hours ago









                mkrieger1mkrieger1

                1,4131824




                1,4131824






















                    Juan C is a new contributor. Be nice, and check out our Code of Conduct.










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                    Juan C is a new contributor. Be nice, and check out our Code of Conduct.












                    Juan C is a new contributor. Be nice, and check out our Code of Conduct.
















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