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Verify two numbers from a list will add up to a number in the same list using Python



The Next CEO of Stack OverflowPrime factorization of a numberGenerator to Tuple to List?Calculating parsimonies using Python; performance is lackingDynamic Programming for printing additive numbers up to digits nProject Euler: #36 Binary and Base 10 PalindromesFinding all divisors of an integerGenerate set of random numbers and remove lowestFind two distinct subsets of a list that add up to the same valueFirstDuplicate FinderSearch father (x) of the number D(x) and measure time and memory consumed










0












$begingroup$


I am trying to solve this question:



  • Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.

I came up with:



def func(l):
sum_set = set() # using a set to store sums of two numbers.
for index in range(len(l)-1):
counter = 1
while counter + index < len(l):
sum = l[index] + l[index+counter]
sum_set.add(sum)
counter += 1
for number in l:
if number in sum_set:
return True
return False

print(func([1,2,3,4]))
print(func([1,2,1,4]))
print(func([1,1,1]))
print(func([1]))
print(func([1,2,5,9]))
print(func([2,1,1]))
print(func([1,-1,0]))


From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:



  • I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g. l = [1,1,1,1,2] should be optimised to l = [1,1,2]

Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?









share









$endgroup$
















    0












    $begingroup$


    I am trying to solve this question:



    • Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.

    I came up with:



    def func(l):
    sum_set = set() # using a set to store sums of two numbers.
    for index in range(len(l)-1):
    counter = 1
    while counter + index < len(l):
    sum = l[index] + l[index+counter]
    sum_set.add(sum)
    counter += 1
    for number in l:
    if number in sum_set:
    return True
    return False

    print(func([1,2,3,4]))
    print(func([1,2,1,4]))
    print(func([1,1,1]))
    print(func([1]))
    print(func([1,2,5,9]))
    print(func([2,1,1]))
    print(func([1,-1,0]))


    From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:



    • I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g. l = [1,1,1,1,2] should be optimised to l = [1,1,2]

    Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?









    share









    $endgroup$














      0












      0








      0





      $begingroup$


      I am trying to solve this question:



      • Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.

      I came up with:



      def func(l):
      sum_set = set() # using a set to store sums of two numbers.
      for index in range(len(l)-1):
      counter = 1
      while counter + index < len(l):
      sum = l[index] + l[index+counter]
      sum_set.add(sum)
      counter += 1
      for number in l:
      if number in sum_set:
      return True
      return False

      print(func([1,2,3,4]))
      print(func([1,2,1,4]))
      print(func([1,1,1]))
      print(func([1]))
      print(func([1,2,5,9]))
      print(func([2,1,1]))
      print(func([1,-1,0]))


      From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:



      • I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g. l = [1,1,1,1,2] should be optimised to l = [1,1,2]

      Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?









      share









      $endgroup$




      I am trying to solve this question:



      • Given a list of integers, verifies if any two numbers from this list add up to another number in the same list.

      I came up with:



      def func(l):
      sum_set = set() # using a set to store sums of two numbers.
      for index in range(len(l)-1):
      counter = 1
      while counter + index < len(l):
      sum = l[index] + l[index+counter]
      sum_set.add(sum)
      counter += 1
      for number in l:
      if number in sum_set:
      return True
      return False

      print(func([1,2,3,4]))
      print(func([1,2,1,4]))
      print(func([1,1,1]))
      print(func([1]))
      print(func([1,2,5,9]))
      print(func([2,1,1]))
      print(func([1,-1,0]))


      From the tests I ran above, my function is working. But there is at least one way I can think of my approach is lacking of:



      • I should filter the original list to get rid of duplicate numbers. If a number appears more than 2 times, its rest of appearance should be disregarded. E.g. l = [1,1,1,1,2] should be optimised to l = [1,1,2]

      Is there any other aspect that I can improve my code snippet on to make it more Pythonic / more efficient?







      python





      share












      share










      share



      share










      asked 3 mins ago









      Yu ZhangYu Zhang

      302312




      302312




















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