Most time/space efficient way to check if all elements in list of integers are 0












0















This is an implementation question for Python 2.7



Say I have a list of integers called nums, and I need to check if all values in nums are equal to zero. nums contains many elements (i.e. more than 10000), with many repeating values.



Using all():



if all(n == 0 for n in set(nums)):  # I assume this conversion from list to set helps?

    # do something


Using set subtraction:



if set(nums) - {0} == set():

    # do something


Edit: better way to do the above approach, courtesy of user U9-Forward



if set(nums) == {0}:

    # do something


How do the time and space complexities compare for each of these approaches? Is there a more efficient way to check this?



Note: for this case, I am trying to avoid using numpy/pandas.






python python-2.7 python-2.x







share|improve this question




















  • 1





    Second could be nicer: set(nums)=={0}

    – U9-Forward
    Nov 26 '18 at 7:48
















0















This is an implementation question for Python 2.7



Say I have a list of integers called nums, and I need to check if all values in nums are equal to zero. nums contains many elements (i.e. more than 10000), with many repeating values.



Using all():



if all(n == 0 for n in set(nums)):  # I assume this conversion from list to set helps?

    # do something


Using set subtraction:



if set(nums) - {0} == set():

    # do something


Edit: better way to do the above approach, courtesy of user U9-Forward



if set(nums) == {0}:

    # do something


How do the time and space complexities compare for each of these approaches? Is there a more efficient way to check this?



Note: for this case, I am trying to avoid using numpy/pandas.






python python-2.7 python-2.x







share|improve this question




















  • 1





    Second could be nicer: set(nums)=={0}

    – U9-Forward
    Nov 26 '18 at 7:48














0












0








0








This is an implementation question for Python 2.7



Say I have a list of integers called nums, and I need to check if all values in nums are equal to zero. nums contains many elements (i.e. more than 10000), with many repeating values.



Using all():



if all(n == 0 for n in set(nums)):  # I assume this conversion from list to set helps?

    # do something


Using set subtraction:



if set(nums) - {0} == set():

    # do something


Edit: better way to do the above approach, courtesy of user U9-Forward



if set(nums) == {0}:

    # do something


How do the time and space complexities compare for each of these approaches? Is there a more efficient way to check this?



Note: for this case, I am trying to avoid using numpy/pandas.






python python-2.7 python-2.x







share|improve this question
















This is an implementation question for Python 2.7



Say I have a list of integers called nums, and I need to check if all values in nums are equal to zero. nums contains many elements (i.e. more than 10000), with many repeating values.



Using all():



if all(n == 0 for n in set(nums)):  # I assume this conversion from list to set helps?

    # do something


Using set subtraction:



if set(nums) - {0} == set():

    # do something


Edit: better way to do the above approach, courtesy of user U9-Forward



if set(nums) == {0}:

    # do something


How do the time and space complexities compare for each of these approaches? Is there a more efficient way to check this?



Note: for this case, I am trying to avoid using numpy/pandas.






python python-2.7 python-2.x




python python-2.7 python-2.x






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 26 '18 at 7:51







ericph

















asked Nov 26 '18 at 7:46









ericphericph

32




32








  • 1





    Second could be nicer: set(nums)=={0}

    – U9-Forward
    Nov 26 '18 at 7:48














  • 1





    Second could be nicer: set(nums)=={0}

    – U9-Forward
    Nov 26 '18 at 7:48








1




1





Second could be nicer: set(nums)=={0}

– U9-Forward
Nov 26 '18 at 7:48





Second could be nicer: set(nums)=={0}

– U9-Forward
Nov 26 '18 at 7:48












4 Answers
4






active

oldest

votes


















2














Any set conversion of nums won't help as it will iterate the entire list:



if all(n == 0 for n in nums):

    # ...


is just fine as it stops at the first non-zero element, disregarding the remainder.



Asymptotically, all these approaches are linear with random data.
Implementational details (no repeated function calls on the generator) makes not any(nums) even faster, but that relies on the absence of any other falsy elements but0, e.g. '' or None.






share|improve this answer


























  • You can get a x10 speedup by eliminating the explicit comparison and the generator.

    – DYZ
    Nov 26 '18 at 7:58











  • How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

    – ericph
    Nov 26 '18 at 8:12





















2














not any(nums) is probably the fastest because it will stop when/if it finds any non-zero element.



Performance comparison:



a = range(10000)

b = [0] * 10000

%timeit not any(a) # 72 ns, fastest for non-zero lists

%timeit not any(b) # 33 ns, fastest for zero lists

%timeit all(n == 0 for n in a) # 365 ns

%timeit all(n == 0 for n in b) # 350 µs

%timeit set(a)=={0} # 228 µs

%timeit set(b)=={0} # 58 µs





share|improve this answer





















  • 1





    And so will all.

    – schwobaseggl
    Nov 26 '18 at 7:52











  • @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

    – DYZ
    Nov 26 '18 at 7:56













  • That's a valid point, but it doesn't make your argument any more true.

    – schwobaseggl
    Nov 26 '18 at 7:58






  • 2





    Let us continue this discussion in chat.

    – schwobaseggl
    Nov 26 '18 at 8:02






  • 1





    +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

    – jimhark
    Nov 26 '18 at 8:41





















0














If you can use numpy, then (np.array(nums) == 0).all() should do it.






share|improve this answer



















  • 1





    That's unlikely to help since the list needs to be read fully to just create the array.

    – Hannes Ovrén
    Nov 26 '18 at 7:51



















0














Additionally to @schwobaseggl's answer, second example could be even better:



if set(nums)=={0}:

    # do something





share|improve this answer























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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Any set conversion of nums won't help as it will iterate the entire list:



    if all(n == 0 for n in nums):
    
    # ...


    is just fine as it stops at the first non-zero element, disregarding the remainder.



    Asymptotically, all these approaches are linear with random data.
    Implementational details (no repeated function calls on the generator) makes not any(nums) even faster, but that relies on the absence of any other falsy elements but0, e.g. '' or None.






    share|improve this answer


























    • You can get a x10 speedup by eliminating the explicit comparison and the generator.

      – DYZ
      Nov 26 '18 at 7:58











    • How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

      – ericph
      Nov 26 '18 at 8:12


















    2














    Any set conversion of nums won't help as it will iterate the entire list:



    if all(n == 0 for n in nums):
    
    # ...


    is just fine as it stops at the first non-zero element, disregarding the remainder.



    Asymptotically, all these approaches are linear with random data.
    Implementational details (no repeated function calls on the generator) makes not any(nums) even faster, but that relies on the absence of any other falsy elements but0, e.g. '' or None.






    share|improve this answer


























    • You can get a x10 speedup by eliminating the explicit comparison and the generator.

      – DYZ
      Nov 26 '18 at 7:58











    • How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

      – ericph
      Nov 26 '18 at 8:12
















    2












    2








    2







    Any set conversion of nums won't help as it will iterate the entire list:



    if all(n == 0 for n in nums):
    
    # ...


    is just fine as it stops at the first non-zero element, disregarding the remainder.



    Asymptotically, all these approaches are linear with random data.
    Implementational details (no repeated function calls on the generator) makes not any(nums) even faster, but that relies on the absence of any other falsy elements but0, e.g. '' or None.






    share|improve this answer















    Any set conversion of nums won't help as it will iterate the entire list:



    if all(n == 0 for n in nums):
    
    # ...


    is just fine as it stops at the first non-zero element, disregarding the remainder.



    Asymptotically, all these approaches are linear with random data.
    Implementational details (no repeated function calls on the generator) makes not any(nums) even faster, but that relies on the absence of any other falsy elements but0, e.g. '' or None.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 26 '18 at 8:06

























    answered Nov 26 '18 at 7:48









    schwobasegglschwobaseggl

    37.2k32442




    37.2k32442













    • You can get a x10 speedup by eliminating the explicit comparison and the generator.

      – DYZ
      Nov 26 '18 at 7:58











    • How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

      – ericph
      Nov 26 '18 at 8:12





















    • You can get a x10 speedup by eliminating the explicit comparison and the generator.

      – DYZ
      Nov 26 '18 at 7:58











    • How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

      – ericph
      Nov 26 '18 at 8:12



















    You can get a x10 speedup by eliminating the explicit comparison and the generator.

    – DYZ
    Nov 26 '18 at 7:58





    You can get a x10 speedup by eliminating the explicit comparison and the generator.

    – DYZ
    Nov 26 '18 at 7:58













    How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

    – ericph
    Nov 26 '18 at 8:12







    How about a case in which nums contains both 0 (int) and 0.0 (float)? Would not any(nums) and all(nums) still work correctly?

    – ericph
    Nov 26 '18 at 8:12















    2














    not any(nums) is probably the fastest because it will stop when/if it finds any non-zero element.



    Performance comparison:



    a = range(10000)
    
b = [0] * 10000
    
%timeit not any(a) # 72 ns, fastest for non-zero lists
    
%timeit not any(b) # 33 ns, fastest for zero lists
    
%timeit all(n == 0 for n in a) # 365 ns
    
%timeit all(n == 0 for n in b) # 350 µs
    
%timeit set(a)=={0} # 228 µs
    
%timeit set(b)=={0} # 58 µs





    share|improve this answer





















    • 1





      And so will all.

      – schwobaseggl
      Nov 26 '18 at 7:52











    • @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

      – DYZ
      Nov 26 '18 at 7:56













    • That's a valid point, but it doesn't make your argument any more true.

      – schwobaseggl
      Nov 26 '18 at 7:58






    • 2





      Let us continue this discussion in chat.

      – schwobaseggl
      Nov 26 '18 at 8:02






    • 1





      +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

      – jimhark
      Nov 26 '18 at 8:41


















    2














    not any(nums) is probably the fastest because it will stop when/if it finds any non-zero element.



    Performance comparison:



    a = range(10000)
    
b = [0] * 10000
    
%timeit not any(a) # 72 ns, fastest for non-zero lists
    
%timeit not any(b) # 33 ns, fastest for zero lists
    
%timeit all(n == 0 for n in a) # 365 ns
    
%timeit all(n == 0 for n in b) # 350 µs
    
%timeit set(a)=={0} # 228 µs
    
%timeit set(b)=={0} # 58 µs





    share|improve this answer





















    • 1





      And so will all.

      – schwobaseggl
      Nov 26 '18 at 7:52











    • @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

      – DYZ
      Nov 26 '18 at 7:56













    • That's a valid point, but it doesn't make your argument any more true.

      – schwobaseggl
      Nov 26 '18 at 7:58






    • 2





      Let us continue this discussion in chat.

      – schwobaseggl
      Nov 26 '18 at 8:02






    • 1





      +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

      – jimhark
      Nov 26 '18 at 8:41
















    2












    2








    2







    not any(nums) is probably the fastest because it will stop when/if it finds any non-zero element.



    Performance comparison:



    a = range(10000)
    
b = [0] * 10000
    
%timeit not any(a) # 72 ns, fastest for non-zero lists
    
%timeit not any(b) # 33 ns, fastest for zero lists
    
%timeit all(n == 0 for n in a) # 365 ns
    
%timeit all(n == 0 for n in b) # 350 µs
    
%timeit set(a)=={0} # 228 µs
    
%timeit set(b)=={0} # 58 µs





    share|improve this answer















    not any(nums) is probably the fastest because it will stop when/if it finds any non-zero element.



    Performance comparison:



    a = range(10000)
    
b = [0] * 10000
    
%timeit not any(a) # 72 ns, fastest for non-zero lists
    
%timeit not any(b) # 33 ns, fastest for zero lists
    
%timeit all(n == 0 for n in a) # 365 ns
    
%timeit all(n == 0 for n in b) # 350 µs
    
%timeit set(a)=={0} # 228 µs
    
%timeit set(b)=={0} # 58 µs






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 26 '18 at 8:43

























    answered Nov 26 '18 at 7:51









    DYZDYZ

    26.5k62049




    26.5k62049








    • 1





      And so will all.

      – schwobaseggl
      Nov 26 '18 at 7:52











    • @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

      – DYZ
      Nov 26 '18 at 7:56













    • That's a valid point, but it doesn't make your argument any more true.

      – schwobaseggl
      Nov 26 '18 at 7:58






    • 2





      Let us continue this discussion in chat.

      – schwobaseggl
      Nov 26 '18 at 8:02






    • 1





      +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

      – jimhark
      Nov 26 '18 at 8:41
















    • 1





      And so will all.

      – schwobaseggl
      Nov 26 '18 at 7:52











    • @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

      – DYZ
      Nov 26 '18 at 7:56













    • That's a valid point, but it doesn't make your argument any more true.

      – schwobaseggl
      Nov 26 '18 at 7:58






    • 2





      Let us continue this discussion in chat.

      – schwobaseggl
      Nov 26 '18 at 8:02






    • 1





      +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

      – jimhark
      Nov 26 '18 at 8:41










    1




    1





    And so will all.

    – schwobaseggl
    Nov 26 '18 at 7:52





    And so will all.

    – schwobaseggl
    Nov 26 '18 at 7:52













    @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

    – DYZ
    Nov 26 '18 at 7:56







    @schwobaseggl In fact, your solution is about 10 times slower than mine - because of the generator.

    – DYZ
    Nov 26 '18 at 7:56















    That's a valid point, but it doesn't make your argument any more true.

    – schwobaseggl
    Nov 26 '18 at 7:58





    That's a valid point, but it doesn't make your argument any more true.

    – schwobaseggl
    Nov 26 '18 at 7:58




    2




    2





    Let us continue this discussion in chat.

    – schwobaseggl
    Nov 26 '18 at 8:02





    Let us continue this discussion in chat.

    – schwobaseggl
    Nov 26 '18 at 8:02




    1




    1





    +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

    – jimhark
    Nov 26 '18 at 8:41







    +1 Because you're timings look good. (I was running timings myself, but you beat me too it). But 'all()' doesn't provide a correct solution. all([0]) returns false where all elements are zero, true is correct. You can't just add a 'not' in front because all([0,1]) also returns false which is correct. If you think about it, you will realize 'all' is not the right test. All your other tests look good.

    – jimhark
    Nov 26 '18 at 8:41













    0














    If you can use numpy, then (np.array(nums) == 0).all() should do it.






    share|improve this answer



















    • 1





      That's unlikely to help since the list needs to be read fully to just create the array.

      – Hannes Ovrén
      Nov 26 '18 at 7:51
















    0














    If you can use numpy, then (np.array(nums) == 0).all() should do it.






    share|improve this answer



















    • 1





      That's unlikely to help since the list needs to be read fully to just create the array.

      – Hannes Ovrén
      Nov 26 '18 at 7:51














    0












    0








    0







    If you can use numpy, then (np.array(nums) == 0).all() should do it.






    share|improve this answer













    If you can use numpy, then (np.array(nums) == 0).all() should do it.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 26 '18 at 7:49









    Thomas LangThomas Lang

    44627




    44627








    • 1





      That's unlikely to help since the list needs to be read fully to just create the array.

      – Hannes Ovrén
      Nov 26 '18 at 7:51














    • 1





      That's unlikely to help since the list needs to be read fully to just create the array.

      – Hannes Ovrén
      Nov 26 '18 at 7:51








    1




    1





    That's unlikely to help since the list needs to be read fully to just create the array.

    – Hannes Ovrén
    Nov 26 '18 at 7:51





    That's unlikely to help since the list needs to be read fully to just create the array.

    – Hannes Ovrén
    Nov 26 '18 at 7:51











    0














    Additionally to @schwobaseggl's answer, second example could be even better:



    if set(nums)=={0}:
    
    # do something





    share|improve this answer




























      0














      Additionally to @schwobaseggl's answer, second example could be even better:



      if set(nums)=={0}:
      
    # do something





      share|improve this answer


























        0












        0








        0







        Additionally to @schwobaseggl's answer, second example could be even better:



        if set(nums)=={0}:
        
    # do something





        share|improve this answer













        Additionally to @schwobaseggl's answer, second example could be even better:



        if set(nums)=={0}:
        
    # do something






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 26 '18 at 7:50









        U9-ForwardU9-Forward

        15.1k41438




        15.1k41438






























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            Lallio

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            Coordenadas: 45° 40' N 9° 38' E Lallio      Comuna    Lallio Localização de Lallio na Itália Coordenadas 45° 40' N 9° 38' E Região Lombardia Província Bérgamo Área - Total 2 km² Altitude 216 m População  - Total 3 826     • Densidade 1 913 hab./km² Outros dados Comunas limítrofes Bergamo, Dalmine, Stezzano, Treviolo Código ISTAT 016123 Código cadastral E422 Código postal 24040 Prefixo telefônico 035 Website www.comune.lallio.bg.it Lallio é uma comuna italiana da região da Lombardia, província de Bérgamo, com cerca de 3.826 habitantes. Estende-se por uma área de 2 km², tendo uma densidade populacional de 1913 hab/km². Faz fronteira com Bergamo, Dalmine, Stezzano, Treviolo. [ 1 ] [ 2 ] [ 3 ] Demografia | Variação demográfica do município entre 1861 e 2011 [ 3 ] Fonte : Istituto Nazionale di Statistica (ISTAT) - Elabor...
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            Futebolista

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            Ronaldo primeiro jogador a ser eleito por três vezes o melhor jogador do mundo pela FIFA. Ronaldinho Gaúcho atuando pelo Barcelona onde foi duas vezes eleito o melhor jogador do mundo pela FIFA. Kaká atuando pelo Milan onde foi eleito em 2007 o melhor jogador do mundo pela FIFA. Lothar Matthäus primeiro jogador a ser eleito o melhor jogador do mundo pela FIFA em 1991 pela Inter de Milão. Neymar durante partida contra a Escócia, em 27 de março de 2011. Crianças praticando futebol. Robinho, futebolista brasileiro, em 15 de novembro de 2006. Atletas do Chelsea, da Inglaterra. A venda de camisas com nome e número dos jogadores é uma grande fonte de arrecadação. Jürgen Klinsmann atuou como jogador e foi treinador da Alemanha. Estádio de futebol, principal local onde laboram os jogadores. O jogador de futebol , ou Futebolista , é um atleta profissional de futebol, e cuja prática do desporto coletivo é fundamental. Estes trabalhadores sã...
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            Jornalista

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            Uma repórter de televisão segurando um microfone na frente de um cinegrafista. Fotojornalistas no Campeonato Mundial de Atletismo 2013 Uma repórter entrevistando Boris Johnson, Prefeito de Londres. Jornalista é o profissional formado em Jornalismo. É a pessoa responsável pela apuração, investigação e apresentação de notícias, reportagens, entrevistas ou distribuição de notícias ou outra informação de interesse coletivo. O trabalho do jornalista é chamado jornalismo. Um jornalista pode trabalhar com questões gerais ou especializar-se em determinadas áreas. No entanto, a maioria dos jornalistas tendem a se especializar, e cooperando com outros jornalistas, produzir publicações que abrangem muitos tópicos. [ 1 ] Por exemplo, um jornalista esportivo cobre notícias dentro do mundo dos esportes, mas este jornalista pode ser uma parte de um jornal que cobre diversos temas. O exercício do Jornalismo é privativo de jornalista. Entre as áreas em que o jornalista trabalha estão o...
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