Repeat values from a sequence up to n












0















Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:



seq = np.array([0,1])

a = np.tile(seq, math.ceil(n/2))[:n]


Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:



n = 6

seq = np.array([1,2,3,4])

np.tile(seq, math.ceil(n/2))[:n]

array([1, 2, 3, 4, 1, 2])





python list numpy







share|improve this question



























    0















    Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:



    seq = np.array([0,1])
    
a = np.tile(seq, math.ceil(n/2))[:n]


    Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:



    n = 6
    
seq = np.array([1,2,3,4])
    
np.tile(seq, math.ceil(n/2))[:n]
    
array([1, 2, 3, 4, 1, 2])





    python list numpy







    share|improve this question

























      0












      0








      0








      Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:



      seq = np.array([0,1])
      
a = np.tile(seq, math.ceil(n/2))[:n]


      Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:



      n = 6
      
seq = np.array([1,2,3,4])
      
np.tile(seq, math.ceil(n/2))[:n]
      
array([1, 2, 3, 4, 1, 2])





      python list numpy







      share|improve this question














      Say I want to repeat a sequence of consecutive 0s and 1s up to n. One way I can think of is:



      seq = np.array([0,1])
      
a = np.tile(seq, math.ceil(n/2))[:n]


      Where I use math.ceil(n/2) so that only an extra number is generated in the case of having an odd n. But is there a more concise way of doing this? This should ideally be extendable to any given sequence, for example:



      n = 6
      
seq = np.array([1,2,3,4])
      
np.tile(seq, math.ceil(n/2))[:n]
      
array([1, 2, 3, 4, 1, 2])





      python list numpy




      python list numpy






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 1 at 21:20









      yatuyatu

      13.4k31341




      13.4k31341
























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          np.resize may work for you.



          In [43]: seq = np.array([1,2,3,4])
          
In [44]: np.resize(seq, 6)
          
Out[44]: array([1, 2, 3, 4, 1, 2])


          We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.






          share|improve this answer
























          • Nice :) Didn't know it repeated it if the new size is greater

            – yatu
            Jan 1 at 21:28











          • The .resize method behaves differently.

            – hpaulj
            Jan 1 at 21:30











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          1














          np.resize may work for you.



          In [43]: seq = np.array([1,2,3,4])
          
In [44]: np.resize(seq, 6)
          
Out[44]: array([1, 2, 3, 4, 1, 2])


          We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.






          share|improve this answer
























          • Nice :) Didn't know it repeated it if the new size is greater

            – yatu
            Jan 1 at 21:28











          • The .resize method behaves differently.

            – hpaulj
            Jan 1 at 21:30
















          1














          np.resize may work for you.



          In [43]: seq = np.array([1,2,3,4])
          
In [44]: np.resize(seq, 6)
          
Out[44]: array([1, 2, 3, 4, 1, 2])


          We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.






          share|improve this answer
























          • Nice :) Didn't know it repeated it if the new size is greater

            – yatu
            Jan 1 at 21:28











          • The .resize method behaves differently.

            – hpaulj
            Jan 1 at 21:30














          1












          1








          1







          np.resize may work for you.



          In [43]: seq = np.array([1,2,3,4])
          
In [44]: np.resize(seq, 6)
          
Out[44]: array([1, 2, 3, 4, 1, 2])


          We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.






          share|improve this answer













          np.resize may work for you.



          In [43]: seq = np.array([1,2,3,4])
          
In [44]: np.resize(seq, 6)
          
Out[44]: array([1, 2, 3, 4, 1, 2])


          We don't use resize (function or method) that often, but in this case the fill pattern for the function version meets your needs.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 1 at 21:25









          hpauljhpaulj

          116k785156




          116k785156













          • Nice :) Didn't know it repeated it if the new size is greater

            – yatu
            Jan 1 at 21:28











          • The .resize method behaves differently.

            – hpaulj
            Jan 1 at 21:30



















          • Nice :) Didn't know it repeated it if the new size is greater

            – yatu
            Jan 1 at 21:28











          • The .resize method behaves differently.

            – hpaulj
            Jan 1 at 21:30

















          Nice :) Didn't know it repeated it if the new size is greater

          – yatu
          Jan 1 at 21:28





          Nice :) Didn't know it repeated it if the new size is greater

          – yatu
          Jan 1 at 21:28













          The .resize method behaves differently.

          – hpaulj
          Jan 1 at 21:30





          The .resize method behaves differently.

          – hpaulj
          Jan 1 at 21:30




















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