Make all sublists same length by prepending nan












0














How can I make all of my sublists the same length (the length of the longest sublist) by prepending np.nan's on each sublist?



import random

[list(range(0,random.randint(1,5))) for x in range(n)]


So if the output is:




[[0, 1], [0], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3]]




It should look like:




[[nan,nan,nan,0, 1], [nan,nan,nan,nan,0], [0, 1, 2, 3, 4], [0, 1, 2,
3, 4], [nan,0, 1, 2, 3]]







python







share|improve this question


















  • 1




    What have you tried so far?
    – Idlehands
    Nov 20 '18 at 19:18










  • @Idlehands I've tried reversing first and then .extend() and then reverse again. But this doesn't work if the list is empty
    – jchaykow
    Nov 20 '18 at 19:22










  • You might be overthinking it... how is the list of sublists supposed to be generated? Is it actual data or is it created with the nan paddings from the beginning?
    – Idlehands
    Nov 20 '18 at 19:25






  • 1




    Saying 'everything' is not useful. Show us the actual code that you tried, which gives us either a starting point or a hint what we should not try.
    – Bram Vanroy
    Nov 20 '18 at 19:25










  • @BramVanroy it's a joke, I wrote two comments
    – jchaykow
    Nov 20 '18 at 19:26
















0














How can I make all of my sublists the same length (the length of the longest sublist) by prepending np.nan's on each sublist?



import random

[list(range(0,random.randint(1,5))) for x in range(n)]


So if the output is:




[[0, 1], [0], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3]]




It should look like:




[[nan,nan,nan,0, 1], [nan,nan,nan,nan,0], [0, 1, 2, 3, 4], [0, 1, 2,
3, 4], [nan,0, 1, 2, 3]]







python







share|improve this question


















  • 1




    What have you tried so far?
    – Idlehands
    Nov 20 '18 at 19:18










  • @Idlehands I've tried reversing first and then .extend() and then reverse again. But this doesn't work if the list is empty
    – jchaykow
    Nov 20 '18 at 19:22










  • You might be overthinking it... how is the list of sublists supposed to be generated? Is it actual data or is it created with the nan paddings from the beginning?
    – Idlehands
    Nov 20 '18 at 19:25






  • 1




    Saying 'everything' is not useful. Show us the actual code that you tried, which gives us either a starting point or a hint what we should not try.
    – Bram Vanroy
    Nov 20 '18 at 19:25










  • @BramVanroy it's a joke, I wrote two comments
    – jchaykow
    Nov 20 '18 at 19:26














0












0








0







How can I make all of my sublists the same length (the length of the longest sublist) by prepending np.nan's on each sublist?



import random

[list(range(0,random.randint(1,5))) for x in range(n)]


So if the output is:




[[0, 1], [0], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3]]




It should look like:




[[nan,nan,nan,0, 1], [nan,nan,nan,nan,0], [0, 1, 2, 3, 4], [0, 1, 2,
3, 4], [nan,0, 1, 2, 3]]







python







share|improve this question













How can I make all of my sublists the same length (the length of the longest sublist) by prepending np.nan's on each sublist?



import random

[list(range(0,random.randint(1,5))) for x in range(n)]


So if the output is:




[[0, 1], [0], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4], [0, 1, 2, 3]]




It should look like:




[[nan,nan,nan,0, 1], [nan,nan,nan,nan,0], [0, 1, 2, 3, 4], [0, 1, 2,
3, 4], [nan,0, 1, 2, 3]]







python




python






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 20 '18 at 19:16









jchaykow

497318




497318








  • 1




    What have you tried so far?
    – Idlehands
    Nov 20 '18 at 19:18










  • @Idlehands I've tried reversing first and then .extend() and then reverse again. But this doesn't work if the list is empty
    – jchaykow
    Nov 20 '18 at 19:22










  • You might be overthinking it... how is the list of sublists supposed to be generated? Is it actual data or is it created with the nan paddings from the beginning?
    – Idlehands
    Nov 20 '18 at 19:25






  • 1




    Saying 'everything' is not useful. Show us the actual code that you tried, which gives us either a starting point or a hint what we should not try.
    – Bram Vanroy
    Nov 20 '18 at 19:25










  • @BramVanroy it's a joke, I wrote two comments
    – jchaykow
    Nov 20 '18 at 19:26














  • 1




    What have you tried so far?
    – Idlehands
    Nov 20 '18 at 19:18










  • @Idlehands I've tried reversing first and then .extend() and then reverse again. But this doesn't work if the list is empty
    – jchaykow
    Nov 20 '18 at 19:22










  • You might be overthinking it... how is the list of sublists supposed to be generated? Is it actual data or is it created with the nan paddings from the beginning?
    – Idlehands
    Nov 20 '18 at 19:25






  • 1




    Saying 'everything' is not useful. Show us the actual code that you tried, which gives us either a starting point or a hint what we should not try.
    – Bram Vanroy
    Nov 20 '18 at 19:25










  • @BramVanroy it's a joke, I wrote two comments
    – jchaykow
    Nov 20 '18 at 19:26








1




1




What have you tried so far?
– Idlehands
Nov 20 '18 at 19:18




What have you tried so far?
– Idlehands
Nov 20 '18 at 19:18












@Idlehands I've tried reversing first and then .extend() and then reverse again. But this doesn't work if the list is empty
– jchaykow
Nov 20 '18 at 19:22




@Idlehands I've tried reversing first and then .extend() and then reverse again. But this doesn't work if the list is empty
– jchaykow
Nov 20 '18 at 19:22












You might be overthinking it... how is the list of sublists supposed to be generated? Is it actual data or is it created with the nan paddings from the beginning?
– Idlehands
Nov 20 '18 at 19:25




You might be overthinking it... how is the list of sublists supposed to be generated? Is it actual data or is it created with the nan paddings from the beginning?
– Idlehands
Nov 20 '18 at 19:25




1




1




Saying 'everything' is not useful. Show us the actual code that you tried, which gives us either a starting point or a hint what we should not try.
– Bram Vanroy
Nov 20 '18 at 19:25




Saying 'everything' is not useful. Show us the actual code that you tried, which gives us either a starting point or a hint what we should not try.
– Bram Vanroy
Nov 20 '18 at 19:25












@BramVanroy it's a joke, I wrote two comments
– jchaykow
Nov 20 '18 at 19:26




@BramVanroy it's a joke, I wrote two comments
– jchaykow
Nov 20 '18 at 19:26












4 Answers
4






active

oldest

votes


















3














First, find the length of the longest sublist using max(). Then, for each sublist, use a slice assignment to replace the content of that list with the right number of NaNs followed by the original list.



import random, math



n = 5

lists = [list(range(0,random.randint(1,5))) for x in range(n)]



# get the maximum length

maxlen = len(max(lists, key=len))



# pad left of each sublist with NaN to make it as long as the longest

for sublist in lists:

    sublist[:] = [math.nan] * (maxlen - len(sublist)) + sublist





share|improve this answer





















  • Accepting this one because it's just barely faster than @Torin May
    – jchaykow
    Nov 20 '18 at 19:38










  • It'll get a lot faster as your n goes up.
    – kindall
    Nov 20 '18 at 19:50



















3














import random

import numpy as np



n = 5

a = [list(range(0,random.randint(1,5))) for x in range(n)]

for c in a:

    while len(c) < n:

        c.insert(0, np.nan)

print(a)





[[nan, nan, nan, nan, 0], [nan, nan, nan, nan, 0], [nan, nan, nan, 0, 1], [nan, nan, 0, 1, 2], [nan, nan, nan, 0, 1]]






This is an example of what I think you are looking for.






share|improve this answer





























    0














    Not the most pythonic code but I think it does the trick:



    longest_list_length = 0
    
for sublist in output:
    
  if len(sublist) > longest_list_length:
    
    longest_list_length = len(sublist)
    

    
for sublist in output:
    
  nans_to_prepend = ['nan'] * (longest_list_length - len(sublist))
    
  sublist = nans_to_prepend + sublist





    share|improve this answer





















    • replacing the value of sublist won't change the value in the original list.
      – kindall
      Nov 20 '18 at 19:28



















    0














    If all you care about is generating the np.nan padding during the random creation, this simple one liner will suffice:



    [[np.nan]*(5-j) + list(range(j)) for j in (random.randint(1, 5) for x in range(5))]
    

    
# [[nan, 0, 1, 2, 3], 
    
#  [nan, nan, 0, 1, 2], 
    
#  [nan, nan, nan, 0, 1], 
    
#  [nan, 0, 1, 2, 3],
    
#  [0, 1, 2, 3, 4]]


    You can of course replace 5 with n and determine your max length.






    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









      3














      First, find the length of the longest sublist using max(). Then, for each sublist, use a slice assignment to replace the content of that list with the right number of NaNs followed by the original list.



      import random, math
      

      
n = 5
      
lists = [list(range(0,random.randint(1,5))) for x in range(n)]
      

      
# get the maximum length
      
maxlen = len(max(lists, key=len))
      

      
# pad left of each sublist with NaN to make it as long as the longest
      
for sublist in lists:
      
    sublist[:] = [math.nan] * (maxlen - len(sublist)) + sublist





      share|improve this answer





















      • Accepting this one because it's just barely faster than @Torin May
        – jchaykow
        Nov 20 '18 at 19:38










      • It'll get a lot faster as your n goes up.
        – kindall
        Nov 20 '18 at 19:50
















      3














      First, find the length of the longest sublist using max(). Then, for each sublist, use a slice assignment to replace the content of that list with the right number of NaNs followed by the original list.



      import random, math
      

      
n = 5
      
lists = [list(range(0,random.randint(1,5))) for x in range(n)]
      

      
# get the maximum length
      
maxlen = len(max(lists, key=len))
      

      
# pad left of each sublist with NaN to make it as long as the longest
      
for sublist in lists:
      
    sublist[:] = [math.nan] * (maxlen - len(sublist)) + sublist





      share|improve this answer





















      • Accepting this one because it's just barely faster than @Torin May
        – jchaykow
        Nov 20 '18 at 19:38










      • It'll get a lot faster as your n goes up.
        – kindall
        Nov 20 '18 at 19:50














      3












      3








      3






      First, find the length of the longest sublist using max(). Then, for each sublist, use a slice assignment to replace the content of that list with the right number of NaNs followed by the original list.



      import random, math
      

      
n = 5
      
lists = [list(range(0,random.randint(1,5))) for x in range(n)]
      

      
# get the maximum length
      
maxlen = len(max(lists, key=len))
      

      
# pad left of each sublist with NaN to make it as long as the longest
      
for sublist in lists:
      
    sublist[:] = [math.nan] * (maxlen - len(sublist)) + sublist





      share|improve this answer












      First, find the length of the longest sublist using max(). Then, for each sublist, use a slice assignment to replace the content of that list with the right number of NaNs followed by the original list.



      import random, math
      

      
n = 5
      
lists = [list(range(0,random.randint(1,5))) for x in range(n)]
      

      
# get the maximum length
      
maxlen = len(max(lists, key=len))
      

      
# pad left of each sublist with NaN to make it as long as the longest
      
for sublist in lists:
      
    sublist[:] = [math.nan] * (maxlen - len(sublist)) + sublist






      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Nov 20 '18 at 19:27









      kindall

      127k17192241




      127k17192241












      • Accepting this one because it's just barely faster than @Torin May
        – jchaykow
        Nov 20 '18 at 19:38










      • It'll get a lot faster as your n goes up.
        – kindall
        Nov 20 '18 at 19:50


















      • Accepting this one because it's just barely faster than @Torin May
        – jchaykow
        Nov 20 '18 at 19:38










      • It'll get a lot faster as your n goes up.
        – kindall
        Nov 20 '18 at 19:50
















      Accepting this one because it's just barely faster than @Torin May
      – jchaykow
      Nov 20 '18 at 19:38




      Accepting this one because it's just barely faster than @Torin May
      – jchaykow
      Nov 20 '18 at 19:38












      It'll get a lot faster as your n goes up.
      – kindall
      Nov 20 '18 at 19:50




      It'll get a lot faster as your n goes up.
      – kindall
      Nov 20 '18 at 19:50













      3














      import random
      
import numpy as np
      

      
n = 5
      
a = [list(range(0,random.randint(1,5))) for x in range(n)]
      
for c in a:
      
    while len(c) < n:
      
        c.insert(0, np.nan)
      
print(a)





      [[nan, nan, nan, nan, 0], [nan, nan, nan, nan, 0], [nan, nan, nan, 0, 1], [nan, nan, 0, 1, 2], [nan, nan, nan, 0, 1]]






      This is an example of what I think you are looking for.






      share|improve this answer


























        3














        import random
        
import numpy as np
        

        
n = 5
        
a = [list(range(0,random.randint(1,5))) for x in range(n)]
        
for c in a:
        
    while len(c) < n:
        
        c.insert(0, np.nan)
        
print(a)





        [[nan, nan, nan, nan, 0], [nan, nan, nan, nan, 0], [nan, nan, nan, 0, 1], [nan, nan, 0, 1, 2], [nan, nan, nan, 0, 1]]






        This is an example of what I think you are looking for.






        share|improve this answer
























          3












          3








          3






          import random
          
import numpy as np
          

          
n = 5
          
a = [list(range(0,random.randint(1,5))) for x in range(n)]
          
for c in a:
          
    while len(c) < n:
          
        c.insert(0, np.nan)
          
print(a)





          [[nan, nan, nan, nan, 0], [nan, nan, nan, nan, 0], [nan, nan, nan, 0, 1], [nan, nan, 0, 1, 2], [nan, nan, nan, 0, 1]]






          This is an example of what I think you are looking for.






          share|improve this answer












          import random
          
import numpy as np
          

          
n = 5
          
a = [list(range(0,random.randint(1,5))) for x in range(n)]
          
for c in a:
          
    while len(c) < n:
          
        c.insert(0, np.nan)
          
print(a)





          [[nan, nan, nan, nan, 0], [nan, nan, nan, nan, 0], [nan, nan, nan, 0, 1], [nan, nan, 0, 1, 2], [nan, nan, nan, 0, 1]]






          This is an example of what I think you are looking for.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 20 '18 at 19:24









          Torin May

          303214




          303214























              0














              Not the most pythonic code but I think it does the trick:



              longest_list_length = 0
              
for sublist in output:
              
  if len(sublist) > longest_list_length:
              
    longest_list_length = len(sublist)
              

              
for sublist in output:
              
  nans_to_prepend = ['nan'] * (longest_list_length - len(sublist))
              
  sublist = nans_to_prepend + sublist





              share|improve this answer





















              • replacing the value of sublist won't change the value in the original list.
                – kindall
                Nov 20 '18 at 19:28
















              0














              Not the most pythonic code but I think it does the trick:



              longest_list_length = 0
              
for sublist in output:
              
  if len(sublist) > longest_list_length:
              
    longest_list_length = len(sublist)
              

              
for sublist in output:
              
  nans_to_prepend = ['nan'] * (longest_list_length - len(sublist))
              
  sublist = nans_to_prepend + sublist





              share|improve this answer





















              • replacing the value of sublist won't change the value in the original list.
                – kindall
                Nov 20 '18 at 19:28














              0












              0








              0






              Not the most pythonic code but I think it does the trick:



              longest_list_length = 0
              
for sublist in output:
              
  if len(sublist) > longest_list_length:
              
    longest_list_length = len(sublist)
              

              
for sublist in output:
              
  nans_to_prepend = ['nan'] * (longest_list_length - len(sublist))
              
  sublist = nans_to_prepend + sublist





              share|improve this answer












              Not the most pythonic code but I think it does the trick:



              longest_list_length = 0
              
for sublist in output:
              
  if len(sublist) > longest_list_length:
              
    longest_list_length = len(sublist)
              

              
for sublist in output:
              
  nans_to_prepend = ['nan'] * (longest_list_length - len(sublist))
              
  sublist = nans_to_prepend + sublist






              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 20 '18 at 19:26









              ritratt

              74031328




              74031328












              • replacing the value of sublist won't change the value in the original list.
                – kindall
                Nov 20 '18 at 19:28


















              • replacing the value of sublist won't change the value in the original list.
                – kindall
                Nov 20 '18 at 19:28
















              replacing the value of sublist won't change the value in the original list.
              – kindall
              Nov 20 '18 at 19:28




              replacing the value of sublist won't change the value in the original list.
              – kindall
              Nov 20 '18 at 19:28











              0














              If all you care about is generating the np.nan padding during the random creation, this simple one liner will suffice:



              [[np.nan]*(5-j) + list(range(j)) for j in (random.randint(1, 5) for x in range(5))]
              

              
# [[nan, 0, 1, 2, 3], 
              
#  [nan, nan, 0, 1, 2], 
              
#  [nan, nan, nan, 0, 1], 
              
#  [nan, 0, 1, 2, 3],
              
#  [0, 1, 2, 3, 4]]


              You can of course replace 5 with n and determine your max length.






              share|improve this answer


























                0














                If all you care about is generating the np.nan padding during the random creation, this simple one liner will suffice:



                [[np.nan]*(5-j) + list(range(j)) for j in (random.randint(1, 5) for x in range(5))]
                

                
# [[nan, 0, 1, 2, 3], 
                
#  [nan, nan, 0, 1, 2], 
                
#  [nan, nan, nan, 0, 1], 
                
#  [nan, 0, 1, 2, 3],
                
#  [0, 1, 2, 3, 4]]


                You can of course replace 5 with n and determine your max length.






                share|improve this answer
























                  0












                  0








                  0






                  If all you care about is generating the np.nan padding during the random creation, this simple one liner will suffice:



                  [[np.nan]*(5-j) + list(range(j)) for j in (random.randint(1, 5) for x in range(5))]
                  

                  
# [[nan, 0, 1, 2, 3], 
                  
#  [nan, nan, 0, 1, 2], 
                  
#  [nan, nan, nan, 0, 1], 
                  
#  [nan, 0, 1, 2, 3],
                  
#  [0, 1, 2, 3, 4]]


                  You can of course replace 5 with n and determine your max length.






                  share|improve this answer












                  If all you care about is generating the np.nan padding during the random creation, this simple one liner will suffice:



                  [[np.nan]*(5-j) + list(range(j)) for j in (random.randint(1, 5) for x in range(5))]
                  

                  
# [[nan, 0, 1, 2, 3], 
                  
#  [nan, nan, 0, 1, 2], 
                  
#  [nan, nan, nan, 0, 1], 
                  
#  [nan, 0, 1, 2, 3],
                  
#  [0, 1, 2, 3, 4]]


                  You can of course replace 5 with n and determine your max length.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 20 '18 at 19:36









                  Idlehands

                  3,9871417




                  3,9871417






























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                      Costa Masnaga

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                      .mw-parser-output .avviso .mbox-text-div>div,.mw-parser-output .avviso .mbox-text-full-div>div{font-size:90%}.mw-parser-output .avviso .mbox-image div{width:52px}.mw-parser-output .avviso .mbox-text-full-div .hide-when-compact{display:block} Questa voce o sezione sull'argomento Lombardia non cita le fonti necessarie o quelle presenti sono insufficienti . Puoi migliorare questa voce aggiungendo citazioni da fonti attendibili secondo le linee guida sull'uso delle fonti. Segui i suggerimenti del progetto di riferimento. Costa Masnaga comune Localizzazione Stato   Italia Regione Lombardia Provincia Lecco Amministrazione Sindaco Sabina Panzeri (Al centro il cittadino) dall'08/06/2009 Territorio Coordinate 45°46′N 9°17′E  /  45.766667°N 9.283333°E 45.766667; 9.283333  ( Costa Masnaga ) Coordinate: 45°46′N 9°17′E  /  45.766667°N 9.283333°E 45.766667; 9.283333  ( Costa Masnaga ) Altitudine 318 m s.l.m
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                      Fotorealismo

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                      Guerrino Boatto, Venice's St. Barnaba Church and canal Il fotorealismo è un genere di pittura basato sull'uso di una o più fotografie dalle quali l'artista prende le informazioni necessarie da utilizzare poi nel processo della creazione dell'opera, in genere ad olio o ad acrilico, al fine di avere un aspetto finale il più simile possibile alla fotografia. Indice 1 Storia 2 Lista dei fotorealisti 3 Note 4 Voci correlate 5 Altri progetti Storia | Il fotorealismo è un movimento che nasce negli Stati Uniti d'America alla fine degli anni sessanta [1] sulla scia della Pop Art [2] [3] [4] , e insieme condividono il carattere reazionario verso la travolgente ascesa dei media, che alla metà del ventesimo secolo si trasforma in un fenomeno di massa talmente imponente da minacciare il valore dell'immagine nell'arte. [5] [6] Ne consegue così che, colto il gesto d'inizio che fu della Pop Art, i fotorealisti aprono ad un
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                      Sidney Franklin

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                      Questa voce sull'argomento registi statunitensi è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sidney Franklin, foto di Carl Van Vechten Oscar alla memoria Irving G. Thalberg 1943 Sidney Franklin , all'anagrafe Sidney Arnold Franklin , (San Francisco, 21 marzo 1893 – Santa Monica, 18 maggio 1972), è stato un regista, produttore cinematografico e attore statunitense. Il 20 giugno 1963, si sposò con l'attrice australiana Enid Bennett. Il matrimonio durò fino alla morte dell'attrice, il 14 maggio 1969 [1] Indice 1 Filmografia 1.1 Regista 1.2 Aiuto regia (parziale) 1.3 Produttore 1.4 Attore (parziale) 2 Note 3 Altri progetti 4 Collegamenti esterni Filmografia | Regista | The Baby , co-regia Chester M. Franklin (1915) The Rivals , co-regia Chester M. Franklin (1915) Little Dick's First Case , co-regia Chester M. Franklin (1915) Her Filmland Hero , co-
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