Displaying elements of a list, in columns












1












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I have a list of strings. I need to list them in rows and columns. Each row should not have more than "cols" number of values. Each of the values in a given row should be "step" away from the previous value. The values should appear only once in the output. Here is what I have. Any better way to write this code?



cols = 4
step = 10
vlist = ["Value" + str(i+1) for i in range(100)]

vlen = len(vlist)
start = 0
while start < vlen and start < step:
num = 0
for idx in range(start, vlen, step):
if num < cols:
print(vlist[idx], end=", ")
num += 1
print("n")
start += 1









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  • $begingroup$
    Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
    $endgroup$
    – greybeard
    5 hours ago
















1












$begingroup$


I have a list of strings. I need to list them in rows and columns. Each row should not have more than "cols" number of values. Each of the values in a given row should be "step" away from the previous value. The values should appear only once in the output. Here is what I have. Any better way to write this code?



cols = 4
step = 10
vlist = ["Value" + str(i+1) for i in range(100)]

vlen = len(vlist)
start = 0
while start < vlen and start < step:
num = 0
for idx in range(start, vlen, step):
if num < cols:
print(vlist[idx], end=", ")
num += 1
print("n")
start += 1









share|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
    $endgroup$
    – greybeard
    5 hours ago














1












1








1





$begingroup$


I have a list of strings. I need to list them in rows and columns. Each row should not have more than "cols" number of values. Each of the values in a given row should be "step" away from the previous value. The values should appear only once in the output. Here is what I have. Any better way to write this code?



cols = 4
step = 10
vlist = ["Value" + str(i+1) for i in range(100)]

vlen = len(vlist)
start = 0
while start < vlen and start < step:
num = 0
for idx in range(start, vlen, step):
if num < cols:
print(vlist[idx], end=", ")
num += 1
print("n")
start += 1









share|improve this question









New contributor




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







$endgroup$




I have a list of strings. I need to list them in rows and columns. Each row should not have more than "cols" number of values. Each of the values in a given row should be "step" away from the previous value. The values should appear only once in the output. Here is what I have. Any better way to write this code?



cols = 4
step = 10
vlist = ["Value" + str(i+1) for i in range(100)]

vlen = len(vlist)
start = 0
while start < vlen and start < step:
num = 0
for idx in range(start, vlen, step):
if num < cols:
print(vlist[idx], end=", ")
num += 1
print("n")
start += 1






python python-3.x formatting






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RebornCodeLover 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









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Check out our Code of Conduct.









share|improve this question




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edited 5 hours ago









200_success

130k17153419




130k17153419






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asked 6 hours ago









RebornCodeLoverRebornCodeLover

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New contributor





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






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Check out our Code of Conduct.












  • $begingroup$
    Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
    $endgroup$
    – greybeard
    5 hours ago


















  • $begingroup$
    Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
    $endgroup$
    – greybeard
    5 hours ago
















$begingroup$
Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
$endgroup$
– greybeard
5 hours ago




$begingroup$
Welcome to Code Review! Is Each [value] should be "step" away from the previous value an "external" requirement? ((Block-) Quote the specification of the result to achieve.) Another interpretation is In a monospace font, each value shall be output 10 places to the right of the preceding one, the advantage being all values getting displayed.
$endgroup$
– greybeard
5 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.



from itertools import islice

cols = 4
step = 10
vlist = ["Value" + str(i+1) for i in range(100)]

for row in (islice(vlist, i, cols * step, step) for i in range(step)):
print(', '.join(row), end=", nn")





share|improve this answer











$endgroup$





















    0












    $begingroup$

    The code can be made more understandable by:




    • introducing row and column indices

    • replace the while loop with a for loop

    • calculate the index for vlist from the values of the row/col indices


    This reduces the number of help variables needed and could result in something like this:



    vlist = ["Value" + str(i+1) for i in range(100)]

    cols = 4
    rows = 10
    for row_idx in range(rows):
    for col_idx in range(cols):

    idx = row_idx + rows * col_idx
    print(vlist[idx], end=", ")

    print("n")





    share|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.



      from itertools import islice

      cols = 4
      step = 10
      vlist = ["Value" + str(i+1) for i in range(100)]

      for row in (islice(vlist, i, cols * step, step) for i in range(step)):
      print(', '.join(row), end=", nn")





      share|improve this answer











      $endgroup$


















        2












        $begingroup$

        Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.



        from itertools import islice

        cols = 4
        step = 10
        vlist = ["Value" + str(i+1) for i in range(100)]

        for row in (islice(vlist, i, cols * step, step) for i in range(step)):
        print(', '.join(row), end=", nn")





        share|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.



          from itertools import islice

          cols = 4
          step = 10
          vlist = ["Value" + str(i+1) for i in range(100)]

          for row in (islice(vlist, i, cols * step, step) for i in range(step)):
          print(', '.join(row), end=", nn")





          share|improve this answer











          $endgroup$



          Fancy iteration in Python is often made easier using the itertools module. For this case, itertools.islice() can help pick out the values for each row.



          from itertools import islice

          cols = 4
          step = 10
          vlist = ["Value" + str(i+1) for i in range(100)]

          for row in (islice(vlist, i, cols * step, step) for i in range(step)):
          print(', '.join(row), end=", nn")






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 4 hours ago

























          answered 4 hours ago









          200_success200_success

          130k17153419




          130k17153419

























              0












              $begingroup$

              The code can be made more understandable by:




              • introducing row and column indices

              • replace the while loop with a for loop

              • calculate the index for vlist from the values of the row/col indices


              This reduces the number of help variables needed and could result in something like this:



              vlist = ["Value" + str(i+1) for i in range(100)]

              cols = 4
              rows = 10
              for row_idx in range(rows):
              for col_idx in range(cols):

              idx = row_idx + rows * col_idx
              print(vlist[idx], end=", ")

              print("n")





              share|improve this answer









              $endgroup$


















                0












                $begingroup$

                The code can be made more understandable by:




                • introducing row and column indices

                • replace the while loop with a for loop

                • calculate the index for vlist from the values of the row/col indices


                This reduces the number of help variables needed and could result in something like this:



                vlist = ["Value" + str(i+1) for i in range(100)]

                cols = 4
                rows = 10
                for row_idx in range(rows):
                for col_idx in range(cols):

                idx = row_idx + rows * col_idx
                print(vlist[idx], end=", ")

                print("n")





                share|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The code can be made more understandable by:




                  • introducing row and column indices

                  • replace the while loop with a for loop

                  • calculate the index for vlist from the values of the row/col indices


                  This reduces the number of help variables needed and could result in something like this:



                  vlist = ["Value" + str(i+1) for i in range(100)]

                  cols = 4
                  rows = 10
                  for row_idx in range(rows):
                  for col_idx in range(cols):

                  idx = row_idx + rows * col_idx
                  print(vlist[idx], end=", ")

                  print("n")





                  share|improve this answer









                  $endgroup$



                  The code can be made more understandable by:




                  • introducing row and column indices

                  • replace the while loop with a for loop

                  • calculate the index for vlist from the values of the row/col indices


                  This reduces the number of help variables needed and could result in something like this:



                  vlist = ["Value" + str(i+1) for i in range(100)]

                  cols = 4
                  rows = 10
                  for row_idx in range(rows):
                  for col_idx in range(cols):

                  idx = row_idx + rows * col_idx
                  print(vlist[idx], end=", ")

                  print("n")






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 5 hours ago









                  Jan KuikenJan Kuiken

                  86838




                  86838






















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