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How to loop to make many .csv files?

P: 27
Hey guy!
I have a couple questions, which I don't think are too advanced. I created an algorithm called "Best" for simplicity I will not include the mathematical details of it. I ask the user to input 6 values: Stock, min,win,Stop,GO, and end. Looking at the last 5 inputs, they are all integer inputs eg: 1,2,3.
This code will take my algorithm and spit out a csv file called "boom.csv" with two columns of numbers based on the input values. My questions are as follow:

-Can a loop be made so i input 3 integers for each user input, then spit out one .csv file for each combination of user input values?
I know I would need to some how create a way to name the .csv files for each combination of user inputs, but how would I do that?
Also how would I make a loop that can look at each combination of the user input values?
So with 3 integers per 5 input values, that should be 3^5 outputs.

Thank you for any help!
Ron Parker

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  1. #!/usr/bin/python
  2. import  numpy
  3. import csv
  4.  
  5.  
  6.  
  7. def Best(seq):
  8.     ALGORITHM
  9.  
  10. Stock=raw_input('Stock Symbol: ' )
  11.  
  12. min=raw_input('Start time (min away from open) : ' )
  13. win=raw_input('Window (min): ' )
  14. Stop=raw_input('SL: ' )
  15. Go=raw_input('PL: ' )
  16. end=raw_input('End time (min away from open) : ' )
  17.  
  18.  
  19. Entry=[ Best(something1[ i:i+120]) for i in range(0, len(something1), 120)]
  20. Entry2=[float(item[0]) for item in Entry]
  21. Entry3=[float(item[1]) for item in Entry]
  22.  
  23.  
  24.  
  25. myFile= open( "Boom.csv", "wb" )
  26. wtr= csv.writer( myFile )
  27. for Start, Exit, in zip(Entry2,Entry3):
  28.     aRow=Start
  29.     bRow=Exit
  30.     cRow=[aRow,bRow]
  31.     wtr.writerow(cRow)
  32.  
  33. myFile.close()
  34.  
Jan 11 '11 #1
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1 Reply


bvdet
Expert Mod 2.5K+
P: 2,851
I think you need a way to calculate all the possible permutations. A function that returns the permutations can be found here.

If you have a list of letters, a file name can be constructed thus:
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  1. >>> ext = "csv"
  2. >>> seq = ['a','b','c','d','e']
  3. >>> '.'.join(["".join(seq), ext])
  4. 'abcde.csv'
  5. >>> 
OR
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  1. >>> "%s.%s" % ("".join(seq), ext)
  2. 'abcde.csv'
  3. >>> 
Jan 12 '11 #2

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