Harvard:Biophysics 101/2007/Notebook:Katie Fifer/2007-2-8: Difference between revisions

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== Script ==
   #!/usr/bin/env python
   #!/usr/bin/env python
    
    
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== Output ==
== Output ==
* program run to generate 10,000 strings
   H 49831
   H 49831
   HH 22372
   HH 22372
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== Testing ==
== Testing ==
* Output for just 5 strings so you can double check by hand
  all strings: ['HHTTHTHTTT', 'THTTTTTTTH', 'TTTHHTHTHT', 'HTHTHTHTTH', 'HTTHHHTTTT']
  H 19
  HH 4
  HHH 1
  HHHH 0
  HHHHH 0
  HHHHHH 0
  HHHHHHH 0
  HHHHHHHH 0
  HHHHHHHHH 0
  HHHHHHHHHH 0
  T 31
  TT 16
  TTT 9
  TTTT 5
  TTTTT 3
  TTTTTT 2
  TTTTTTT 1
  TTTTTTTT 0
  TTTTTTTTT 0
  TTTTTTTTTT0

Revision as of 15:19, 7 February 2007

Script

 #!/usr/bin/env python
 
 # Katie Fifer 
 # asst2.py 
 # 2/7/07 
 # Description: A script to generate 10,000 strings of 10 random
 # coinflips (H or T) and outputs the tally of contiguous (overlapping
 # stretches of 2,3,4,5,6,7,8,9, and 10 H's or T's in that set of
 # 10,000 10-mers
 
 import random
 
 # set constants
 num_strings = 10000
 num_flips = 10
 max_repeat = 10
 all_strings = [ ]
 
 # random number generation
 
 # generate a list of new strings
 for i in range(num_strings):
     new_string = .join([random.choice(['H','T']) for n in range (num_flips)])
     all_strings.append(new_string)
 
 # figure out how many overlapping stretches of H's there are. will do
 # this for each string for each substring. in other words will find
 # all instances of 'HH' in each of the strings, and then all instances
 # of 'HHH' in each of the strings etc.
 
 def analyze (letter):
     for i in range(max_repeat):
 	  # generate the substring to search for. the i + 1 is to account
 	  # for the fact that i starts at 0
 	  substr = .join([letter for n in range (i + 1)])
 	  # for each of the strings in the list, find the number of
 	  # instances of the substring just set (overlapping)
 	  total = 0
 	  for j in range(num_strings):
 	      curr_string = all_strings[j]
 	      count = 0
 	      pos = curr_string.find(substr, 0)
 	      while not pos == -1:
 		  count = count + 1
 		  total = total + 1
 		  pos = curr_string.find(substr, pos + 1)
 	  print substr, total
   
 analyze('H')
 analyze('T')


Output

  • program run to generate 10,000 strings
 H 49831
 HH 22372
 HHH 9860
 HHHH 4232
 HHHHH 1813
 HHHHHH 754
 HHHHHHH 313
 HHHHHHHH 127
 HHHHHHHHH 44
 HHHHHHHHHH 8
 T 50169
 TT 22622
 TTT 10065
 TTTT 4401
 TTTTT 1937
 TTTTTT 824
 TTTTTTT 341
 TTTTTTTT 122
 TTTTTTTTT 37
 TTTTTTTTTT 7

Testing

  • Output for just 5 strings so you can double check by hand
 all strings: ['HHTTHTHTTT', 'THTTTTTTTH', 'TTTHHTHTHT', 'HTHTHTHTTH', 'HTTHHHTTTT']
 H 19
 HH 4
 HHH 1
 HHHH 0
 HHHHH 0
 HHHHHH 0
 HHHHHHH 0
 HHHHHHHH 0
 HHHHHHHHH 0
 HHHHHHHHHH 0
 T 31
 TT 16
 TTT 9
 TTTT 5
 TTTTT 3
 TTTTTT 2
 TTTTTTT 1
 TTTTTTTT 0
 TTTTTTTTT 0
 TTTTTTTTTT0