Introduction to Random Strings (ROSALIND PROB)

If I want to find the propobility that string s and some random string are equal, it'll 0.25 to the power of s length. Here I have not a random string, and it's not 0.25, but some other number, which is easy to find from gc-content.
Hint in this problem is very useful.

Given: A DNA string s of length at most 100 bp and an array A
containing at most 20 numbers between 0 and 1.
Return: An array B
having the same length as A in which B[k] represents the common logarithm of the probability that a random string constructed with the GC-content found in A[k] will match s exactly.
Hint: One property of the logarithm function is that for any positive numbers x and y, log10(x⋅y)=log10(x)+log10(y).


from __future__ import division
import re
import sys
import math

def atcg_prob(x):
 cg_prob = float(x)
 at_prob = 1 - cg_prob
 atcg_prob = {}
 atcg_prob['A'] = at_prob / 2
 atcg_prob['T'] = atcg_prob['A']
 atcg_prob['C'] = cg_prob / 2
 atcg_prob['G'] = atcg_prob['C']
 return atcg_prob

def main():
 if len(sys.argv) > 1:
  res = ''
  i = 2
  while i < len(sys.argv):
   cont = atcg_prob(sys.argv[i])
   prob = 0
   for nuk in sys.argv[1]:
    prob = prob + math.log(cont[nuk], 10)
   #res = res + str(math.log(prob, 10)) + ' '
   res = res + str(round(prob, 3)) + ' ' 
   i += 1
  print res
 else:
  print 'Enter datas!'

if __name__ == '__main__':
 main()

Хожу на химфак, слушаю лекции.

Хожу на химический факультет, слушаю органику. Многое понимаю, но сделать ничего не могу. Хотя было бы странно, если бы что-то могла, позанимавшись химией всего 60 часов. На занятиях очень интересно, прям как в кино. Тут пример того, что рассказывают.
Как получить фенолфталеин из фенола и фталевого ангидрида.
Вообще не очень понятно, почему атакуется только верхний кислород, а нижний нет. Но такая уж она химия - волшебная наука.

А тут показано почему фенолфталеин меняет окраску в щелочной среде. Как видно, от щелочи появляется много двойных связей по всей молекуле. Они то и меняет оптические свойства, так что щелочной среде он становится малиновым. Кстати, в кислой он розовый. 

Overlap Graphs (ROSALIND GRPH)

Given: A collection of DNA strings in FASTA format having total length at most 10 kbp.
Return: The adjacency list corresponding to O3. You may return edges in any order.

import re

f = open('12.txt', 'r')
reads = re.findall(r'(Rosalind_[0-9]+)\n(([A-T]+\n)+)', f.read())
suffix = {}
prefix = {}
for s in reads:
 string = s[1].replace('\n', '')
 if len(string) > 3:
  head = string[:3]
  if head in prefix:
   prefix[head].append(s[0])
  else:
   prefix[head] = [s[0]]
  tail = string[-3:]
  if tail in suffix:
   suffix[tail].append(s[0])
  else:
   suffix[tail] = [s[0]]
for rec in suffix:
 if rec in prefix:
  i = 0
  while i < len(suffix[rec]):
   j = 0
   while j < len(prefix[rec]):
    if suffix[rec][i] != prefix[rec][j]:
     print suffix[rec][i] + ' ' + prefix[rec][j]
    j += 1
   i += 1

Consensus and Profile (ROSALIND CONS)

numpy helps!
Given: A collection of at most 10 DNA strings of equal length (at most 1 kbp) in FASTA format.
Return: A consensus string and profile matrix for the collection. (If several possible consensus strings exist, then you may return any one of them.)

import numpy as np
import re

f = open('11.txt', 'r')
strings = re.findall(r'(>Rosalind_[0-9]+)\n(([A-T]+\n)+)', f.read())
str_len = len(strings[0][1].replace('\n', ''))
profile = np.zeros((4, str_len))
for s in strings:
 counter = 0
 str_data = np.zeros((4, str_len))
 st = s[1].replace('\n', '')
 for i in st:
  if i == 'A':
   str_data[0,counter] = 1
  elif i == 'C':
   str_data[1,counter] = 1
  elif i == 'G':
   str_data[2,counter] = 1
  elif i == 'T':
   str_data[3,counter] = 1
  counter += 1
 profile = profile + str_data
consensus = ''
position = 0
while position < str_len:
 column = profile[:, position]
 column_max = column.max()
 nucleotides = ['A', 'C', 'G', 'T']
 i = 0
 while i < 4:
  if column[i] == column_max:
   consensus = consensus + nucleotides[i]
   break
  i += 1
 position += 1
print consensus
A_line = ''
C_line = ''
G_line = ''
T_line = ''
j = 0
while j < str_len:
 A_line = A_line + str(int(profile[0, j])) + ' '
 C_line = C_line + str(int(profile[1, j])) + ' '
 G_line = G_line + str(int(profile[2, j])) + ' '
 T_line = T_line + str(int(profile[3, j])) + ' '
 j += 1
print 'A: ' + A_line
print 'C: ' + C_line
print 'G: ' + G_line
print 'T: ' + T_line

Translating RNA into Protein (ROSALIND PROT)

Given: An RNA string s corresponding to a strand of mRNA (of length at most 10 kbp).
Return: The protein string encoded by s.

import sys

def codon_table():
 nukaa = {}
 nukaa['UUU'] = 'F'
 nukaa['CUU'] = 'L'
 nukaa['AUU'] = 'I'
 nukaa['GUU'] = 'V'
 nukaa['UUC'] = 'F'
 nukaa['CUC'] = 'L'
 nukaa['AUC'] = 'I'
 nukaa['GUC'] = 'V'
 nukaa['UUA'] = 'L'
 nukaa['CUA'] = 'L'
 nukaa['AUA'] = 'I'
 nukaa['GUA'] = 'V'
 nukaa['UUG'] = 'L'
 nukaa['CUG'] = 'L'
 nukaa['AUG'] = 'M'
 nukaa['GUG'] = 'V'
 nukaa['UCU'] = 'S'
 nukaa['CCU'] = 'P'
 nukaa['ACU'] = 'T'
 nukaa['GCU'] = 'A'
 nukaa['UCC'] = 'S'
 nukaa['CCC'] = 'P'
 nukaa['ACC'] = 'T'
 nukaa['GCC'] = 'A'
 nukaa['UCA'] = 'S'
 nukaa['CCA'] = 'P'
 nukaa['ACA'] = 'T'
 nukaa['GCA'] = 'A'
 nukaa['UCG'] = 'S'
 nukaa['CCG'] = 'P'
 nukaa['ACG'] = 'T'
 nukaa['GCG'] = 'A'
 nukaa['UAU'] = 'Y'
 nukaa['CAU'] = 'H'
 nukaa['AAU'] = 'N'
 nukaa['GAU'] = 'D'
 nukaa['UAC'] = 'Y'
 nukaa['CAC'] = 'H'
 nukaa['AAC'] = 'N'
 nukaa['GAC'] = 'D'
 nukaa['CAA'] = 'Q'
 nukaa['AAA'] = 'K'
 nukaa['GAA'] = 'E'
 nukaa['CAG'] = 'Q'
 nukaa['AAG'] = 'K'
 nukaa['GAG'] = 'E'
 nukaa['UGU'] = 'C'
 nukaa['CGU'] = 'R'
 nukaa['AGU'] = 'S'
 nukaa['GGU'] = 'G'
 nukaa['UGC'] = 'C'
 nukaa['CGC'] = 'R'
 nukaa['AGC'] = 'S'
 nukaa['GGC'] = 'G'
 nukaa['CGA'] = 'R'
 nukaa['AGA'] = 'R'
 nukaa['GGA'] = 'G'
 nukaa['UGG'] = 'W'
 nukaa['CGG'] = 'R'
 nukaa['AGG'] = 'R'
 nukaa['GGG'] = 'G'
 return nukaa


def main():
 if len(sys.argv) > 1:
  rna_string = sys.argv[1]
  nukaa = codon_table()
  nucl_str_len = len(rna_string) // 3
  nucl_string = ''
  i = 0
  while i < nucl_str_len:
   codon = rna_string[3*i:3*(i+1)]
   if codon=='UAA' or codon=='UAG' or codon=='UGA':
    break
   else:
     nucl_string += nukaa[codon]
   i+=1
  print nucl_string
 else:
  print 'Enter RNA sequence.'

if __name__ == '__main__':
 main()

Finding a Motif in DNA (ROSALIND SUBS)

Given: Two DNA strings s and t (each of length at most 1 kbp).
Return: All locations of t as a substring of s.

s = 'TCACTCGATTGGAACCGA'
t = 'CACTCGACA'
occurence = []
start_position = 0
while True:
 start_position = s.find(t, start_position+1)
 if start_position == -1:
  break
 occurence.append(start_position+1)
res = ''
for i in occurence:
 res += str(i) + ' '
print res

Mendel's First Love (ROSALIND IPRB)

Given: Three positive integers k, m, and n, representing a population containing k+m+n organisms: k individuals are homozygous dominant for a factor, m are heterozygous, and n are homozygous recessive.
Return: The probability that two randomly selected mating organisms will produce an individual possessing a dominant allele (and thus displaying the dominant phenotype). Assume that any two organisms can mate.

import sys

def main():
 if len(sys.argv) > 1:
  k = int(sys.argv[1])
  m = int(sys.argv[2])
  n = int(sys.argv[3])
  T = k + m + n
  Z = T*(T - 1)
  print round(1-(n*(n-1) + 0.25*m*(m-1) + m*n)/Z, 5)
 else:
  print 'Enter k, m, n.'

if __name__ == '__main__':
 main()