#!/usr/bin/env python
# -*- coding: utf-8 -*-
"""
On dispose d'un exemple de chaîne ADN (constituée des symboles 'A', 'C', 'G', 'T')
Le programme utilise plusieurs techniques pour donner les nombres d'occurrences respectifs des différentes bases
"""
adn = "AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC"

# utilisation d'une liste et de la méthode .count()
bases = ["A","C","G","T"]
for base in bases:
    print(adn.count(base),)
print()

# Variante :
for c in 'ACGT':
    print(adn.count(c),)
print()

# variante un peu moins lisible
out = []
for c in 'ACGT':
    out.append(str(adn.count(c)))
print(' '.join(out))

# utilisation de la technique "list comprehension"
count = [adn.count(c) for c in 'ACGT']
for val in count:
    print(val,)
print()

# autre "list comprehension", avec impression formatée → version "one line"
print("%d %d %d %d" % tuple([adn.count(X) for X in "ACGT"]))

# count "à la main", sans utilisation de fonctions/librairie
ACGT = "ACGT"
count = [0,0,0,0]
for c in adn:
    for i in range(len(ACGT)):
        if c == ACGT[i]:
            count[i] +=1
for val in count:
    print(val,)
print()

# count "à la main", avec .index()
ACGT = "ACGT"
count = [0,0,0,0]
for c in adn:
    count[ACGT.index(c)] += 1
for val in count:
    print(val,)
print()

# utilisation de la librairie collections
from collections import defaultdict
ncount = defaultdict(int)
for c in adn:
    ncount[c] += 1
print(ncount['A'], ncount['C'], ncount['G'], ncount['T'])

# collections.Counter
from collections import Counter
for k,v in sorted(Counter(adn).items()):
    print(v,)
print()

# avec un dictionnaire
freq = {'A': 0, 'C': 0, 'G': 0, 'T': 0}
for c in adn:
    freq[c] += 1
print(freq['A'], freq['C'], freq['G'], freq['T'])

# avec un dictionnaire et count(), impression différente
dico={}
for base in bases:
    dico[base] = adn.count(base)
for key,val in dico.items():
    print("{} = {}".format(key, val))