Dictionnaires et ensembles en Python

Dictionnaires Python : associations clé-valeur

d = {'apple': 3, 'banana': 5}
print(d['apple'])   # 3
d['cherry'] = 7
print(len(d))       # 3
print('banana' in d)  # True
del d['apple']
print(d)            # {'banana': 5, 'cherry': 7}

Recherche sûre avec .get()

freq = {}
words = ['the', 'cat', 'sat', 'on', 'the', 'mat']
for w in words:
    freq[w] = freq.get(w, 0) + 1
print(freq)
# {'the': 2, 'cat': 1, 'sat': 1, 'on': 1, 'mat': 1}

print(freq.get('dog', 0))  # 0  (no KeyError)

defaultdict pour regrouper plus simplement

from collections import defaultdict

groups = defaultdict(list)
words = ['eat', 'tea', 'tan', 'ate', 'nat', 'bat']
for w in words:
    key = ''.join(sorted(w))  # canonical anagram key
    groups[key].append(w)

print(list(groups.values()))
# [['eat','tea','ate'], ['tan','nat'], ['bat']]

Counter : tables de fréquences rapides

from collections import Counter

c = Counter('abracadabra')
print(c)           # Counter({'a':5,'b':2,'r':2,'c':1,'d':1})
print(c.most_common(2))  # [('a', 5), ('b', 2)]

# Valid anagram check
def is_anagram(s, t):
    return Counter(s) == Counter(t)

print(is_anagram('anagram', 'nagaram'))  # True

Ensembles Python : collections uniques non ordonnées

seen = set()
nums = [1, 2, 3, 2, 1, 4]
duplicates = []
for n in nums:
    if n in seen:          # O(1) check
        duplicates.append(n)
    seen.add(n)
print(duplicates)  # [2, 1]
print(len(seen))   # 4  (unique values)

Opérations sur les ensembles pour les entretiens

a = {1, 2, 3, 4}
b = {3, 4, 5, 6}

print(a | b)  # {1, 2, 3, 4, 5, 6}  union
print(a & b)  # {3, 4}              intersection
print(a - b)  # {1, 2}              difference
print(a ^ b)  # {1, 2, 5, 6}        symmetric diff

Vérification d’appartenance : liste ou ensemble

word_list = ['apple', 'banana', 'cherry', 'date']
word_set  = set(word_list)

# O(n) per check
print('banana' in word_list)  # True

# O(1) per check
print('banana' in word_set)   # True

# Practical example: find common elements
a = [1, 2, 3, 4, 5]
b = [3, 4, 5, 6, 7]
common = [x for x in a if x in set(b)]
print(common)  # [3, 4, 5]

Parcourir les dictionnaires : clés, valeurs et éléments

scores = {'Alice': 90, 'Bob': 75, 'Carol': 88}

for name, score in scores.items():
    print(f'{name}: {score}')

# Find key with max value
best = max(scores, key=scores.get)
print(best)  # Alice

# Safe deletion
to_del = [k for k, v in scores.items() if v < 80]
for k in to_del:
    del scores[k]
print(scores)  # {'Alice': 90, 'Carol': 88}

Frozenset : ensembles hachables

from collections import defaultdict

words = ['eat', 'tea', 'tan', 'ate', 'nat', 'bat']
groups = defaultdict(list)
for w in words:
    key = frozenset(w)  # hashable; 'eat','tea','ate' all share same key
    groups[key].append(w)

print([sorted(g) for g in groups.values()])
# [['ate','eat','tea'], ['nat','tan'], ['bat']]

Compréhensions de dictionnaires pour transformer les données

# Invert a dict
original = {'a': 1, 'b': 2, 'c': 3}
inverted = {v: k for k, v in original.items()}
print(inverted)  # {1:'a', 2:'b', 3:'c'}

# Filter by value
scores = {'Alice': 90, 'Bob': 55, 'Carol': 78}
passing = {k: v for k, v in scores.items() if v >= 60}
print(passing)  # {'Alice': 90, 'Carol': 78}

Plus longue séquence consécutive

def longest_consecutive(nums):
    num_set = set(nums)
    best = 0
    for n in num_set:
        if n - 1 not in num_set:  # start of sequence
            cur = n
            streak = 1
            while cur + 1 in num_set:
                cur += 1
                streak += 1
            best = max(best, streak)
    return best

print(longest_consecutive([100,4,200,1,3,2]))  # 4 (1,2,3,4)

Vérification rapide

Récapitulatif de la leçon