Itertools 고급 패턴: Python에서 조합 반복자 함수의 잠재력 발휘하기

            
import itertools

base_colors = ['red', 'blue', 'green']
trim_colors = ['silver', 'gold']
accent_colors = ['white', 'black']

color_combinations = list(itertools.product(base_colors, trim_colors, accent_colors))

print(color_combinations)

            

              
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[('red', 'silver', 'white'), ('red', 'silver', 'black'), ('red', 'gold', 'white'), ('red', 'gold', 'black'), ('blue', 'silver', 'white'), ('blue', 'silver', 'black'), ('blue', 'gold', 'white'), ('blue', 'gold', 'black'), ('green', 'silver', 'white'), ('green', 'silver', 'black'), ('green', 'gold', 'white'), ('green', 'gold', 'black')]

            

              
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import itertools

players = ['Alice', 'Bob', 'Charlie', 'David']

team_lineups = list(itertools.permutations(players))

for lineup in team_lineups:
 print(lineup)

            

              
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('Alice', 'Bob', 'Charlie', 'David')
('Alice', 'Bob', 'David', 'Charlie')
('Alice', 'Charlie', 'Bob', 'David')
('Alice', 'Charlie', 'David', 'Bob')
('Alice', 'David', 'Bob', 'Charlie')
('Alice', 'David', 'Charlie', 'Bob')
('Bob', 'Alice', 'Charlie', 'David')
('Bob', 'Alice', 'David', 'Charlie')
('Bob', 'Charlie', 'Alice', 'David')
('Bob', 'Charlie', 'David', 'Alice')
('Bob', 'David', 'Alice', 'Charlie')
('Bob', 'David', 'Charlie', 'Alice')
('Charlie', 'Alice', 'Bob', 'David')
('Charlie', 'Alice', 'David', 'Bob')
('Charlie', 'Bob', 'Alice', 'David')
('Charlie', 'Bob', 'David', 'Alice')
('Charlie', 'David', 'Alice', 'Bob')
('Charlie', 'David', 'Bob', 'Alice')
('David', 'Alice', 'Bob', 'Charlie')
('David', 'Alice', 'Charlie', 'Bob')
('David', 'Bob', 'Alice', 'Charlie')
('David', 'Bob', 'Charlie', 'Alice')
('David', 'Charlie', 'Alice', 'Bob')
('David', 'Charlie', 'Bob', 'Alice')

            

              
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first_three_batters = list(itertools.permutations(players, 3))

for lineup in first_three_batters:
 print(lineup)

            

              
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import itertools

candidates = ['A', 'B', 'C', 'D', 'E']

committee_combinations = list(itertools.combinations(candidates, 3))

for committee in committee_combinations:
 print(committee)

            

              
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('A', 'B', 'C')
('A', 'B', 'D')
('A', 'B', 'E')
('A', 'C', 'D')
('A', 'C', 'E')
('A', 'D', 'E')
('B', 'C', 'D')
('B', 'C', 'E')
('B', 'D', 'E')
('C', 'D', 'E')

            

              
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import itertools

flavors = ['chocolate', 'vanilla', 'strawberry']

scoop_combinations = list(itertools.combinations_with_replacement(flavors, 2))

for combination in scoop_combinations:
 print(combination)

            

              
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('chocolate', 'chocolate')
('chocolate', 'vanilla')
('chocolate', 'strawberry')
('vanilla', 'vanilla')
('vanilla', 'strawberry')
('strawberry', 'strawberry')

            

              
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import itertools

currencies = ['USD', 'EUR', 'JPY', 'GBP', 'AUD']

exchange_pairs = list(itertools.combinations(currencies, 2))

for pair in exchange_pairs:
 print(pair)

            

              
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import itertools

# This is a simplified example, route optimization usually involves distance calculations
cities = ['London', 'Tokyo', 'New York', 'Sydney']

possible_routes = list(itertools.permutations(cities))

# In a real-world scenario, you would calculate the total distance for each route
# and select the shortest one.

for route in possible_routes:
 print(route)

            

              
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import itertools

# Example product configuration options
regions = ['North America', 'Europe', 'Asia']
languages = ['English', 'French', 'Japanese']
currencies = ['USD', 'EUR', 'JPY']

product_configurations = list(itertools.product(regions, languages, currencies))

for config in product_configurations:
 print(config)

            

              
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import itertools

numbers = [1, 2, 3, 4]

# Calculate the sum of squares for each combination of two numbers
def sum_of_squares(x, y):
 return x**2 + y**2

combinations = itertools.combinations(numbers, 2)

results = list(itertools.starmap(sum_of_squares, combinations))

print(results)

            

              
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import itertools

numbers = [1, 2, 3, 4, 5, 6]

# Generate combinations of three numbers where the sum is greater than 10
combinations = itertools.combinations(numbers, 3)

filtered_combinations = [comb for comb in combinations if sum(comb) > 10]

print(filtered_combinations)

            

              
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import itertools

# Process combinations in chunks
def process_combinations(data, chunk_size):
 iterator = itertools.combinations(data, 2)
 while True:
 chunk = list(itertools.islice(iterator, chunk_size))
 if not chunk:
 break
 # Process the chunk
 for combination in chunk:
 print(combination)

large_data = range(1000)
process_combinations(large_data, 100)

            

              
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