컴프리헨션과 내장 함수

리스트 컴프리헨션: 간결한 필터링

# Traditional loop
squares = []
for n in range(1, 6):
    squares.append(n * n)
print(squares)  # [1, 4, 9, 16, 25]

# List comprehension
squares = [n * n for n in range(1, 6)]
print(squares)  # [1, 4, 9, 16, 25]

# With filter
evens = [n for n in range(10) if n % 2 == 0]
print(evens)    # [0, 2, 4, 6, 8]

2차원 격자를 위한 중첩 컴프리헨션

# WRONG: all rows are the same object!
bad = [[0] * 3] * 3
bad[0][0] = 9
print(bad)  # [[9,0,0],[9,0,0],[9,0,0]]  oops!

# CORRECT: each row is a separate list
good = [[0] * 3 for _ in range(3)]
good[0][0] = 9
print(good)  # [[9,0,0],[0,0,0],[0,0,0]]

딕셔너리와 집합 컴프리헨션

# Dict comprehension: square lookup
sq_map = {n: n**2 for n in range(1, 6)}
print(sq_map)  # {1:1, 2:4, 3:9, 4:16, 5:25}

# Set comprehension: unique lengths
words = ['cat', 'dog', 'elephant', 'ant']
unique_lengths = {len(w) for w in words}
print(unique_lengths)  # {3, 8}  (order varies)

제너레이터 표현식: 메모리 효율적 사용

# List comprehension builds all values at once
total = sum([n**2 for n in range(1_000_000)])

# Generator yields one at a time — lower memory
total = sum(n**2 for n in range(1_000_000))
print(total)  # 333332833333500000

# any/all with generators short-circuit early
nums = [4, 6, 8, 3, 10]
has_odd = any(n % 2 == 1 for n in nums)
print(has_odd)  # True  (stops at 3)

map()과 filter(): 함수형 방식

nums = [1, 2, 3, 4, 5]

# map: apply function to each element
doubled = list(map(lambda n: n * 2, nums))
print(doubled)  # [2, 4, 6, 8, 10]

# filter: keep elements passing predicate
evens = list(filter(lambda n: n % 2 == 0, nums))
print(evens)    # [2, 4]

# Equivalent comprehensions (often preferred)
doubled = [n * 2 for n in nums]
evens   = [n for n in nums if n % 2 == 0]

zip(): 시퀀스 쌍 만들기

keys   = ['a', 'b', 'c']
values = [1, 2, 3]

pairs = list(zip(keys, values))
print(pairs)  # [('a',1), ('b',2), ('c',3)]

# Build dict from two lists
d = dict(zip(keys, values))
print(d)      # {'a':1, 'b':2, 'c':3}

# Transpose a matrix
matrix = [[1,2,3],[4,5,6],[7,8,9]]
transposed = [list(row) for row in zip(*matrix)]
print(transposed)  # [[1,4,7],[2,5,8],[3,6,9]]

enumerate(): 인덱스와 값 함께 사용하기

fruits = ['apple', 'banana', 'cherry']

# Instead of: for i in range(len(fruits)):
for i, fruit in enumerate(fruits):
    print(i, fruit)
# 0 apple / 1 banana / 2 cherry

# Start from 1
for i, fruit in enumerate(fruits, 1):
    print(f'{i}. {fruit}')
# 1. apple / 2. banana / 3. cherry

키 함수를 사용한 sorted()

# Sort by second element of tuple
intervals = [(1,3),(2,1),(0,5)]
print(sorted(intervals, key=lambda x: x[1]))
# [(2,1),(1,3),(0,5)]

# Sort strings case-insensitively
words = ['Banana', 'apple', 'Cherry']
print(sorted(words, key=str.lower))
# ['apple', 'Banana', 'Cherry']

# Sort by multiple keys: first by length, then alphabetically
words = ['fig', 'apple', 'ant', 'kiwi']
print(sorted(words, key=lambda w: (len(w), w)))
# ['ant', 'fig', 'kiwi', 'apple']

키를 사용한 min()과 max()

words = ['banana', 'fig', 'strawberry', 'kiwi']

longest = max(words, key=len)
print(longest)   # strawberry

shortest = min(words, key=len)
print(shortest)  # fig

# Find interval with earliest end
intervals = [(2,6),(1,3),(4,5)]
earlist_end = min(intervals, key=lambda x: x[1])
print(earlist_end)  # (1, 3)

빠른 단락 평가 검사를 위한 any()와 all()

nums = [2, 4, 6, 7, 8]

all_even = all(n % 2 == 0 for n in nums)
print(all_even)  # False  (7 is odd)

has_large = any(n > 5 for n in nums)
print(has_large) # True  (6 qualifies, stops there)

# Practical: check if sudoku row has no duplicates
row = [1, 2, 3, 4, 5, 6, 7, 8, 9]
valid = all(1 <= n <= 9 for n in row) and len(set(row)) == 9
print(valid)  # True

sum(), abs(), divmod()

# sum with generator
print(sum(n**2 for n in range(1, 6)))  # 55

# abs for distance problems
print(abs(-7))   # 7

# divmod for digit extraction
num = 1234
digits = []
while num:
    num, d = divmod(num, 10)
    digits.append(d)
digits.reverse()
print(digits)  # [1, 2, 3, 4]

간단한 확인

레슨 요약