Python列表推导式

print('\n'.join([''.join(['%s*%s=%-2s '%(y,x,x*y)for y in range(1,x+1)])for x in range(1,10)]))

print('\n'.join([' '.join([f"{j}x{i}={i*j}" for j in range(1, i + 1)]) for i in range(1, 10)]))

# 计算0~9所有偶数平方
squares = []
for x in range(10):
    if x%2 == 0:
        squares.append(x**2)

squares

[0, 4, 16, 36, 64]

squares = [x**2 for x in range(10) if x%2==0]

squares

[0, 4, 16, 36, 64]

# 注意，这里的x和y没有关系
[(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]

[(1, 3), (1, 4), (2, 3), (2, 1), (2, 4), (3, 1), (3, 4)]

names = [['Tom', 'Billy', 'Jefferson', 'Andrew', 'Wesley', 'Steven', 'Joe'],
         ['Alice', 'Jill', 'Ana', 'Wendy', 'Jennifer', 'Sherry', 'Eva']]

print([name for lst in names for name in lst if name.count('e') >= 2])  # 注意遍历顺序，这是实现的关键

[(str(x), x**2) for x in range(6)]

[('0', 0), ('1', 1), ('2', 4), ('3', 9), ('4', 16), ('5', 25)]

# 注意，这里的x和y是有关系的
vec = [[1,2,3], [4,5,6], [7,8,9]]
[y for x in vec for y in x]

[1, 2, 3, 4, 5, 6, 7, 8, 9]

result = ["偶" if i % 2 == 0 else "奇" for i in range(1, 10+1) ]
print(result)

# 返回的是去重元素的集合
a = {x for x in 'abracadabra' if x not in 'abc'}

a

{'d', 'r'}

# 返回的是一个新的字典
{str(x): x**2 for x in (2, 4, 6)}

{'2': 4, '4': 16, '6': 36}

mcase = {'a': 10, 'b': 34}
mcase_frequency = {mcase[k]: k for k in mcase}
print(mcase_frequency)