• while for循环


    x = 'spam'
    while x:
    print(x, end=' ')
    x = x[1:] # spam pam am m

    break 跳出最近所在的循环

    continue 跳到所在循环的开头处

    pass 啥都不做

    def func():
    pass

    def func():
    ...

    continue 跳到所在循环的开头处

    x = 10
    while x:
    x = x-1
    if x % 2 ==0:continue
    print(x, end=' ') # 9 7 5 3 1

    判断是不是质数

    x = y //2

    while x > 1:

    if y % x == 0:

    print(y, 'has factor', x)

    break

    x -= 1

    else:

    print(y, 'is prime')

    for i in 'spam':
    print(i, end='') # spam

    sum = 0
    for i in [1, 2, 3]:
    sum += i
    print(sum) # 6

    sum = 1

    for i in [1, 2, 3, 4]:sum *= i
    print(sum) # 24

    for循环中的元组赋值和解包

    for (i, j) in ((1, 2),(3, 4)): # 1 2
    print(i, j) # 3 4

    d = {'a':1,'b':2,'c':3} # 'a' 1
    for i in d: # 'b' 2
    print(i, d[i]) # 'c' 3

    for (i, j) in d.items(): # 'a' 1
    print(i, j) # 'b' 2
    # 'c' 3

    t = [(1, 2), (3, 4), (5, 6)]
    for both in t: # 1, 2
    a, b = both # 3, 4
    print(a, b) # 5, 6

    for ((a, b), c) in [((1, 2),3), ((4, 5), 6)]: # 1 2 3
    print(a, b, c) # 4 5 6

    for ((a, b), c) in [((1, 2),3), ('xy', 6)]: # 1 2 3
    print(a, b, c) # x y 6

    a, *b, c = (1, 2, 3, 4)
    print(a, b, c) # 1 [2, 3] 4

    for (a, *b, c) in [(1, 2, 3, 4), (5, 6, 7, 8)]:
    print(a, b, c)

    1 [2, 3] 4

    5 [6, 7] 8

    items = ['aaa', 111, (4, 5), 2.01]
    tests = [(4, 5), 3.14]
    for key in tests:
    for item in items:
    if item == key:
    print(key, 'was found')
    break
    else:
    print(key, 'not found')

    (4, 5) was found

    3.14 not found

    for key in tests:
    if key in items:
    print(key, 'was found')
    else:
    print(key, 'not found')

    (4, 5) was found

    3.14 not found

    编写循环技巧

    for比while容易些,执行也比较快。

    灵活使用range, range在py3 中是迭代器

    print(range(5)) # range(0, 5)
    print(list(range(5))) # [0, 1, 2, 3, 4]

    l = [1, 2, 3, 4, 5]
    for x in l:
    x += 1
    print(l) # [1, 2, 3, 4, 5]

    这或许和你想象的不太一样,因为修改的是循环变量x, 而不是列表。

    每次经过循环时, x会引用从列表中取出来的下一个整数。

    比如第一轮迭代 x是1, 下次一次迭代, 循环主体把x设为不同的对象,

    也就是2, 但是没更新1所来自的那个列表

    l = [1, 2, 3, 4, 5]
    for i in range(len(l)):
    l[i] += 1

    print(l) # [2, 3, 4, 5, 6]

    更简便的便是用列表解析, 并且运行的更快

    print([i+1 for i in [1, 2, 3, 4, 5]])

    [2, 3, 4, 5, 6]

    并行遍历,zip 和 map

    zip会取得一个或多个序列为参数,然后返回元组的列表。将这些序列中的并排的

    元素配成对

    l1 = [1, 2, 3, 4]
    l2 = [5, 6, 7, 8]

    print(zip(l1, l2)) # <zip object at 0x00000000032C34C8>
    print(list(zip(l1, l2))) # [(1, 5), (2, 6), (3, 7), (4, 8)]

    for (x, y) in zip(l1, l2):
    print(x, y, '-->', x+y)

    1 5 --> 6

    2 6 --> 8

    3 7 --> 10

    4 8 --> 12

    s1 = 'abc'
    s2 = 'xyz123'

    print(list(map(None, s1, s2)))

    python2 中运行将会为较短的序列用None来补齐

    [('a', 'x'), ('b', 'y'), ('c', 'z'), (None, '1'), (None, '2'), (None, '3')]

    print(list(map(ord, 'spam'))) # [115, 112, 97, 109]

    用zip构造字典

    k = ['spam', 'eggs', 'toast']
    v = [1, 3, 5]
    print(list(zip(k, v)))
    d = {}

    [('spam', 1), ('eggs', 3), ('toast', 5)]

    for i, j in zip(k, v):
    d[i] = j
    print(d)

    {'spam': 1, 'eggs': 3, 'toast': 5}

    其实也可以用zip来创建字典

    k = ['spam', 'eggs', 'toast']
    v = [1, 3, 5]
    d = dict(zip(k, v))
    print(d)
    {'spam': 1, 'eggs': 3, 'toast': 5}

    用enumerate来产生偏移和元素

    s = 'spam'
    for i, j in enumerate(s):
    print(i,'-->',j)

    0 --> s

    1 --> p

    2 --> a

    3 --> m

    enumerate函数返回一个生成器对象,可以通过next调用。

    e = enumerate('spam')
    print(e)

    <enumerate object at 0x000000000301E0D8>

    这个对象有个__next__方法,可以在每次迭代的时候返回一个元组,

    可以在for中循环通过元组赋值将运算解包

    print(next(e))
    print(next(e))
    print(next(e))

    (0, 's')

    (1, 'p')

    (2, 'a')

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  • 原文地址:https://www.cnblogs.com/jibandefeng/p/11198679.html
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