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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‘)

python if循环

标签：prime   cto   not   参数   更新   prim   调用   元素   strong   

原文地址：https://blog.51cto.com/14418623/2420528