python cookbook 迭代器与生成器
代理迭代
a = [1, 2, 3] for i in iter(a): print(i) for i in a.__iter__(): print(i)
这里的两个方法是一样的,调用iter()其实就是简单的调用了对象的__iter__()方法。
使用生成器创建新的迭代器
def frange(start, stop, increment): x = start while x < stop: yield x x += increment for n in frange(0, 4, 0.5): print(n) print(list(frange(0, 4, 0.5)))
看下面这个
>>> def countdown (n): ... print ('Starting to count from', n) ... while n > 0: ... yield n ... n -= 1 ... print ('Done!') ... >>> # Create the generator, notice no output appears >>> c = countdown(3) >>> c <generator object countdown at 0x1006a0af0>>>> next(c) Starting to count from 3 3>>> next(c) 2>>> next(c) 1>>> next(c) Done! Traceback (most recent call last): File "<stdin>", line 1, in < module > StopIteration
一个生成器函数主要特征是它只会回应在迭代中使用到的 next 操作。 一旦生成器函数返回退出,迭代终止。我们在迭代中通常使用的for语句会自动处理这些细节,所以你无需担心。
def frange(start, stop, increment): x = start while x < stop: yield x x += increment for n in frange(0, 4, 0.5): print(n) print(list(frange(0, 4, 0.5)))
反向迭代
a = [1, 2, 3, 4] for x in reversed(a): print(x, end=' ')
主要想说的是,对象重写__reversed__()方法即可调用reversed()进行反向迭代
itertools的一些使用
def count(n): while True: yield n n += 1 c = count(0) import itertools for x in itertools.islice(c, 10, 20): print(x, end=" ") with open('test.txt') as f: for line in f: print(line, end='') print(format('开始部分的注释不显示', '*>30')) with open('test.txt') as f: for line in itertools.dropwhile(lambda line: line.startswith('#'), f): print(line, end='') items = ['a', 'b', 'c'] #items的所有可能组合,不包含相同元素 for p in itertools.permutations(items): print(p, end=' ') print() #指定长度的所有排序 for p in itertools.permutations(items, 2): print(p, end=' ')
如果遇到一些复杂的迭代器的使用,可以先看看itertools里有没有可用的方法。
同时迭代多个序列
a = [1, 2, 3] b = ['w', 'x', 'y', 'z'] for i in zip(a,b): print(i)
输出:
(1, 'w') (2, 'x') (3, 'y') for i in itertools.zip_longest(a, b): print(i)
输出:
(1, 'w') (2, 'x') (3, 'y') (None, 'z')
不同序列上的迭代
import itertools a = [1, 2, 3, 4] b = ['x', 'y', 'z'] for x in itertools.chain(a, b): print(x)
这样比 a + b 在进行迭代要好很多
展开嵌套的序列
from collections import Iterable def flattern(items, ignore_types=(str, bytes)): for x in items: if isinstance(x, Iterable) and not isinstance(x, ignore_types): yield from flattern(x) else: yield x items = [1, 2, [3, 4, [5, 'hello world'], 7], 8] for x in flattern(items): print(x, end=' ')
def count(n): while True: yield n n += 1 c = count(0) import itertools for x in itertools.islice(c, 10, 20): print(x, end=" ") with open('test.txt') as f: for line in f: print(line, end='') print(format('开始部分的注释不显示', '*>30')) with open('test.txt') as f: for line in itertools.dropwhile(lambda line: line.startswith('#'), f): print(line, end='') items = ['a', 'b', 'c'] #items的所有可能组合,不包含相同元素 for p in itertools.permutations(items): print(p, end=' ') print() #指定长度的所有排序 for p in itertools.permutations(items, 2): print(p, end=' ')