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python iterator and itertools module

Feb 22, 2017 am 10:28 AM

Iterator In python, the iterator protocol is to implement the __iter() method and next() method of the object, where the former returns the object itself , which returns the next element of the container. Objects that implement these two methods are iterable objects. Iterators are lazy and are only generated when used, which provides benefits for processing large amounts of data, unlike writing all data to memory at once. Below I wrote an iterator myself. You can see that you can use a for loop to process the iterator you wrote. For objects that implement the iterator protocol, you can use any iterator tool similar to a for loop. However, looking at the output below, the second output is empty. Why is this? When we use a list, we can output the same object multiple times. What is the difference between this and an object that implements its own iterator protocol?

class it(object):

def __init__(self, n):

self.a = 0
self.n = n

def __iter__(self):
<br/>
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return self<br/><br/> def next(self):<br/> if self.a < self.n:<br/> self.a += 1<br/> return self.a<br/> else:<br/> raise StopIteration<br/><br/>i=it(5)<br/>for j in i:<br/> print j,<br/>print ''<br/>print '------'<br/>for j in i:<br/> print j<br/># 1 2 3 4 5<br/># ------<br/> After studying, I learned that list and other types of iterators return an iterator object, not an iterator object. itself. Then I wrote the following code for testing. As you can see from the printout, objects like TestIt can be used repeatedly. So there is another question. Is an object that does not implement the next() method still an iterator object? This is because when using the it class, an iterator object is returned, and the iteration function is implemented using the it iterator, which is equivalent to implementing the iterator protocol. The iterator protocol is very useful in python. There is a module itertools in python about iterators. Now I will learn about the itertools module and see what surprises there are! <br/>
class TestIt(object):
def __init__(self, a):
self.a = a 
def __iter__(self):
return it(self.a)
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itertools

Infinite iterator

1 count(), accepts two parameters, the first The first is the starting number, the second is the stride, starting from 0 by default, the usage is as follows

import itertools as it

c = it.count(10, 2)
for i in c:
    if i > 20:
        break
    print i,
# 10 12 14 16 18 20
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2 cycle(), accepts one parameter, which is an iterator object (list , string, etc.), it will loop to generate the elements in the iterator

c = it.cycle([1, 2, 3])
i = 1
for j in c:
    if i > 7:
        break
    print j,
    i += 1
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3 repeat(), accepts two parameters and is used to generate the first parameter n times

for j in it.repeat([1, 2, 3], 4):
print j
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Limited iterator, choose the one you feel you commonly use and introduce it

1 chain(), accepts multiple iterator objects as parameters and connects them Up chain('abc', [1, 2, 3])

2 compress(data, selectors), filter the previous parameters according to the latter parameters, both parameters need to be Iterator object

3

dropwhile(pre, iterable), the pre parameter is a function, When pre(i) is True, return this item and all following items

#4 groupby(iterable[, keyfunc]), where ##iterable is an iterable object, keyfunc is a grouping function, used to The consecutive items in iterable are grouped. If not specified, the consecutive identical items in iterable are grouped by default and an iterator of (key, sub-iterator) is returned.

5 ifilter(function or None, sequence),将 iterable 中 function(item) 为 True 的元素组成一个迭代器返回,如果 function 是 None,则返回 iterable 中所有计算为 True 的项

6 tee(iterable [,n]),tee 用于从 iterable 创建 n 个独立的迭代器,以元组的形式返回,n 的默认值是 2。

<br/>

 for j in it.tee(&#39;abc&#39;, 4):
     print list(j)
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<br/>

组合生成器

1 permutations(iterable[, r]),用于生成一个排列,r是生成排列的元素长度,不指定则为默认长度

<br/>

  list(it.permutations(  list(it.permutations(, 2
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<br/>

2 combinations(iterable, r), 求序列的组合,其中,r 指定生成组合的元素的长度,是必需的参数

3 combinations_with_replacement(iterable, r),生成的组合包含自身元素

 print list(it.combinations_with_replacement(&#39;abc&#39;, 2))
 # [(&#39;a&#39;, &#39;a&#39;), (&#39;a&#39;, &#39;b&#39;), (&#39;a&#39;, &#39;c&#39;), (&#39;b&#39;, &#39;b&#39;), (&#39;b&#39;, &#39;c&#39;), (&#39;c&#39;, &#39;c&#39;)]
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