pythonic-迭代器函数-itertools

认识

Python 的itertools模块提供了很多节省内存的高效迭代器, 尤其解决了一些关于数据量太大而导致内存溢出(outofmemory)的场景.

我们平时用的循环绝大多数是这样的.

# while 循环: 求1+2+...100
s, i = 0, 1
while i <= 100:
    s += i
    i += 1
print('while-loop: the some of 1+2+..100 is:', s)

# for 循环
s = 0
for i in range(101):
    s += i
print('for-loop: the some of 1+2+..100 is:', s)

while-loop: the some of 1+2+..100 is: 5050
for-loop: the some of 1+2+..100 is: 5050

但如果数据量特别大的话就凉凉了, 所以引入了itertools,迭代器, 类似于懒加载的思想

常用API

• chain()
• groupby()
• accumulate()
• compress()
• takewhile()
• islice()
• repeat()

chain 拼接元素

• 把一组迭代对象串联起来，形成一个更大的迭代器：

# join / split
s = "If you please draw me a sheep?"

s1 = s.split()

s2 = "-".join(s1)

print("split->:", s1)
print("join->:", s2)

split->: ['If', 'you', 'please', 'draw', 'me', 'a', 'sheep?']
join->: If-you-please-draw-me-a-sheep?

import itertools

# chain
s = itertools.chain(['if', 'you'], ['please draw', 'me', 'a'], 'shape')
s

<itertools.chain at 0x1d883602240>

list(s)

['if', 'you', 'please draw', 'me', 'a', 's', 'h', 'a', 'p', 'e']

不难发现, 这就是迭代器嘛, 真的没啥.跟join差不多. 那么它是如何节省内存的呢, 其实就是一个简单的迭代器思想, 一次读取一个元素进内存,这样就高效节约内存了呀

def chain(*iterables):
    for iter_ in iterables:
        for elem in iter_:
            yield elem

groupby 相邻元素

• 把迭代器中相邻的重复元素挑出来放在一

# 只要作用于函数的两个元素返回的值相等，这两个元素就被认为是在一组的，而函数返回值作为组的key
for key, group in itertools.groupby('AAABBBCCAAAdde'):
    print(key, list(group))

A ['A', 'A', 'A']
B ['B', 'B', 'B']
C ['C', 'C']
A ['A', 'A', 'A']
d ['d', 'd']
e ['e']

# 忽略大小写
for key, group in itertools.groupby('AaaBBbcCAAa', lambda c: c.upper()):
    print(key, list(group))

A ['A', 'a', 'a']
B ['B', 'B', 'b']
C ['c', 'C']
A ['A', 'A', 'a']

accumulate 累积汇总

list(itertools.accumulate([1,2,3,4,5], lambda x,y: x*y))

[1, 2, 6, 24, 120]

# 伪代码
def accumulate(iterable, func=None, *, initial=None):
    iter_ = iter(iterable)
    ret = initial
    # 循环迭代
    if initial is None:
        try:
            ret = next(iter_)
        except StopIteration:
            return
    yield ret
    # 遍历每个元素, 调用传入的函数去处理
    for elem in iter_:
        ret = func(elem)
        yield ret

compress 过滤

list(itertools.compress('youge', [1,0,True,3]))

['y', 'u', 'g']

def compress(data, selectors):
    for d, s in zip(data, selectors):
        if s:
            return d

# demo
for data, key in zip([1,2], 'abcd'):
    print(data,key)
    if key:
        print(data)

1 a
1
2 b
2

# Pythonic
def compress(data, selectors):
    return (d for d, s in zip(data, selectors) if s)

# tset
ret = compress(['love', 'you', 'forever'], ['love', None, 'dd', 'forever'])
print(ret)
print(list(ret))

<generator object compress.<locals>.<genexpr> at 0x000001D8831498E0>
['love', 'forever']

生成器

• 在类中实现了iter()方法和next()方法的对象即生成器
• 代码上有两种形式: 元组生成器 或者 函数中出现 yield 关键字

zip

• 对应位置进行元素拼接, 当最短的匹配上了, 则停止, 也被称为"拉长函数"

take-while

• takewhile: 依次迭代, 满足条件则返回, 继续迭代, 一旦不满足条件则退出

# takewhile
s1 = list(itertools.takewhile(lambda x:x<=2, [0,3,2,1,-1,3,0]))
print(s1)

s2 = list(itertools.takewhile(lambda x:x<5, [1,4,6,4,1,3]))
print(s2)

# dropwhile
s3 = list(itertools.filterfalse(lambda x:x%2==0, range(10)))
print(s3)

[0]
[1, 4]
[1, 3, 5, 7, 9]

def take_while(condition, iter_obj):
    for elem in iter_obj:
        if conditon(elem):
            yield elem
        else:
            break

dropwhile: 不满足条件的则返回

islice 切片

# 普通的切片,也是要先全部读入内存
# 注意是深拷贝的哦
l = [1,2,3,4,5]
print(l[::--1])

# generator 方式
# 默认的 start, stop, step, 只能传0或正数, 但可以自己改写的呀
list(itertools.islice(l, 0,3,1))

s = slice(3,4,5) # 只接收3个参数
s.start
s.stop

[1, 2, 3, 4, 5]

[1, 2, 3]

3

4

import sys

def slice(iter_obj, *args):
    s = slice(*args)

    start = s.start or 0
    stop = s.stop or sys.maxsize # 很大的常量
    step = s.step or 1
    # 构成可迭代的对象(下标)
    iter_ = iter(range(start, stop, step))
    try:
        next_i = next(iter_)
    except StopIteration:
#         for i, elem n zip(range(start), iter_obj):
            pass
        return
    try:
        i, elem in enumerate(iter_obj):
            if i == next_i:
                yield elem
                next_i = next(elem)
    except StopIteration:
        pass

[1, 2, 3, 4, 5]

repeat

list(itertools.repeat(['youge'], 3))

[['youge'], ['youge'], ['youge']]

def repeat(obj, times=None):
    if times is None:
        while True:  # 一直返回
            yield obj
    else:
        for i in range(times):
            yield obj