使用 itertools.product 并希望播种一个值 [英] Using itertools.product and want to seed a value

问题描述

所以我写了一个小脚本来从网站下载图片.它通过一个 7 alpha 字符值，其中第一个字符始终是一个数字.问题是如果我想停止脚本并重新启动它，我必须重新开始.

So I've wrote a small script to download pictures from a website. It goes through a 7 alpha charactor value, where the first char is always a number. The problem is if I want to stop the script and start it up again I have to start all over.

我可以用我得到的最后一个值以某种方式播种 itertools.product 吗，这样我就不必再次遍历它们了.

Can I seed itertools.product somehow with the last value I got so I don't have to go through them all again.

感谢您的任何意见.

这是代码的一部分:

numbers = '0123456789'
alnum = numbers + 'abcdefghijklmnopqrstuvwxyz'

len7 = itertools.product(numbers, alnum, alnum, alnum, alnum, alnum, alnum) # length 7

for p in itertools.chain(len7):
    currentid = ''.join(p) 

    #semi static vars
    url = 'http://mysite.com/images/'
    url += currentid

    #Need to get the real url cause the redirect
    print "Trying " + url
    req = urllib2.Request(url)
    res = openaurl(req)
    if res == "continue": continue
    finalurl = res.geturl()

    #ok we have the full url now time to if it is real
    try: file = urllib2.urlopen(finalurl)
    except urllib2.HTTPError, e:
        print e.code

    im = cStringIO.StringIO(file.read())
    img = Image.open(im)
    writeimage(img)

推荐答案

这里是基于 pypy 的库代码的解决方案(感谢 agf 在评论中的建议).

here's a solution based on pypy's library code (thanks to agf's suggestion in the comments).

状态可通过 .state 属性获得，并且可以通过 .goto(state) 重置，其中 state 是序列(从 0 开始).最后有一个演示(恐怕你需要向下滚动).

the state is available via the .state attribute and can be reset via .goto(state) where state is an index into the sequence (starting at 0). there's a demo at the end (you need to scroll down, i'm afraid).

这比丢弃值要快得多.

> cat prod.py 

class product(object):

    def __init__(self, *args, **kw):
        if len(kw) > 1:
            raise TypeError("product() takes at most 1 argument (%d given)" %
                             len(kw))
        self.repeat = kw.get('repeat', 1)
        self.gears = [x for x in args] * self.repeat
        self.num_gears = len(self.gears)
        self.reset()

    def reset(self):
        # initialization of indicies to loop over
        self.indicies = [(0, len(self.gears[x]))
                         for x in range(0, self.num_gears)]
        self.cont = True
        self.state = 0

    def goto(self, n):
        self.reset()
        self.state = n
        x = self.num_gears
        while n > 0 and x > 0:
            x -= 1
            n, m = divmod(n, len(self.gears[x]))
            self.indicies[x] = (m, self.indicies[x][1])
        if n > 0:
            self.reset()
            raise ValueError("state exceeded")

    def roll_gears(self):
        # Starting from the end of the gear indicies work to the front
        # incrementing the gear until the limit is reached. When the limit
        # is reached carry operation to the next gear
        self.state += 1
        should_carry = True
        for n in range(0, self.num_gears):
            nth_gear = self.num_gears - n - 1
            if should_carry:
                count, lim = self.indicies[nth_gear]
                count += 1
                if count == lim and nth_gear == 0:
                    self.cont = False
                if count == lim:
                    should_carry = True
                    count = 0
                else:
                    should_carry = False
                self.indicies[nth_gear] = (count, lim)
            else:
                break

    def __iter__(self):
        return self

    def next(self):
        if not self.cont:
            raise StopIteration
        l = []
        for x in range(0, self.num_gears):
            index, limit = self.indicies[x]
            l.append(self.gears[x][index])
        self.roll_gears()
        return tuple(l)

p = product('abc', '12')
print list(p)
p.reset()
print list(p)
p.goto(2)
print list(p)
p.goto(4)
print list(p)
> python prod.py 
[('a', '1'), ('a', '2'), ('b', '1'), ('b', '2'), ('c', '1'), ('c', '2')]
[('a', '1'), ('a', '2'), ('b', '1'), ('b', '2'), ('c', '1'), ('c', '2')]
[('b', '1'), ('b', '2'), ('c', '1'), ('c', '2')]
[('c', '1'), ('c', '2')]

你应该更多地测试它——我可能犯了一个愚蠢的错误——但这个想法很简单，所以你应该能够修复它:o)你可以自由使用我的更改；不知道原始 pypy 许可证是什么.

you should test it more - i may have made a dumb mistake - but the idea is quite simple, so you should be able to fix it :o) you're free to use my changes; no idea what the original pypy licence is.

还有 state 并不是真正的完整状态——它不包括原始参数——它只是序列中的一个索引.也许称其为索引会更好，但是代码中已经有索引了...

also state isn't really the full state - it doesn't include the original arguments - it's just an index into the sequence. maybe it would have been better to call it index, but there are already indici[sic]es in the code...

更新

这是一个更简单的版本，它的想法相同，但通过转换数字序列来工作.所以你只需 imap 它在 count(n) 上得到 n 的序列偏移量.

here's a simpler version that is the same idea but works by transforming a sequence of numbers. so you just imap it over count(n) to get the sequence offset by n.

> cat prod2.py 

from itertools import count, imap

def make_product(*values):
    def fold((n, l), v):
        (n, m) = divmod(n, len(v))
        return (n, l + [v[m]])
    def product(n):
        (n, l) = reduce(fold, values, (n, []))
        if n > 0: raise StopIteration
        return tuple(l)
    return product

print list(imap(make_product(['a','b','c'], [1,2,3]), count()))
print list(imap(make_product(['a','b','c'], [1,2,3]), count(3)))

def product_from(n, *values):
    return imap(make_product(*values), count(n))

print list(product_from(4, ['a','b','c'], [1,2,3]))

> python prod2.py 
[('a', 1), ('b', 1), ('c', 1), ('a', 2), ('b', 2), ('c', 2), ('a', 3), ('b', 3), ('c', 3)]
[('a', 2), ('b', 2), ('c', 2), ('a', 3), ('b', 3), ('c', 3)]
[('b', 2), ('c', 2), ('a', 3), ('b', 3), ('c', 3)]

(这里的缺点是如果你想停止和重新启动，你需要自己跟踪你使用了多少)

(the downside here is that if you want to stop and restart you need to have kept track yourself of how many you have used)

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