python yield and stopiteration in one loop?

不问归期 提交于 2019-12-03 04:57:56
Ben

You seem to be overcomplicating this quite a bit:

>>> q = [1, 2, 3, 4]
>>> def gen(header='something', footer='anything'):
        yield header
        for thing in q:
            yield thing
        yield footer


>>> for tmp in gen():
        print(tmp)


something
1
2
3
4
anything

StopIteration will automatically be raised when a generator stops yielding. It's part of the protocol of how generators work. Unless you're doing something very complex, you don't need to (and shouldn't) deal with StopIteration at all. Just yield each value you want to return from the generator in turn, then let the function return.

Here's a code in which use of StopIteration isn't required, a break is enough:

li = [12,51,98,4,36,99,33,1,125,78,9,369,48,47,214,4]

def gen( cont, header='something', footer='anything' ):
    yield header
    for x in cont:
        if x<100:
            yield x
        else:
            yield footer
            break

for y in gen(li):
    print '1 or 2 digits only:',y

result

1 or 2 digits only: something
1 or 2 digits only: 12
1 or 2 digits only: 51
1 or 2 digits only: 98
1 or 2 digits only: 4
1 or 2 digits only: 36
1 or 2 digits only: 99
1 or 2 digits only: 33
1 or 2 digits only: 1
1 or 2 digits only: anything

Now, here's a moderately complex code in which it seems to me that we can't do without use of StopIteration. Does this interest you ?

import Queue
q = Queue.Queue()

li = [12,51,98,4,36,99,33,1,125,78,9,369,48,47,214,4]

def gen( cont, header='something', footer='anything' ):
    def qput(ili = [0]):
        eli = li[ili[0]]
        q.put(eli)
        ili[0] = ili[0] + 1
        return eli

    qput()
    qput()
    qput()
    qput()
    qput()
    yield header

    while True:
        try:
            print '\nq.qsize() first is %s' % q.qsize()

            el = q.get(None)

            if el>9:
                print 'el==',el
                yield 1000+el
                qput()
            else:
                print 'el==%s   el//3==%s' % (el,el//3)
                print 'there are %s items in q and q is emptied %s times :' % (q.qsize(),el//3)
                for emp in xrange(el//3):
                    print '%s is removed from q' % q.get(None)
                    if q.qsize()==0 and emp<el//3:
                        print 'ah !! q is now completely empty, no more emptying is possible !'

            print 'q.qsize() second is %s' % q.qsize()

        except Queue.Empty:
            yield footer
            raise StopIteration


print 'li == %s\n' % li
for i,nb in enumerate(gen(li)):
    print ' * obtained from enumerate(gen(li)) : %s - %s' % (i,nb)

result

li == [12, 51, 98, 4, 36, 99, 33, 1, 125, 78, 9, 369, 48, 47, 214, 4]

 * obtained from enumerate(gen(li)) : 0 - something

q.qsize() first is 5
el== 12
 * obtained from enumerate(gen(li)) : 1 - 1012
q.qsize() second is 5

q.qsize() first is 5
el== 51
 * obtained from enumerate(gen(li)) : 2 - 1051
q.qsize() second is 5

q.qsize() first is 5
el== 98
 * obtained from enumerate(gen(li)) : 3 - 1098
q.qsize() second is 5

q.qsize() first is 5
el==4   el//3==1
there are 4 items in q and q is emptied 1 times :
36 is removed from q
q.qsize() second is 3

q.qsize() first is 3
el== 99
 * obtained from enumerate(gen(li)) : 4 - 1099
q.qsize() second is 3

q.qsize() first is 3
el== 33
 * obtained from enumerate(gen(li)) : 5 - 1033
q.qsize() second is 3

q.qsize() first is 3
el==1   el//3==0
there are 2 items in q and q is emptied 0 times :
q.qsize() second is 2

q.qsize() first is 2
el== 125
 * obtained from enumerate(gen(li)) : 6 - 1125
q.qsize() second is 2

q.qsize() first is 2
el== 78
 * obtained from enumerate(gen(li)) : 7 - 1078
q.qsize() second is 2

q.qsize() first is 2
el==9   el//3==3
there are 1 items in q and q is emptied 3 times :
369 is removed from q
ah !! q is now completely empty, no more emptying is possible !
 * obtained from enumerate(gen(li)) : 8 - anything

Note that this program runs correctly only with q.get(None) , not q.get()

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