-í é ¶>> def f(): ... yield 1 ... yield 2 >>> for i in f(): ... print i 1 2 >>> g = f() >>> g.next() 1 >>> g.next() 2 "Falling off the end" stops the generator: >>> g.next() Traceback (most recent call last): File "", line 1, in ? File "", line 2, in g StopIteration "return" also stops the generator: >>> def f(): ... yield 1 ... return ... yield 2 # never reached ... >>> g = f() >>> g.next() 1 >>> g.next() Traceback (most recent call last): File "", line 1, in ? File "", line 3, in f StopIteration >>> g.next() # once stopped, can't be resumed Traceback (most recent call last): File "", line 1, in ? StopIteration "raise StopIteration" stops the generator too: >>> def f(): ... yield 1 ... raise StopIteration ... yield 2 # never reached ... >>> g = f() >>> g.next() 1 >>> g.next() Traceback (most recent call last): File "", line 1, in ? StopIteration >>> g.next() Traceback (most recent call last): File "", line 1, in ? StopIteration However, they are not exactly equivalent: >>> def g1(): ... try: ... return ... except: ... yield 1 ... >>> list(g1()) [] >>> def g2(): ... try: ... raise StopIteration ... except: ... yield 42 >>> print list(g2()) [42] This may be surprising at first: >>> def g3(): ... try: ... return ... finally: ... yield 1 ... >>> list(g3()) [1] Let's create an alternate range() function implemented as a generator: >>> def yrange(n): ... for i in range(n): ... yield i ... >>> list(yrange(5)) [0, 1, 2, 3, 4] Generators always return to the most recent caller: >>> def creator(): ... r = yrange(5) ... print "creator", r.next() ... return r ... >>> def caller(): ... r = creator() ... for i in r: ... print "caller", i ... >>> caller() creator 0 caller 1 caller 2 caller 3 caller 4 Generators can call other generators: >>> def zrange(n): ... for i in yrange(n): ... yield i ... >>> list(zrange(5)) [0, 1, 2, 3, 4] sm Specification: Yield Restriction: A generator cannot be resumed while it is actively running: >>> def g(): ... i = me.next() ... yield i >>> me = g() >>> me.next() Traceback (most recent call last): ... File "", line 2, in g ValueError: generator already executing Specification: Return Note that return isn't always equivalent to raising StopIteration: the difference lies in how enclosing try/except constructs are treated. For example, >>> def f1(): ... try: ... return ... except: ... yield 1 >>> print list(f1()) [] because, as in any function, return simply exits, but >>> def f2(): ... try: ... raise StopIteration ... except: ... yield 42 >>> print list(f2()) [42] because StopIteration is captured by a bare "except", as is any exception. Specification: Generators and Exception Propagation >>> def f(): ... return 1//0 >>> def g(): ... yield f() # the zero division exception propagates ... yield 42 # and we'll never get here >>> k = g() >>> k.next() Traceback (most recent call last): File "", line 1, in ? File "", line 2, in g File "", line 2, in f ZeroDivisionError: integer division or modulo by zero >>> k.next() # and the generator cannot be resumed Traceback (most recent call last): File "", line 1, in ? StopIteration >>> Specification: Try/Except/Finally >>> def f(): ... try: ... yield 1 ... try: ... yield 2 ... 1//0 ... yield 3 # never get here ... except ZeroDivisionError: ... yield 4 ... yield 5 ... raise ... except: ... yield 6 ... yield 7 # the "raise" above stops this ... except: ... yield 8 ... yield 9 ... try: ... x = 12 ... finally: ... yield 10 ... yield 11 >>> print list(f()) [1, 2, 4, 5, 8, 9, 10, 11] >>> Guido's binary tree example. >>> # A binary tree class. >>> class Tree: ... ... def __init__(self, label, left=None, right=None): ... self.label = label ... self.left = left ... self.right = right ... ... def __repr__(self, level=0, indent=" "): ... s = level*indent + `self.label` ... if self.left: ... s = s + "\n" + self.left.__repr__(level+1, indent) ... if self.right: ... s = s + "\n" + self.right.__repr__(level+1, indent) ... return s ... ... def __iter__(self): ... return inorder(self) >>> # Create a Tree from a list. >>> def tree(list): ... n = len(list) ... if n == 0: ... return [] ... i = n // 2 ... return Tree(list[i], tree(list[:i]), tree(list[i+1:])) >>> # Show it off: create a tree. >>> t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ") >>> # A recursive generator that generates Tree leaves in in-order. >>> def inorder(t): ... if t: ... for x in inorder(t.left): ... yield x ... yield t.label ... for x in inorder(t.right): ... yield x >>> # Show it off: create a tree. ... t = tree("ABCDEFGHIJKLMNOPQRSTUVWXYZ") ... # Print the nodes of the tree in in-order. ... for x in t: ... print x, A B C D E F G H I J K L M N O P Q R S T U V W X Y Z >>> # A non-recursive generator. >>> def inorder(node): ... stack = [] ... while node: ... while node.left: ... stack.append(node) ... node = node.left ... yield node.label ... while not node.right: ... try: ... node = stack.pop() ... except IndexError: ... return ... yield node.label ... node = node.right >>> # Exercise the non-recursive generator. >>> for x in t: ... print x, A B C D E F G H I J K L M N O P Q R S T U V W X Y Z s¥ The difference between yielding None and returning it. >>> def g(): ... for i in range(3): ... yield None ... yield None ... return >>> list(g()) [None, None, None, None] Ensure that explicitly raising StopIteration acts like any other exception in try/except, not like a return. >>> def g(): ... yield 1 ... try: ... raise StopIteration ... except: ... yield 2 ... yield 3 >>> list(g()) [1, 2, 3] Next one was posted to c.l.py. >>> def gcomb(x, k): ... "Generate all combinations of k elements from list x." ... ... if k > len(x): ... return ... if k == 0: ... yield [] ... else: ... first, rest = x[0], x[1:] ... # A combination does or doesn't contain first. ... # If it does, the remainder is a k-1 comb of rest. ... for c in gcomb(rest, k-1): ... c.insert(0, first) ... yield c ... # If it doesn't contain first, it's a k comb of rest. ... for c in gcomb(rest, k): ... yield c >>> seq = range(1, 5) >>> for k in range(len(seq) + 2): ... print "%d-combs of %s:" % (k, seq) ... for c in gcomb(seq, k): ... print " ", c 0-combs of [1, 2, 3, 4]: [] 1-combs of [1, 2, 3, 4]: [1] [2] [3] [4] 2-combs of [1, 2, 3, 4]: [1, 2] [1, 3] [1, 4] [2, 3] [2, 4] [3, 4] 3-combs of [1, 2, 3, 4]: [1, 2, 3] [1, 2, 4] [1, 3, 4] [2, 3, 4] 4-combs of [1, 2, 3, 4]: [1, 2, 3, 4] 5-combs of [1, 2, 3, 4]: From the Iterators list, about the types of these things. >>> def g(): ... yield 1 ... >>> type(g) >>> i = g() >>> type(i) >>> [s for s in dir(i) if not s.startswith('_')] ['gi_frame', 'gi_running', 'next'] >>> print i.next.__doc__ x.next() -> the next value, or raise StopIteration >>> iter(i) is i 1 >>> import types >>> isinstance(i, types.GeneratorType) 1 And more, added later. >>> i.gi_running 0 >>> type(i.gi_frame) >>> i.gi_running = 42 Traceback (most recent call last): ... TypeError: readonly attribute >>> def g(): ... yield me.gi_running >>> me = g() >>> me.gi_running 0 >>> me.next() 1 >>> me.gi_running 0 A clever union-find implementation from c.l.py, due to David Eppstein. Sent: Friday, June 29, 2001 12:16 PM To: python-list@python.org Subject: Re: PEP 255: Simple Generators >>> class disjointSet: ... def __init__(self, name): ... self.name = name ... self.parent = None ... self.generator = self.generate() ... ... def generate(self): ... while not self.parent: ... yield self ... for x in self.parent.generator: ... yield x ... ... def find(self): ... return self.generator.next() ... ... def union(self, parent): ... if self.parent: ... raise ValueError("Sorry, I'm not a root!") ... self.parent = parent ... ... def __str__(self): ... return self.name >>> names = "ABCDEFGHIJKLM" >>> sets = [disjointSet(name) for name in names] >>> roots = sets[:] >>> import random >>> random.seed(42) >>> while 1: ... for s in sets: ... print "%s->%s" % (s, s.find()), ... print ... if len(roots) > 1: ... s1 = random.choice(roots) ... roots.remove(s1) ... s2 = random.choice(roots) ... s1.union(s2) ... print "merged", s1, "into", s2 ... else: ... break A->A B->B C->C D->D E->E F->F G->G H->H I->I J->J K->K L->L M->M merged D into G A->A B->B C->C D->G E->E F->F G->G H->H I->I J->J K->K L->L M->M merged C into F A->A B->B C->F D->G E->E F->F G->G H->H I->I J->J K->K L->L M->M merged L into A A->A B->B C->F D->G E->E F->F G->G H->H I->I J->J K->K L->A M->M merged H into E A->A B->B C->F D->G E->E F->F G->G H->E I->I J->J K->K L->A M->M merged B into E A->A B->E C->F D->G E->E F->F G->G H->E I->I J->J K->K L->A M->M merged J into G A->A B->E C->F D->G E->E F->F G->G H->E I->I J->G K->K L->A M->M merged E into G A->A B->G C->F D->G E->G F->F G->G H->G I->I J->G K->K L->A M->M merged M into G A->A B->G C->F D->G E->G F->F G->G H->G I->I J->G K->K L->A M->G merged I into K A->A B->G C->F D->G E->G F->F G->G H->G I->K J->G K->K L->A M->G merged K into A A->A B->G C->F D->G E->G F->F G->G H->G I->A J->G K->A L->A M->G merged F into A A->A B->G C->A D->G E->G F->A G->G H->G I->A J->G K->A L->A M->G merged A into G A->G B->G C->G D->G E->G F->G G->G H->G I->G J->G K->G L->G M->G s/ Build up to a recursive Sieve of Eratosthenes generator. >>> def firstn(g, n): ... return [g.next() for i in range(n)] >>> def intsfrom(i): ... while 1: ... yield i ... i += 1 >>> firstn(intsfrom(5), 7) [5, 6, 7, 8, 9, 10, 11] >>> def exclude_multiples(n, ints): ... for i in ints: ... if i % n: ... yield i >>> firstn(exclude_multiples(3, intsfrom(1)), 6) [1, 2, 4, 5, 7, 8] >>> def sieve(ints): ... prime = ints.next() ... yield prime ... not_divisible_by_prime = exclude_multiples(prime, ints) ... for p in sieve(not_divisible_by_prime): ... yield p >>> primes = sieve(intsfrom(2)) >>> firstn(primes, 20) [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71] Another famous problem: generate all integers of the form 2**i * 3**j * 5**k in increasing order, where i,j,k >= 0. Trickier than it may look at first! Try writing it without generators, and correctly, and without generating 3 internal results for each result output. >>> def times(n, g): ... for i in g: ... yield n * i >>> firstn(times(10, intsfrom(1)), 10) [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] >>> def merge(g, h): ... ng = g.next() ... nh = h.next() ... while 1: ... if ng < nh: ... yield ng ... ng = g.next() ... elif ng > nh: ... yield nh ... nh = h.next() ... else: ... yield ng ... ng = g.next() ... nh = h.next() The following works, but is doing a whale of a lot of redundant work -- it's not clear how to get the internal uses of m235 to share a single generator. Note that me_times2 (etc) each need to see every element in the result sequence. So this is an example where lazy lists are more natural (you can look at the head of a lazy list any number of times). >>> def m235(): ... yield 1 ... me_times2 = times(2, m235()) ... me_times3 = times(3, m235()) ... me_times5 = times(5, m235()) ... for i in merge(merge(me_times2, ... me_times3), ... me_times5): ... yield i Don't print "too many" of these -- the implementation above is extremely inefficient: each call of m235() leads to 3 recursive calls, and in turn each of those 3 more, and so on, and so on, until we've descended enough levels to satisfy the print stmts. Very odd: when I printed 5 lines of results below, this managed to screw up Win98's malloc in "the usual" way, i.e. the heap grew over 4Mb so Win98 started fragmenting address space, and it *looked* like a very slow leak. >>> result = m235() >>> for i in range(3): ... print firstn(result, 15) [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24] [25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80] [81, 90, 96, 100, 108, 120, 125, 128, 135, 144, 150, 160, 162, 180, 192] Heh. Here's one way to get a shared list, complete with an excruciating namespace renaming trick. The *pretty* part is that the times() and merge() functions can be reused as-is, because they only assume their stream arguments are iterable -- a LazyList is the same as a generator to times(). >>> class LazyList: ... def __init__(self, g): ... self.sofar = [] ... self.fetch = g.next ... ... def __getitem__(self, i): ... sofar, fetch = self.sofar, self.fetch ... while i >= len(sofar): ... sofar.append(fetch()) ... return sofar[i] >>> def m235(): ... yield 1 ... # Gack: m235 below actually refers to a LazyList. ... me_times2 = times(2, m235) ... me_times3 = times(3, m235) ... me_times5 = times(5, m235) ... for i in merge(merge(me_times2, ... me_times3), ... me_times5): ... yield i Print as many of these as you like -- *this* implementation is memory- efficient. >>> m235 = LazyList(m235()) >>> for i in range(5): ... print [m235[j] for j in range(15*i, 15*(i+1))] [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24] [25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80] [81, 90, 96, 100, 108, 120, 125, 128, 135, 144, 150, 160, 162, 180, 192] [200, 216, 225, 240, 243, 250, 256, 270, 288, 300, 320, 324, 360, 375, 384] [400, 405, 432, 450, 480, 486, 500, 512, 540, 576, 600, 625, 640, 648, 675] Ye olde Fibonacci generator, LazyList style. >>> def fibgen(a, b): ... ... def sum(g, h): ... while 1: ... yield g.next() + h.next() ... ... def tail(g): ... g.next() # throw first away ... for x in g: ... yield x ... ... yield a ... yield b ... for s in sum(iter(fib), ... tail(iter(fib))): ... yield s >>> fib = LazyList(fibgen(1, 2)) >>> firstn(iter(fib), 17) [1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584] s >>> def f(): ... return 22 ... yield 1 Traceback (most recent call last): ... SyntaxError: 'return' with argument inside generator (, line 2) >>> def f(): ... yield 1 ... return 22 Traceback (most recent call last): ... SyntaxError: 'return' with argument inside generator (, line 3) "return None" is not the same as "return" in a generator: >>> def f(): ... yield 1 ... return None Traceback (most recent call last): ... SyntaxError: 'return' with argument inside generator (, line 3) This one is fine: >>> def f(): ... yield 1 ... return >>> def f(): ... try: ... yield 1 ... finally: ... pass Traceback (most recent call last): ... SyntaxError: 'yield' not allowed in a 'try' block with a 'finally' clause (, line 3) >>> def f(): ... try: ... try: ... 1//0 ... except ZeroDivisionError: ... yield 666 # bad because *outer* try has finally ... except: ... pass ... finally: ... pass Traceback (most recent call last): ... SyntaxError: 'yield' not allowed in a 'try' block with a 'finally' clause (, line 6) But this is fine: >>> def f(): ... try: ... try: ... yield 12 ... 1//0 ... except ZeroDivisionError: ... yield 666 ... except: ... try: ... x = 12 ... finally: ... yield 12 ... except: ... return >>> list(f()) [12, 666] >>> def f(): ... yield Traceback (most recent call last): SyntaxError: invalid syntax >>> def f(): ... if 0: ... yield Traceback (most recent call last): SyntaxError: invalid syntax >>> def f(): ... if 0: ... yield 1 >>> type(f()) >>> def f(): ... if "": ... yield None >>> type(f()) >>> def f(): ... return ... try: ... if x==4: ... pass ... elif 0: ... try: ... 1//0 ... except SyntaxError: ... pass ... else: ... if 0: ... while 12: ... x += 1 ... yield 2 # don't blink ... f(a, b, c, d, e) ... else: ... pass ... except: ... x = 1 ... return >>> type(f()) >>> def f(): ... if 0: ... def g(): ... yield 1 ... >>> type(f()) >>> def f(): ... if 0: ... class C: ... def __init__(self): ... yield 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This illustrates dumbest- possible use of conjoin, just to generate the full cross-product. >>> for c in conjoin([lambda: iter((0, 1))] * 3): ... print c [0, 0, 0] [0, 0, 1] [0, 1, 0] [0, 1, 1] [1, 0, 0] [1, 0, 1] [1, 1, 0] [1, 1, 1] For efficiency in typical backtracking apps, conjoin() yields the same list object each time. So if you want to save away a full account of its generated sequence, you need to copy its results. >>> def gencopy(iterator): ... for x in iterator: ... yield x[:] >>> for n in range(10): ... all = list(gencopy(conjoin([lambda: iter((0, 1))] * n))) ... print n, len(all), all[0] == [0] * n, all[-1] == [1] * n 0 1 1 1 1 2 1 1 2 4 1 1 3 8 1 1 4 16 1 1 5 32 1 1 6 64 1 1 7 128 1 1 8 256 1 1 9 512 1 1 And run an 8-queens solver. >>> q = Queens(8) >>> LIMIT = 2 >>> count = 0 >>> for row2col in q.solve(): ... count += 1 ... if count <= LIMIT: ... print "Solution", count ... q.printsolution(row2col) Solution 1 +-+-+-+-+-+-+-+-+ |Q| | | | | | | | +-+-+-+-+-+-+-+-+ | | | | |Q| | | | +-+-+-+-+-+-+-+-+ | | | | | | | |Q| +-+-+-+-+-+-+-+-+ | | | | | |Q| | | +-+-+-+-+-+-+-+-+ | | |Q| | | | | | +-+-+-+-+-+-+-+-+ | | | | | | |Q| | +-+-+-+-+-+-+-+-+ | |Q| | | | | | | +-+-+-+-+-+-+-+-+ | | | |Q| | | | | +-+-+-+-+-+-+-+-+ Solution 2 +-+-+-+-+-+-+-+-+ |Q| | | | | | | | +-+-+-+-+-+-+-+-+ | | | | | |Q| | | +-+-+-+-+-+-+-+-+ | | | | | | | |Q| +-+-+-+-+-+-+-+-+ | | |Q| | | | | | +-+-+-+-+-+-+-+-+ | | | | | | |Q| | +-+-+-+-+-+-+-+-+ | | | |Q| | | | | +-+-+-+-+-+-+-+-+ | |Q| | | | | | | +-+-+-+-+-+-+-+-+ | | | | |Q| | | | +-+-+-+-+-+-+-+-+ >>> print count, "solutions in all." 92 solutions in all. And run a Knight's Tour on a 10x10 board. Note that there are about 20,000 solutions even on a 6x6 board, so don't dare run this to exhaustion. >>> k = Knights(10, 10) >>> LIMIT = 2 >>> count = 0 >>> for x in k.solve(): ... count += 1 ... if count <= LIMIT: ... print "Solution", count ... k.printsolution(x) ... else: ... break Solution 1 +---+---+---+---+---+---+---+---+---+---+ | 1| 58| 27| 34| 3| 40| 29| 10| 5| 8| +---+---+---+---+---+---+---+---+---+---+ | 26| 35| 2| 57| 28| 33| 4| 7| 30| 11| +---+---+---+---+---+---+---+---+---+---+ | 59|100| 73| 36| 41| 56| 39| 32| 9| 6| +---+---+---+---+---+---+---+---+---+---+ | 74| 25| 60| 55| 72| 37| 42| 49| 12| 31| +---+---+---+---+---+---+---+---+---+---+ | 61| 86| 99| 76| 63| 52| 47| 38| 43| 50| +---+---+---+---+---+---+---+---+---+---+ | 24| 75| 62| 85| 54| 71| 64| 51| 48| 13| +---+---+---+---+---+---+---+---+---+---+ | 87| 98| 91| 80| 77| 84| 53| 46| 65| 44| +---+---+---+---+---+---+---+---+---+---+ | 90| 23| 88| 95| 70| 79| 68| 83| 14| 17| +---+---+---+---+---+---+---+---+---+---+ | 97| 92| 21| 78| 81| 94| 19| 16| 45| 66| +---+---+---+---+---+---+---+---+---+---+ | 22| 89| 96| 93| 20| 69| 82| 67| 18| 15| +---+---+---+---+---+---+---+---+---+---+ Solution 2 +---+---+---+---+---+---+---+---+---+---+ | 1| 58| 27| 34| 3| 40| 29| 10| 5| 8| +---+---+---+---+---+---+---+---+---+---+ | 26| 35| 2| 57| 28| 33| 4| 7| 30| 11| +---+---+---+---+---+---+---+---+---+---+ | 59|100| 73| 36| 41| 56| 39| 32| 9| 6| +---+---+---+---+---+---+---+---+---+---+ | 74| 25| 60| 55| 72| 37| 42| 49| 12| 31| +---+---+---+---+---+---+---+---+---+---+ | 61| 86| 99| 76| 63| 52| 47| 38| 43| 50| +---+---+---+---+---+---+---+---+---+---+ | 24| 75| 62| 85| 54| 71| 64| 51| 48| 13| +---+---+---+---+---+---+---+---+---+---+ | 87| 98| 89| 80| 77| 84| 53| 46| 65| 44| +---+---+---+---+---+---+---+---+---+---+ | 90| 23| 92| 95| 70| 79| 68| 83| 14| 17| +---+---+---+---+---+---+---+---+---+---+ | 97| 88| 21| 78| 81| 94| 19| 16| 45| 66| +---+---+---+---+---+---+---+---+---+---+ | 22| 91| 96| 93| 20| 69| 82| 67| 18| 15| +---+---+---+---+---+---+---+---+---+---+ stutspepsemailsfunssyntaxsconjoincs;JKdk}dk}dk}LR|i||ƒdS(N(sdocteststest_supportstest_generatorssrun_doctestsverbose(sverbosestest_supportstest_generatorssdoctest((s*/usr/lib/python2.2/test/test_generators.pys test_mainJss__main__iN(s __future__s generatorsstutorial_testss pep_testssemail_testss fun_testsssyntax_testssconjoinsflat_conjoinsQueenssKnightss conjoin_testss__test__sNones test_mains__name__( s fun_testssconjoinssyntax_testss conjoin_testssemail_testsstutorial_testssflat_conjoins pep_testss test_mains__test__sKnightssQueenss generators((s*/usr/lib/python2.2/test/test_generators.pys?s ‡ ¥ ¼ µ=$7¿ “?