Re: My Generator Paradox!
I believe I understand now. the yield keyword is sort of like a cousin to return. return will bring back an object I can work with and so does yield *but* yield's object will most likely support the .next() method. So, if I worked with a function that ends with the return keyword and it returns a list, I can run list operations and list methods on it. if a function ends with the yield keyword a generator should return. So, calling the function by it's name will always reset and initialize the generator. Whereas assigning to the functions yielded return grants access to the real generator in which I can use the next() method. Maybe I've explained it wrong *but* it does make sense to me now. I just couldn't grasp it because I am still new to the keyword yield and didn't know it sort of works like return. I really wish to thank you fellas so much for your examples and explanations! I think I got it! I thank you all again! -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
vbgunz wrote: I believe I understand now. the yield keyword is sort of like a cousin to return. return will bring back an object I can work with and so does yield *but* yield's object will most likely support the .next() method. No, that's not really how it works. When a generator function is called, it returns the generator object immediately. None of the code inside is executed. Every time you call that generator function, you get a new generator object with the initial state. The objects that are yielded inside the code don't show up yet. The code inside the generator gets executed only when the generator object is iterated over (or its .next() method is called). The objects that are yielded are the results of calling the .next() method. -- Robert Kern [EMAIL PROTECTED] I have come to believe that the whole world is an enigma, a harmless enigma that is made terrible by our own mad attempt to interpret it as though it had an underlying truth. -- Umberto Eco -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
Robert Kern wrote: vbgunz wrote: I believe I understand now. the yield keyword is sort of like a cousin to return. return will bring back an object I can work with and so does yield *but* yield's object will most likely support the .next() method. No, that's not really how it works. When a generator function is called, it returns the generator object immediately. None of the code inside is executed. Every time you call that generator function, you get a new generator object with the initial state. The objects that are yielded inside the code don't show up yet. The code inside the generator gets executed only when the generator object is iterated over (or its .next() method is called). The objects that are yielded are the results of calling the .next() method. Maybe this will clarify it further. def gen(n): ... while 1: ... print 'before yield' ... yield n ... print 'after yield' ... g = gen('hello') g.next() before yield 'hello' g.next() after yield before yield 'hello' g.next() after yield before yield 'hello' When the next() method is called the generator runs until it reaches a yield. At which point it's rests until the next() method is called again. Although there are times when I wish it could run (as a thread) until it reaches a yield and then continue after the next() method is called until it reaches the next yield. Cheers, Ron -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
OK. I hope my understanding of the yield keyword and generators in a general sense are now better understood. When a generator function is assigned to an identifier, no code is executed and a generator is immediately returned. When the next() method is called on the new generator, code from top to bottom executes within the generator until it reaches it's first yield. Many yields can appear within one generator. When this is the case a next method call will execute code from yield to yield. Code that appears in a loop after a yield keyword is executed on the next() method call. I hope I got it right. I love you guys for your patience and examples. It is greatly appreciated and means very much to me! Thank you fellas! -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
vbgunz wrote: OK. I hope my understanding of the yield keyword and generators in a general sense are now better understood. When a generator function is assigned to an identifier, no code is executed and a generator is immediately returned. When the next() method is called on the new generator, code from top to bottom executes within the generator until it reaches it's first yield. Many yields can appear within one generator. When this is the case a next method call will execute code from yield to yield. Code that appears in a loop after a yield keyword is executed on the next() method call. I hope I got it right. I love you guys for your patience and examples. It is greatly appreciated and means very much to me! Thank you fellas! Yep, looks like you have it. ;-) Only need to add what happens if a generator exits the end after the yield(s). If it were a function it would return a None object even if it didn't have a return at the end. But a generator raises a StopIteration Exception. def gen(): ... yield 'hello' ... print 'all done' ... g = gen() g.next() 'hello' g.next() all done Traceback (most recent call last): File stdin, line 1, in ? StopIteration This is the signal to indicate iteration is finished. You don't see it when you are using generators as iterators because it's usually caught by the object or statement using the generator. Cheers, Ron -- http://mail.python.org/mailman/listinfo/python-list
My Generator Paradox!
I am afraid that this is the first time in which I would probably need something explained to me as if I were a little child. I am having a hard time getting this through my thick skull. What in the world is wrong with this!? ''' ### ''' def generatorFunction(sequence=['item1', 'item2', 'item3']): for item in sequence: yield item yieldedValue = generatorFunction() '''this seems to work perfectly.''' print '-' * 32 print yieldedValue # generator object at 0xb723014c print yieldedValue.next() # item1 print yieldedValue.next() # item2 print yieldedValue.next() # item3 '''this is where things don't make any sense!''' print '-' * 32 print generatorFunction() # generator object at 0xb723022c print generatorFunction().next()# item1 print generatorFunction().next()# item1 print generatorFunction().next()# item1 ''' ### ''' the first set of calls assigned to yieldedValue work but the second set without assignment don't. I asked for help on this at #python (I love those people in there!) and was told the following... generatorFunction() is a call (obvious) when calling the second set, I am resetting the iteration and this explains why I only and always get item1. ok. *but* why in the world does the first set of calls work? technically, isn't yieldedValue == generatorFunction() on a name basis? I mean isn't the following technically the same? generatorFunction() yieldedValue = generatorFunction() aren't they both the same? To me they should be but obviously this creates the point of this paradox. I don't understand what is happening here... Can someone care to explain why the assignment works but not the direct call? In a sense shouldn't the assignment yield the same results as the direct call and vice versa? I am confused :( Thank you for any help on this! -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
Em Qui, 2006-03-16 às 16:17 -0800, vbgunz escreveu: print generatorFunction() # generator object at 0xb723022c print generatorFunction().next()# item1 print generatorFunction().next()# item1 print generatorFunction().next()# item1 Each time you say generatorFunction(), it gives you a new generator, thus returning the first item again. -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
vbgunz wrote: def generatorFunction(sequence=['item1', 'item2', 'item3']): for item in sequence: yield item yieldedValue = generatorFunction() You're creating an iterator here and binding it to name yieldedValue (which is bogus, it should be named valueGenerator or sth like that). '''this seems to work perfectly.''' print '-' * 32 print yieldedValue # generator object at 0xb723014c print yieldedValue.next() # item1 print yieldedValue.next() # item2 print yieldedValue.next() # item3 You're calling your iterator's next() method getting all of values, as expected. '''this is where things don't make any sense!''' print '-' * 32 print generatorFunction() # generator object at 0xb723022c You're creating a new iterator here. print generatorFunction().next()# item1 Another anonymous iterator gets created here. Instantly its next() method is called, yielding first value. print generatorFunction().next()# item1 And so on... generatorFunction() is a call (obvious) when calling the second set, I am resetting the iteration and this explains why I only and always get item1. ok. *but* why in the world does the first set of calls work? technically, isn't yieldedValue == generatorFunction() on a name basis? I mean isn't the following technically the same? generatorFunction() yieldedValue = generatorFunction() Well, first statement creates new iterator which is garbage collected right away (as it has no bindings). Second statement creates an iterator and binds it to name yieldedValue. Then it can be used as typical iterator. Calling yieldedValue.next() just calls method of the same iterator you've created a moment ago. It's still the same object. The difference is like the difference between following two lines: list() # creating a new list new_list = list() # creating a new list and binding its name So, rewriting your example from generator to dictionary objects: alist = [1, 2, 3, 4, 5] print alist # [1, 2, 3, 4, 5] print alist.pop() # 5 print alist.pop() # 4 print alist.pop() # 3 print [1, 2, 3, 4, 5] # [1, 2, 3, 4, 5] print [1, 2, 3, 4, 5].pop() # 5 print [1, 2, 3, 4, 5].pop() # 5 print [1, 2, 3, 4, 5].pop() # 5 Remember that generator is an object and you'll be fine. mk -- . o . http://joker.linuxstuff.pl . . o It's easier to get forgiveness for being wrong o o o than forgiveness for being right. -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
vbgunz wrote: I am afraid that this is the first time in which I would probably need something explained to me as if I were a little child. I am having a hard time getting this through my thick skull. What in the world is wrong with this!? ''' ### ''' def generatorFunction(sequence=['item1', 'item2', 'item3']): for item in sequence: yield item yieldedValue = generatorFunction() '''this seems to work perfectly.''' print '-' * 32 print yieldedValue # generator object at 0xb723014c print yieldedValue.next() # item1 print yieldedValue.next() # item2 print yieldedValue.next() # item3 '''this is where things don't make any sense!''' print '-' * 32 print generatorFunction() # generator object at 0xb723022c print generatorFunction().next()# item1 print generatorFunction().next()# item1 print generatorFunction().next()# item1 ''' ### ''' the first set of calls assigned to yieldedValue work but the second set without assignment don't. I asked for help on this at #python (I love those people in there!) and was told the following... generatorFunction() is a call (obvious) when calling the second set, I am resetting the iteration and this explains why I only and always get item1. ok. *but* why in the world does the first set of calls work? technically, isn't yieldedValue == generatorFunction() on a name basis? I mean isn't the following technically the same? generatorFunction() yieldedValue = generatorFunction() aren't they both the same? [snip] In that short example, they happen to be different, but equivalent, objects, but that will not always be the case. Consider this: a = generatorFunction() b = generatorFunction() print a.next() item1 Now a and b are both generators for generatorFunction, but calling a.next() again will return 'item2', while calling b.next() will return 'item1'. The value returned by generatorFunction is an object, which has an internal state that makes it distinct from other objects of the same type. So once your yieldedValue has been altered by calling the next() method, it is no longer equivalent to a fresh instance of the generator. -- David -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
it's easy to explain class X: pass x=X() y=X() x and y are different instances one can put in x x.item = 1 y doesn't even have an attribute item for example similar with generators they are *different* objects of same kind generator def fib(): ... a,b = 1,1 ... while True: ... a,b = b,a+b ... yield a,b ... f1 = fib() f2 = fib() f1 generator object at 0x4042866c f2 generator object at 0x404db42c# different addresses f1 is f2 False f1.next() (1, 2) f1.next() (2, 3) f1.next() (3, 5) f2.next() (1, 2) it's only natural that each objects starts it's own fibonaci serie hth, Daniel -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
vbgunz wrote: I am afraid that this is the first time in which I would probably need something explained to me as if I were a little child. I am having a hard time getting this through my thick skull. What in the world is wrong with this!? ''' ### ''' def generatorFunction(sequence=['item1', 'item2', 'item3']): for item in sequence: yield item yieldedValue = generatorFunction() '''this seems to work perfectly.''' print '-' * 32 print yieldedValue # generator object at 0xb723014c print yieldedValue.next() # item1 print yieldedValue.next() # item2 print yieldedValue.next() # item3 '''this is where things don't make any sense!''' print '-' * 32 print generatorFunction() # generator object at 0xb723022c print generatorFunction().next()# item1 print generatorFunction().next()# item1 print generatorFunction().next()# item1 ''' ### ''' the first set of calls assigned to yieldedValue work but the second set without assignment don't. I asked for help on this at #python (I love those people in there!) and was told the following... generatorFunction() is a call (obvious) when calling the second set, I am resetting the iteration and this explains why I only and always get item1. ok. *but* why in the world does the first set of calls work? technically, isn't yieldedValue == generatorFunction() on a name basis? I mean isn't the following technically the same? generatorFunction() yieldedValue = generatorFunction() No. Look at this a = generatorFunction() b = generatorFunction() a==b False Why aren'y they the same? Here's a clue: generatorFunction() generator object at 0x00B28418 generatorFunction() generator object at 0x00AD24E0 Note the addresses are different. Try this b.next() 'item1' b.next() 'item2' generatorFunction().next() 'item1' Just like your example, the generator re-initailized and printed item1. But that's not the same generator as b b.next() 'item3' aren't they both the same? To me they should be but obviously this creates the point of this paradox. I don't understand what is happening here... Can someone care to explain why the assignment works but not the direct call? In a sense shouldn't the assignment yield the same results as the direct call and vice versa? I am confused :( Thank you for any help on this! -- http://mail.python.org/mailman/listinfo/python-list
Re: My Generator Paradox!
vbgunz wrote: I am afraid that this is the first time in which I would probably need something explained to me as if I were a little child. I am having a hard time getting this through my thick skull. What in the world is wrong with this!? ''' ### ''' def generatorFunction(sequence=['item1', 'item2', 'item3']): for item in sequence: yield item yieldedValue = generatorFunction() '''this seems to work perfectly.''' print '-' * 32 print yieldedValue # generator object at 0xb723014c print yieldedValue.next() # item1 print yieldedValue.next() # item2 print yieldedValue.next() # item3 '''this is where things don't make any sense!''' print '-' * 32 print generatorFunction() # generator object at 0xb723022c print generatorFunction().next()# item1 print generatorFunction().next()# item1 print generatorFunction().next()# item1 ''' ### ''' does the following surprise you too ? f = open(filename) print f.readline() # prints first line print f.readline() # prints second line print f.readline() # prints third line print open(filename).readline() # prints first line print open(filename).readline() # prints first line print open(filename).readline() # prints first line /F -- http://mail.python.org/mailman/listinfo/python-list