On Thu, Jun 11, 2009 at 3:30 PM, <bangpypers-requ...@python.org> wrote: > Send BangPypers mailing list submissions to > bangpyp...@python.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://mail.python.org/mailman/listinfo/bangpypers > or, via email, send a message with subject or body 'help' to > bangpypers-requ...@python.org > > You can reach the person managing the list at > bangpypers-ow...@python.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of BangPypers digest..." > > > Today's Topics: > > 1. [New bie question ] Clarification on "Multiply" operator > applied to list(data structure) (Aman Aggarwal) > 2. Re: [New bie question ] Clarification on "Multiply" operator > applied to list(data structure) (Senthil Kumaran) > 3. Re: [New bie question ] Clarification on "Multiply" operator > applied to list(data structure) (Shivaraj M S) > 4. Re: [New bie question ] Clarification on "Multiply" operator > applied to list(data structure) (Shivaraj M S) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 10 Jun 2009 16:57:39 +0530 > From: Aman Aggarwal <aman....@gmail.com> > To: bangpypers <bangpypers@python.org> > Subject: [BangPypers] [New bie question ] Clarification on "Multiply" > operator applied to list(data structure) > Message-ID: > <882494550906100427x1a99ac60h9540f04cc656a...@mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > Hello there > > I am reading?"How to think like a computer scientist"?which is an > introductory test in Python. > > I wanna clarify the behaviour of multiply operator(*) when applied to > list(data structure). > > Consider the function?make_matrix > > def make_matrix(rows, columns): > """ > ? >>> make_matrix(4, 2) > ? [[0, 0], [0, 0], [0, 0], [0, 0]] > ? >>> m = make_matrix(4, 2) > ? >>> m[1][1] = 7 > ? >>> m > ? [[0, 0], [0, 7], [0, 0], [0, 0]] > """ > return [[0] * columns] * rows > > > > > The actual output is > > [[0, 7], [0, 7], [0, 7], [0, 7]] > > > > > The correct version of?make_matrix?is : > > > > def make_matrix(rows, columns): > """ > ? >>> make_matrix(3, 5) > ? [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]] > ? >>> make_matrix(4, 2) > ? [[0, 0], [0, 0], [0, 0], [0, 0]] > ? >>> m = make_matrix(4, 2) > ? >>> m[1][1] = 7 > ? >>> m > ? [[0, 0], [0, 7], [0, 0], [0, 0]] > """ > matrix = [] > for row in range(rows): > ? ? matrix += [[0] * columns] > return matrix > > > The reason why first version of?make_matrix?fails ( as explained in > the book at 9.8 ) is that > > "...each row is an alias of the other rows..." > > I wonder why > > [[0] * columns] * rows > > causes?"...each row is an alias of the other rows..." > > but not > > [[0] * columns] > > i.e. why each [0] in a row is not an alias of other row element. > > > > > > /* > ?Everything worth doing is worth doing in excess > */ > > > ------------------------------ > > Message: 2 > Date: Wed, 10 Jun 2009 19:45:20 +0530 > From: Senthil Kumaran <orsent...@gmail.com> > To: Bangalore Python Users Group - India <bangpypers@python.org> > Subject: Re: [BangPypers] [New bie question ] Clarification on > "Multiply" operator applied to list(data structure) > Message-ID: <20090610141520.gb14...@ubuntu.ubuntu-domain> > Content-Type: text/plain; charset=iso-8859-1 > > On Wed, Jun 10, 2009 at 04:57:39PM +0530, Aman Aggarwal wrote: >> >> I am reading?"How to think like a computer scientist"?which is an >> introductory test in Python. >> >> I wanna clarify the behaviour of multiply operator(*) when applied to >> list(data structure). >> >> >> I wonder why >> >> [[0] * columns] * rows >> >> causes?"...each row is an alias of the other rows..." >> >> but not >> >> [[0] * columns] >> >> i.e. why each [0] in a row is not an alias of other row element. > > > It is bit tricky in that code.. > You have to understand that in the second option, the matrix is formed > row-by-row, wherein a new row (list) is created as [0] * columns. > > > This is what the second version essentially is doing: > >>>> rows = 2 >>>> cols = 2 >>>> mat = [] >>>> mat = [[0] * cols] # first row >>>> mat = mat + [[0] * cols] # second row >>>> mat > [[0, 0], [0, 0]] >>>> mat[0][0] = 'Spam' >>>> mat > [['Spam', 0], [0, 0]] >>>> > > This is the what the first version of the code was doing. > >>>> rows = 2 >>>> cols = 2 >>>> mat = [] >>>> mat = [[0] * cols] # first row >>>> mat = mat * 2 # second row >>>> mat > [[0, 0], [0, 0]] >>>> mat[0][0] = 'Spam' >>>> mat > [['Spam', 0], ['Spam', 0]] >>>> > > Does this clarify your doubt? > > -- > Senthil > > > > ------------------------------ > > Message: 3 > Date: Wed, 10 Jun 2009 06:07:27 -0700 (PDT) > From: Shivaraj M S <shivraj...@gmail.com> > To: bangpypers@python.org > Subject: Re: [BangPypers] [New bie question ] Clarification on > "Multiply" operator applied to list(data structure) > Message-ID: <23961937.p...@talk.nabble.com> > Content-Type: text/plain; charset=UTF-8 > > > The first one is a reference and second one is not. > case 1: >>>> m = [[0]*2]*4 >>>> m[1][1]=7 >>>> m > [[0, 7], [0, 7], [0, 7], [0, 7]] > > case 2: >>>> m=[] >>>> for r in range(4): > ... m+=[[0]*2] > ... >>>> m > [[0, 0], [0, 0], [0, 0], [0, 0]] >>>> m[1][1]=7 >>>> m > [[0, 0], [0, 7], [0, 0], [0, 0]] > > We can make the second case look like first just by creating a reference to > [[0]*2] which is what was done implicitly when multiplication operator was > used. > > case 2 - modified for reference: >>>> m=[] >>>> mi = [[0]*2] >>>> for r in range(4): > ... m+=mi > ... >>>> m > [[0, 0], [0, 0], [0, 0], [0, 0]] >>>> mi[0][1]=7 >>>> m > [[0, 7], [0, 7], [0, 7], [0, 7]] >>>> m[1][1]=1 >>>> m > [[0, 1], [0, 1], [0, 1], [0, 1]] >>>> mi > [[0, 1]] > > > > > Aman Aggarwal-4 wrote: >> >> Hello there >> >> I am reading?"How to think like a computer scientist"?which is an >> introductory test in Python. >> >> I wanna clarify the behaviour of multiply operator(*) when applied to >> list(data structure). >> >> Consider the function?make_matrix >> >> def make_matrix(rows, columns): >> """ >> ? >>> make_matrix(4, 2) >> ? [[0, 0], [0, 0], [0, 0], [0, 0]] >> ? >>> m = make_matrix(4, 2) >> ? >>> m[1][1] = 7 >> ? >>> m >> ? [[0, 0], [0, 7], [0, 0], [0, 0]] >> """ >> return [[0] * columns] * rows >> >> >> >> >> The actual output is >> >> [[0, 7], [0, 7], [0, 7], [0, 7]] >> >> >> >> >> The correct version of?make_matrix?is : >> >> >> >> def make_matrix(rows, columns): >> """ >> ? >>> make_matrix(3, 5) >> ? [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]] >> ? >>> make_matrix(4, 2) >> ? [[0, 0], [0, 0], [0, 0], [0, 0]] >> ? >>> m = make_matrix(4, 2) >> ? >>> m[1][1] = 7 >> ? >>> m >> ? [[0, 0], [0, 7], [0, 0], [0, 0]] >> """ >> matrix = [] >> for row in range(rows): >> ? ? matrix += [[0] * columns] >> return matrix >> >> >> The reason why first version of?make_matrix?fails ( as explained in >> the book at 9.8 ) is that >> >> "...each row is an alias of the other rows..." >> >> I wonder why >> >> [[0] * columns] * rows >> >> causes?"...each row is an alias of the other rows..." >> >> but not >> >> [[0] * columns] >> >> i.e. why each [0] in a row is not an alias of other row element. >> >> >> >> >> >> /* >> ?Everything worth doing is worth doing in excess >> */ >> _______________________________________________ >> BangPypers mailing list >> BangPypers@python.org >> http://mail.python.org/mailman/listinfo/bangpypers >> >> > > -- > View this message in context: > http://www.nabble.com/-New-bie-question---Clarification-on-%22Multiply%22-operator-applied-to-list%28data-structure%29-tp23960277p23961937.html > Sent from the BangPypers - Bangalore Python Users Group mailing list archive > at Nabble.com. > > > > ------------------------------ > > Message: 4 > Date: Wed, 10 Jun 2009 06:09:16 -0700 (PDT) > From: Shivaraj M S <shivraj...@gmail.com> > To: bangpypers@python.org > Subject: Re: [BangPypers] [New bie question ] Clarification on > "Multiply" operator applied to list(data structure) > Message-ID: <23961977.p...@talk.nabble.com> > Content-Type: text/plain; charset=UTF-8 > > > The first one is a case of reference/alias and second one is not. > case 1: >>>> m = [[0]*2]*4 >>>> m[1][1]=7 >>>> m > [[0, 7], [0, 7], [0, 7], [0, 7]] > > case 2: >>>> m=[] >>>> for r in range(4): > ... m+=[[0]*2] > ... >>>> m > [[0, 0], [0, 0], [0, 0], [0, 0]] >>>> m[1][1]=7 >>>> m > [[0, 0], [0, 7], [0, 0], [0, 0]] > > We can make the second case look like first just by creating a reference to > [[0]*2] which is what was done implicitly when multiplication operator was > used. > > case 2 - modified for reference: >>>> m=[] >>>> mi = [[0]*2] >>>> for r in range(4): > ... m+=mi > ... >>>> m > [[0, 0], [0, 0], [0, 0], [0, 0]] >>>> mi[0][1]=7 >>>> m > [[0, 7], [0, 7], [0, 7], [0, 7]] >>>> m[1][1]=1 >>>> m > [[0, 1], [0, 1], [0, 1], [0, 1]] >>>> mi > [[0, 1]] > > > > > Aman Aggarwal-4 wrote: >> >> Hello there >> >> I am reading?"How to think like a computer scientist"?which is an >> introductory test in Python. >> >> I wanna clarify the behaviour of multiply operator(*) when applied to >> list(data structure). >> >> Consider the function?make_matrix >> >> def make_matrix(rows, columns): >> """ >> ? >>> make_matrix(4, 2) >> ? [[0, 0], [0, 0], [0, 0], [0, 0]] >> ? >>> m = make_matrix(4, 2) >> ? >>> m[1][1] = 7 >> ? >>> m >> ? [[0, 0], [0, 7], [0, 0], [0, 0]] >> """ >> return [[0] * columns] * rows >> >> >> >> >> The actual output is >> >> [[0, 7], [0, 7], [0, 7], [0, 7]] >> >> >> >> >> The correct version of?make_matrix?is : >> >> >> >> def make_matrix(rows, columns): >> """ >> ? >>> make_matrix(3, 5) >> ? [[0, 0, 0, 0, 0], [0, 0, 0, 0, 0], [0, 0, 0, 0, 0]] >> ? >>> make_matrix(4, 2) >> ? [[0, 0], [0, 0], [0, 0], [0, 0]] >> ? >>> m = make_matrix(4, 2) >> ? >>> m[1][1] = 7 >> ? >>> m >> ? [[0, 0], [0, 7], [0, 0], [0, 0]] >> """ >> matrix = [] >> for row in range(rows): >> ? ? matrix += [[0] * columns] >> return matrix >> >> >> The reason why first version of?make_matrix?fails ( as explained in >> the book at 9.8 ) is that >> >> "...each row is an alias of the other rows..." >> >> I wonder why >> >> [[0] * columns] * rows >> >> causes?"...each row is an alias of the other rows..." >> >> but not >> >> [[0] * columns] >> >> i.e. why each [0] in a row is not an alias of other row element. >> >> >> >> >> >> /* >> ?Everything worth doing is worth doing in excess >> */ >> _______________________________________________ >> BangPypers mailing list >> BangPypers@python.org >> http://mail.python.org/mailman/listinfo/bangpypers >> >> > > -- > View this message in context: > http://www.nabble.com/-New-bie-question---Clarification-on-%22Multiply%22-operator-applied-to-list%28data-structure%29-tp23960277p23961977.html > Sent from the BangPypers - Bangalore Python Users Group mailing list archive > at Nabble.com. > > > > ------------------------------ > > _______________________________________________ > BangPypers mailing list > BangPypers@python.org > http://mail.python.org/mailman/listinfo/bangpypers > > > End of BangPypers Digest, Vol 22, Issue 9 > ***************************************** >
Hi there thanks for the response. I posted the same question here : http://stackoverflow.com/questions/974931/multiply-operator-applied-to-listdata-structure The discussions on that page clarify my doubt. Kindly read it. Thanks and kind regards Aman _______________________________________________ BangPypers mailing list BangPypers@python.org http://mail.python.org/mailman/listinfo/bangpypers
