Re: dictionary error: list assignment index out of range
From: Vaduvoiu Tiberiu
> Well, to quote firefox: this is embarrassing. I've realized the dictionary
initialization is wrong, as [] means its a tuple, I should use {}. That's why I
> don't like working nights..it's only in the morning when you start seeing
things better. I apologize for the mail. Cheers
[] is for lists.
() is for tuples.
HTH.
Octavian
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Re: dictionary error: list assignment index out of range
Well, to quote firefox: this is embarrassing. I've realized the dictionary
initialization is wrong, as [] means its a tuple, I should use {}. That's why I
don't like working nights..it's only in the morning when you start seeing
things
better. I apologize for the mail. Cheers
From: Vaduvoiu Tiberiu
To: python-list@python.org
Sent: Fri, January 28, 2011 9:34:57 AM
Subject: dictionary error: list assignment index out of range
Hy everyone, I'm trying to learng python for a week or two and there's a thing
that is really disturbing me as I do not understand what the problem is. I'm
trying to use a dictionary to remember when a user has visited a city. Code is
really basic:
in the class init method I added
self.visited = []
and in the function where i check if city was visited:
cityNumber = 1 #example
if (not cityNumber in self.visited):
#do some stuff
self.visited[cityNumber] = "true"
Apparently the last line causes the error: list assignment index out of range.
I
read that this is the simplest way to assign a value to a
dictionary(dict[key]=value). So why is the error appearing?? Thanks a lot in
advance
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dictionary error: list assignment index out of range
Hy everyone, I'm trying to learng python for a week or two and there's a thing that is really disturbing me as I do not understand what the problem is. I'm trying to use a dictionary to remember when a user has visited a city. Code is really basic: in the class init method I added self.visited = [] and in the function where i check if city was visited: cityNumber = 1 #example if (not cityNumber in self.visited): #do some stuff self.visited[cityNumber] = "true" Apparently the last line causes the error: list assignment index out of range. I read that this is the simplest way to assign a value to a dictionary(dict[key]=value). So why is the error appearing?? Thanks a lot in advance -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment using concatenation "*"
I suggest you should build your list using a list comprehension: >>>a = [[0]*3 for i in range(3)] >>>a [[0, 0, 0], [0, 0, 0], [0, 0, 0]] >>>a[0][1] = 1 [[0, 1, 0], [0, 0, 0], [0, 0, 0]] -- Steve R. Hastings"Vita est" [EMAIL PROTECTED]http://www.blarg.net/~steveha -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment using concatenation "*"
> if I do: > > a = [ [0] * 3 ] * 3 > a[0][1] = 1 > > I get > > a = [[0,1,0],[0,1,0],[0,1,0]] The language reference calls '*' the "repetition" operator. It's not making copies of what it repeats, it is repeating it. Consider the following code: >>> a = [] >>> b = [] >>> a == b True >>> a is b False >>> >>> a = b = [] >>> a is b True >>> a.append(1) >>> a [1] >>> b [1] Each time you use [], you are creating a new list. So the first code sets a and b to two different new lists. The second one, "a = b = []", only creates a single list, and binds both a and b to that same list. In your example, first you create a list containing [0, 0, 0]; then you repeat the same list three times. >>> a = [[0]*3]*3 >>> a [[0, 0, 0], [0, 0, 0], [0, 0, 0]] >>> a[0] is a[1] True >>> a[0] is a[2] True When you run [0]*3 you are repeating 0 three times. But 0 is not mutable. When you modify a[0] to some new value, you are replacing a reference to the immutable 0 with some new reference. Thus, [0]*3 is a safe way to create a list of three 0 values. When you have a list that contains three references to the same mutable, and you change the mutable, you get the results you discovered. -- Steve R. Hastings"Vita est" [EMAIL PROTECTED]http://www.blarg.net/~steveha -- http://mail.python.org/mailman/listinfo/python-list
list assignment using concatenation "*"
If I do: a = [ [0,0,0], [0,0,0], [0,0,0] ] a[0][1] = 1 I get: a = [ [0,1,0],[0,0,0],[0,0,0] ] as expected But if I do: a = [ [0] * 3 ] * 3 a[0][1] = 1 I get a = [[0,1,0],[0,1,0],[0,1,0]] AFAIC, "*" is supposed to generate multiple copies of the given token. Therefore I thought both cases would be the same, but they are not. Can anyone explain to me what exactly is going on in the second case? -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment
Norvell Spearman <[EMAIL PROTECTED]> writes: > Lutz and Ascher have tuple and list assignment as separate entries in > their assignment statement forms table so I was expecting there to be > some difference; thanks for setting me straight. In older Python versions there was a difference between list unpacking and tuple unpacking. The former would only work with lists and the latter with tuples. With Python 1.5, both were unified into a more general sequence unpacking, but for backwards compatibility both syntaxes were kept. Bernhard -- Intevation GmbH http://intevation.de/ Skencil http://skencil.org/ Thuban http://thuban.intevation.org/ -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment
Jeffrey Schwab wrote: > TMTOWTDI, after all. :) A bit ironic that that's the official motto of Perl, don't you think? -- Norvell Spearman -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment
Raymond Hettinger wrote: > It's not different. They are ways of writing the same thing. Lutz and Ascher have tuple and list assignment as separate entries in their assignment statement forms table so I was expecting there to be some difference; thanks for setting me straight. -- Norvell Spearman -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment
Raymond Hettinger wrote: >> [spam, ham] = ['yum', 'YUM'] >> >>I don't see how this is any different than a tuple unpacking assignment: >> >> >>> a, b = 1, 2 > > > It's not different. They are ways of writing the same thing. TMTOWTDI, after all. :) -- http://mail.python.org/mailman/listinfo/python-list
Re: list assignment
> [spam, ham] = ['yum', 'YUM'] > > I don't see how this is any different than a tuple unpacking assignment: > > >>> a, b = 1, 2 It's not different. They are ways of writing the same thing. Raymond Hettinger -- http://mail.python.org/mailman/listinfo/python-list
list assignment
In "Learning Python," by Lutz and Ascher, there's a table showing different assignment statement forms. One form shown is list assignment. The authors give this as an example: [spam, ham] = ['yum', 'YUM'] I don't see how this is any different than a tuple unpacking assignment: >>> a, b = 1, 2 >>> a, b (1, 2) >>> [a, b] = [1, 2] >>> a, b (1, 2) In both instances the names a and b are both mapped to 1 and 2 so why are there two different forms? Thanks for any answers. -- Norvell Spearman -- http://mail.python.org/mailman/listinfo/python-list
