Append to python List

[RAHUL RAJ]

I'm getting same output for both code parts, why not for th code parts in question?

On Thursday, May 9, 2013 1:48:51 PM UTC+5:30, RAHUL RAJ wrote:
> Then what about this code part?
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> [(x, y) for x in [1,2,3] for y in [3,1,4] if x != y]
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> and the following code part:
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> for x in [1,2,3]:
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>   for y in [3,1,4]:
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>     if x != y:
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>       combs.append((x, y))
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> On Thursday, May 9, 2013 12:24:24 PM UTC+5:30, Gary Herron wrote:
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> > On 05/08/2013 11:36 PM, RAHUL RAJ wrote:
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> > 
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> > > Checkout the following code:
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> > >
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> > > sample2 = [x+y for x in range(1,10) for y in range(1,10) if x!=y]
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> > 
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> > > output=[]
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> > 
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> > > output=[x for x in sample2 if x not in output]
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> > 
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> > This statement is not doing what you expect.  It is not building a list 
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> > 
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> > in the variable named output, it is building a list (anonymously) then 
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> > 
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> > binding it to the variable output once it's built.  Therefore output is 
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> > [] for the whole list building operation.
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> > The later operation works, because your *are* building the list in place 
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> > as you go.
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> > >
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> > > the output I get is
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> > > 3 4 5 6 7 8 9 10 3 5 6 7 8 9 10 11 4 5 7 8 9 10 11 12 5 6 7 9 10 11 12 13 6 7 8 9 11 12 13 14 7 8 9 10 11 13 14 15 8 9 10 11 12 13 15 16 9 10 11 12 13 14 15 17 10 11 12 13 14 15 16 17
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> > 
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> > >
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> > > which contains duplicate values.
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> > >
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> > >
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> > >
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> > >
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> > > But if I do like this:
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> > >
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> > > sample2 = [x+y for x in range(1,10) for y in range(1,10) if x!=y]
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> > 
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> > > output=[]
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> > 
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> > > for x in sample2:
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> > >    if x not in output:
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> > 
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> > >       output.append(x)
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> > >
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> > >
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> > > the value of 'output' I get like this:
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> > 
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> > > 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
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> > >
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> > > I know that both the programs have the same functionality, but why do I have different outputs?
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> > >
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> > > Please help!