Re: st: axis bar starting from 0 on graph (not the label)

[Nick Cox <njcoxstata@gmail.com>]

Yes and no. You can always add a line at y = 0 by -yline()- and make
that as emphatic as you like by varying its width and colour.
Making this a light line and the background a little darker can work
as well as making this a dark line against a light background.

You would have to fight Stata every stop of the way if you wanted to
move axis labels, ticks and titles. You'd need to re-create your own.
Stata's bias is that although people do sometimes want axes running
through the graph, they usually end up very messy, so it won't play
willingly. Most statistical software seems to have moved away from the
conventions that the axes must be the lines y = 0, x = 0 as high
school algebra insists and to separate out the functions of titling,
labelling and ticking from showing where the zero lines are.

That does not rule out something clever with the Editor or Stata's
class language, except that I don't know what it is.

If you have access to MS Excel, what you want is I think the default
any way, as is all too evident from work I often see. The essence of
the problem is that if data points or line can be negatives they often
collide with axis stuff.

See also

http://www.stata-journal.com/article.html?article=gr0043

There have been postings on this in the past, but I didn't find them
on a quick search of the archives.

Nick

On Thu, Oct 20, 2011 at 6:11 AM, lreine ycenna <lreine.ycenna@gmail.com> wrote:
> HI,
> Is it possible to have the axis line start from where 0 is on the
> y-axis? (I'm not talking about labeling my x-axis from 0, but the
> x-axis bar) i.e. I have both positive and negative values on my y-axis
> and currently the x-xaxis bar starts from the very bottom end of my
> y-axis which is negative. Here's the command I use.
>
> xtreg Y A B AxB, fe
> generate income=((_n-1)/10)
> matrix b=e(b)
> matrix V=e(V)
> scalar b1=b[1,1]
> scalar b2=b[1,2]
> scalar b3=b[1,3]
> scalar varb1=V[1,1]
> scalar varb2=V[2,2]
> scalar varb3=V[3,3]
> scalar covb1b3=V[1,3]
> scalar covb2b3=V[2,3]
> scalar list b1 b2 b3 varb1 varb2 varb3 covb1b3 covb2b3
> gen lab=b1+b3*income
> gen conse=sqrt(varb1+varb3*(income^2)+2*covb1b3*income)
> gen a=1.96*conse
> gen upper=conb+a
> gen lower=conb-a
> graph twoway  (line lab income, clwidth(medium) clcolor(blue) clcolor(black))

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