

Hello, dear participants!
Could you tip me, is there any simple and nice way to build scatterplot
for three different types of data (, and o and *  signs, for example)
with legend.
Now i can guess only that way:
plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
points(.... etc
What is the simple and nice way?
Thank you very much for your kindness and help.

Evgeniy Kachalin
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Evgeniy Kachalin wrote:
> Hello, dear participants!
>
> Could you tip me, is there any simple and nice way to build scatterplot
> for three different types of data (, and o and *  signs, for example)
> with legend.
>
> Now i can guess only that way:
>
> plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
> points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
> points(.... etc
>
> What is the simple and nice way?
> Thank you very much for your kindness and help.
>
Example:
with(iris,
plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
with(iris,
legend(7, 4.4, legend = unique(as.character(Species)),
pch = unique(as.integer(Species))))
Uwe Ligges
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Uwe Ligges пишет:
> Evgeniy Kachalin wrote:
>
>> Hello, dear participants!
>>
>> Could you tip me, is there any simple and nice way to build
>> scatterplot for three different types of data (, and o and *  signs,
>> for example) with legend.
>>
>> Now i can guess only that way:
>>
>> plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
>> points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
>> points(.... etc
>>
>> What is the simple and nice way?
>> Thank you very much for your kindness and help.
>>
>
>
> Example:
>
>
> with(iris,
> plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
> with(iris,
> legend(7, 4.4, legend = unique(as.character(Species)),
> pch = unique(as.integer(Species))))
>
Uwe, sorry for my stupid question. You mean that when pch=factor , plot
can recycle the factor and use it for subscripts or marks.
Then pch=as.integer(Species) results in c(1,2,3) for 3 factor levels.
And I need symbols 15,16,17 and colors red, blue, green.
So then I do:
iris$Species>spec.symb
iris$Species>spec.col
levels(spec.symb)<c(15,16,17)
levels(spec.col)<c('red','green','blue')
That's the only way?
More of that!!! 'Plot' does not like factors in 'pch'. So it must be so:
plot(x~y,data, pch=as.integer(as.character(spec.symb))).
That's totally crazy...

Evgeniy
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Hi!
Just use your factors for indexing c(15,16,17) and
c("red","green","blue"). So, with the iris data:
>with(iris, plot(Sepal.Length, Sepal.Width,
pch=c(15,16,17)[as.integer(Species)],
col=c("red","green","blue")[as.integer(Species)] ))
Best regards,
Kyosti Kurikka
> > Evgeniy Kachalin wrote:
> >
> >> Hello, dear participants!
> >>
> >> Could you tip me, is there any simple and nice way to build
> >> scatterplot for three different types of data (, and o and *  signs,
> >> for example) with legend.
> >>
> >> Now i can guess only that way:
> >>
> >> plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
> >> points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
> >> points(.... etc
> >>
> >> What is the simple and nice way?
> >> Thank you very much for your kindness and help.
> >>
> >
> >
> > Example:
> >
> >
> > with(iris,
> > plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
> > with(iris,
> > legend(7, 4.4, legend = unique(as.character(Species)),
> > pch = unique(as.integer(Species))))
> >
>
> Uwe, sorry for my stupid question. You mean that when pch=factor , plot
> can recycle the factor and use it for subscripts or marks.
>
> Then pch=as.integer(Species) results in c(1,2,3) for 3 factor levels.
> And I need symbols 15,16,17 and colors red, blue, green.
>
> So then I do:
> iris$Species>spec.symb
> iris$Species>spec.col
> levels(spec.symb)<c(15,16,17)
> levels(spec.col)<c('red','green','blue')
>
> That's the only way?
> More of that!!! 'Plot' does not like factors in 'pch'. So it must be so:
> plot(x~y,data, pch=as.integer(as.character(spec.symb))).
> That's totally crazy...
>
> 
> Evgeniy
>
> ______________________________________________
> [hidden email] mailing list
> https://stat.ethz.ch/mailman/listinfo/rhelp> PLEASE do read the posting guide! http://www.Rproject.org/postingguide.html>
>
______________________________________________
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Evgeniy Kachalin wrote:
> Uwe Ligges пишет:
>
>> Evgeniy Kachalin wrote:
>>
>>> Hello, dear participants!
>>>
>>> Could you tip me, is there any simple and nice way to build
>>> scatterplot for three different types of data (, and o and * 
>>> signs, for example) with legend.
>>>
>>> Now i can guess only that way:
>>>
>>> plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
>>> points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
>>> points(.... etc
>>>
>>> What is the simple and nice way?
>>> Thank you very much for your kindness and help.
>>>
>>
>>
>> Example:
>>
>>
>> with(iris,
>> plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
>> with(iris,
>> legend(7, 4.4, legend = unique(as.character(Species)),
>> pch = unique(as.integer(Species))))
>>
>
> Uwe, sorry for my stupid question. You mean that when pch=factor , plot
> can recycle the factor and use it for subscripts or marks.
Yes, it can recycle, but in the example above it does not recycle but
takes the whole "Species" vector.
> Then pch=as.integer(Species) results in c(1,2,3) for 3 factor levels.
> And I need symbols 15,16,17 and colors red, blue, green.
What about adding 14 as in as.integer(Species)+14, or 1 for the colors,
respectively?
> So then I do:
> iris$Species>spec.symb
> iris$Species>spec.col
> levels(spec.symb)<c(15,16,17)
> levels(spec.col)<c('red','green','blue')
>
> That's the only way?
This is one qay of many.
> More of that!!! 'Plot' does not like factors in 'pch'. So it must be so:
> plot(x~y,data, pch=as.integer(as.character(spec.symb))).
> That's totally crazy...
You can set up your own pch variable of course, if you don't like it
this fast and easy way.
Uwe Ligges
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