Re: [R] ColorBrewer question

2012-01-30 Thread Mario Giesel
It works! Thanks a lot for your explanations, Michael.
 
Good luck,
 Mario



Von: R. Michael Weylandt michael.weyla...@gmail.com

Cc: r-help@r-project.org r-help@r-project.org 
Gesendet: 5:22 Montag, 30.Januar 2012
Betreff: Re: [R] ColorBrewer question

I believe you need to use the scale_fill_brewer since fill is the
color of the bars while color is the outside of the bars in
ggplot2-speak:

E.g., with built-in data (it's polite to provide yours so that your
minimal working example is working):

data(diamonds)
ggplot(diamonds, aes(clarity)) + geom_bar(aes(fill = clarity, color = clarity))

# Note the borders are now changed but the fill is the same
ggplot(diamonds, aes(clarity)) + geom_bar(aes(fill = clarity, color =
clarity)) + scale_color_brewer(pal = Blues)

# Now the fill is changed, but you probably want to drop the border
coloring since it's hideous against the blues
ggplot(diamonds, aes(clarity)) + geom_bar(aes(fill = clarity, color =
clarity)) + scale_fill_brewer(pal = Blues)

# So lovely
ggplot(diamonds, aes(clarity)) + geom_bar(aes(fill = clarity)) +
scale_fill_brewer(pal = Blues)

Michael


 Hello, R friends,

 I'm trying to change colors of my horizontal bars so that they show a 
 sequence.
 I chose the ColorBrewer palette Blues. However the resulting plot doesn't 
 show any changes to the default.
 I tried several places of + scale_colour_brewer(type=seq, pal = Blues) 
 with no effect.
 This is my code:

 p - ggplot(data, aes(x = gender))  + 
 scale_y_continuous(,formatter=percent) + xlab(Gender) + coord_flip() +  
    scale_colour_brewer(type=seq, pal = Blues)
 p+geom_bar(aes(fill=pet),colour='black',position='fill')


 Any ideas welcome.
 Thanks,
  Mario
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[R] Displaying percentages within bars

2012-01-30 Thread Mario Giesel
Hello, R friends,
 
I've got this graph:

p - ggplot(diamonds, aes(x = color))  + scale_fill_brewer(type=seq, pal = 
Blues)
  + scale_y_continuous(,formatter=percent) + coord_flip()
p+geom_bar(aes(fill=cut),colour='black',position='fill')
 
Is it possible to place percentages within each field of the bars?
So, for instance, the dark blue field of color = D would contain a number of 
about 42.0%.
A second question is how to change the size of this number.
 
Any comments are welcome!
Mario
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[R] ColorBrewer question

2012-01-29 Thread Mario Giesel
Hello, R friends,
 
I'm trying to change colors of my horizontal bars so that they show a sequence.
I chose the ColorBrewer palette Blues. However the resulting plot doesn't 
show any changes to the default.
I tried several places of + scale_colour_brewer(type=seq, pal = Blues) 
with no effect.
This is my code:
 
p - ggplot(data, aes(x = gender))  + 
scale_y_continuous(,formatter=percent) + xlab(Gender) + coord_flip() +    
 scale_colour_brewer(type=seq, pal = Blues)
p+geom_bar(aes(fill=pet),colour='black',position='fill') 
 
 
Any ideas welcome.
Thanks,
 Mario
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Re: [R] Horizontal stacked 100% bars with ggplot2

2012-01-29 Thread Mario Giesel
Thanks, Carlos,
 
 your solution looks nice, however, there are 4 bars instead of 2 i.e. bars are 
not stacked.
Meanwhile I got a solution for that task.
So thanks for sharing your version of it!
 
Good luck,
 Mario



Von: Carlos Ortega c...@qualityexcellence.es

Cc: r-help@r-project.org r-help@r-project.org 
Gesendet: 23:02 Samstag, 28.Januar 2012
Betreff: Re: [R] Horizontal stacked 100% bars with ggplot2


Hello,

If it helps...:



Lines - pet gender
dog male
dog female
dog male
cat female
cat female
cat male


d.f - read.table(textConnection(Lines), header=T, as.is = TRUE)

d.tab-table(d.f$pet, d.f$gender)
d.f.tab-as.data.frame(table(d.f$pet, d.f$gender))
names(d.f.tab)-c('pet','gender', 'Freq')


library(lattice)
histogram(
  ~ Freq | pet *gender, data=d.f.tab, 
  groups=gender, stack=T, horizontal=T
  )


Regards,
Carlos Ortega
www.qualityexcellence.es





Hello, R friends,

I'm trying to crack this nut:


Example Data.

pet    gender
dog    male
dog    female
dog    male
cat    female
cat    female
cat    male

Plot Task.

Horizontal 100% bars where
y axis shows gender factor (male vs. female)
and x axis shows percentage of kind of pets (dog vs. cat)
so that % dogs + % cats are stacked in 1 bar and sum up to 100% (for each 
gender group 1 bar).

How can this be done with ggplot2?

[[elided Yahoo spam]]

Mario

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[R] Horizontal stacked 100% bars with ggplot2

2012-01-27 Thread Mario Giesel
Hello, R friends,

I'm trying to crack this nut:


Example Data.

pet    gender
dog    male
dog    female
dog    male
cat    female
cat    female
cat    male

Plot Task.

Horizontal 100% bars where 
y axis shows gender factor (male vs. female) 
and x axis shows percentage of kind of pets (dog vs. cat)
so that % dogs + % cats are stacked in 1 bar and sum up to 100% (for each 
gender group 1 bar).

How can this be done with ggplot2?

Thanks for any comments!

Mario

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[R] pretty(range(data$z),10) Error

2012-01-17 Thread Mario Giesel
Hello, R-List,
I'm getting error messages when adding geom_density2d() [package ggplot2]:

Fehler in pretty(range(data$z), 10) : 
  NA/NaN/Inf in externem Funktionsaufruf (arg 1)
Zusätzlich: Warnmeldungen:
1: Removed 1 rows containing missing values (stat_contour). 
2: In min(x) : kein nicht-fehlendes Argument für min; gebe Inf zurück
3: In max(x) : kein nicht-fehlendes Argument für max; gebe -Inf zurück
4: In min(x) : kein nicht-fehlendes Argument für min; gebe Inf zurück
5: In max(x) : kein nicht-fehlendes Argument für max; gebe -Inf zurück

 
I installed 2.11.1 Patched in the hope to overcome this obstacle but in vain.
I tried reading spss and csv format - no effect.
 
While this one works:
ggplot(sp2, aes(x=Qq04, y=Qq01)) + geom_point() + facet_wrap(~ segment,ncol=3)
This one fails:
ggplot(sp2, aes(x=Qq04, y=Qq01)) + geom_point() + facet_wrap(~ segment,ncol=3) 
+ geom_density2d()
 
But it only fails for 2 out of 10 x-variables, although they do not differ in 
structure.
Data structure looks like this:
 
segment Qq01 Qq02 Qq02a Qq04 Qq05 Qq07 Qq08 Qq10 Qq11 Qq15 Qq17a
A 7 5 5 5 5 4 5 5 3 7 7
A 5 4 4 3 4 3 4 4 4 5 6
B 3 3 3 3 2 3 4 2 4 3 2
B 6 4 5 3 3 3 4 4 3 6 6
C 3 3 3 3 3 4 2 3 3 4 2
C 2 1 4 3 1 4 1 1 3 4 2
D 5 3 4 3 3 3 4 3 3 6 3
D 5 3 4 2 2 4 4 3 3 4 4
E 3 3 3 2 3 4 4 3 2 3 4
E 7 5 5 5 5 4 4 5 1 7 7
F 6 5 4 3 3 3 4 3 3 6 6
F 5 3 4 3 3 4 4 3 3 4 4
G 4 3 3 3 2 1 3 3 3 4 4
G 6 5 5 5 5 4 5 5 3 7 7
H 6 5 5 4 4 4 5 4 3 5 6
H 7 5 5 5 4 4 5 5 3 7 7
I 6 5 5 4 4 4 5 4 3 6 6
I 6 3 4 2 3 4 4 3 4 6 4
J 5 3 4 1 3 4 4 3 3 5 5
J 4 2 4 3 2 2 3 2 3 4 4
K 4 4 4 2 3 4 4 4 3 5 5
K 6 4 3 3 2 3 4 3 3 6 5
L 6 4 5 3 3 4 5 4 3 6 6
L 3 1 2 2 1 3 3 1 3 4 4
...
About 1.800 cases (more than 100 per segment).
Did anybody experience similar problems?
 
Thanks for all comments,
 Mario
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[R] Reply on ggplot2 - tricky problem

2012-01-06 Thread Mario Giesel
That comes really close to what I had in mind.
Thanks a lot for helping out, Justin!

Good luck,
 Mario


Re: [R] ggplot2 - tricky problem 
 
how bout:

dat-data.frame(id=1:4,city=c('berlin','munich'),likeability=c(5,4,6,5),uniqueness=c(3,4,4,4))

ggplot(ddply(melt(dat,
                          id.vars=c('id','city')),
                  .(variable,city),
                  summarise,
                  value=mean(value)),
          aes(x=factor(city),y=value)) +
geom_point() +
facet_wrap(~variable)

the line drawing is a bit more tricky...  Since the x values are factors 
rather than continuous, fitting a line to them is kind of nonsense.  It 
matters which order they are in for example.  If instead you want to 
plot something like:

ggplot(dat,aes(x=likeability,y=uniqueness,colour=city))+geom_point()+geom_smooth(aes(group=city),method='lm')

You could draw fit lines that make a bit more sense.  Forgive me if I'm over 
simplifying your problem!


Justin

On Thu, Jan 5, 2012 at 7:46 AM, Mario Giesel rr.gie...@yahoo.de wrote:

Hello, R friends,

 I've been struggling quite a bit with ggplot2.
Having worked through Hadleys book twice I still wonder how to solve this task.


1. Short example Dataframe:

id    city    Likeability    Uniqueness
1    Berlin    5    3
2    Munich    4    4
3    Berlin    6    4
4    Munich    5    4

2. Task:

a) Facetting plots for each attitude (1 plot for likeability and uniqueness 
each, horizontally on one page)
b) Showing Berlin and Munich together on x axis
c) Showing the means of Berlin and Munich on y axis (means of cities in 
likeability on first plot, means of cities in uniqueness on second plot)
d) Drawing a line through mean points on each plot



Hope I could explain it understandably. Any help is appreciated!

Thanks a lot,
 Mario

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Antwort an: Antwort an Justin Haynes Senden
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[R] ggplot2 - tricky problem

2012-01-05 Thread Mario Giesel
Hello, R friends,

 I've been struggling quite a bit with ggplot2. 
Having worked through Hadleys book twice I still wonder how to solve this task.


1. Short example Dataframe:

id    city    Likeability    Uniqueness
1    Berlin    5    3
2    Munich    4    4
3    Berlin    6    4
4    Munich    5    4

2. Task:

a) Facetting plots for each attitude (1 plot for likeability and uniqueness 
each, horizontally on one page)
b) Showing Berlin and Munich together on x axis
c) Showing the means of Berlin and Munich on y axis (means of cities in 
likeability on first plot, means of cities in uniqueness on second plot)
d) Drawing a line through mean points on each plot



Hope I could explain it understandably. Any help is appreciated!

Thanks a lot,
 Mario

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[R] Filling a variable with unmentioned categories

2011-11-18 Thread Mario Giesel
Hello, list,
 
I've been struggling with this task for a while looking for an efficient way to 
solve it:
There are two variables 'price' and 'mentioned'.
I want to 'enlarge' data so that missing price points within the price range 
are added to variable price.
Also variable 'mentioned' is to receive values 0 in these cases.
Note: Price points in original data can repeat if several persons mentioned 
that price point.
 
Example:
a) Original data:
price mentioned
0,00 1
0,00 1
0,02 1
0,03 1
0,03 1
0,04 1
0,10 1
0,16 1
0,19 1
0,19 1
 
price range is 0,00 - 0,19. Some prices are missing, e.g. 0,01 .
 
b) Supplemented data:
price mentioned
0,00 1
0,00 1
0,01 0
0,02 1
0,03 1
0,03 1
0,04 1
0,05 0
0,06 0
0,07 0
0,08 0
0,09 0
0,10 1
0,11 0
0,12 0
0,13 0
0,14 0
0,15 0
0,16 1
0,17 0
0,18 0
0,19 1
0,19 1
0,20 0
 
Any ideas are welcome.
Thanks a lot,

 Mario
Aachen, Germany

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Re: [R] Numerical Format on axis

2011-11-17 Thread Mario Giesel
Thoiusand thanks to David and Don.
Great help!
 
Mario
Aachen, Germany



Von: MacQueen, Don macque...@llnl.gov

Cc: r-help@r-project.org r-help@r-project.org; David Winsemius 
dwinsem...@comcast.net
Gesendet: 2:30 Donnerstag, 17.November 2011 
Betreff: Re: [R] Numerical Format on axis

To add to what David suggests, and since you're new to R, something like
this:

plot(x,y, yaxt='n')
yticks - pretty(y)
axis(2, at=yticks, labels=sprintf(%1.2f,yticks))

See the help page for par
  ?par
and look for the entry for 'xaxt' to see what the 'yaxt' arg to plot does.

-- 
Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 11/16/11 6:35 AM, David Winsemius dwinsem...@comcast.net wrote:


On Nov 16, 2011, at 7:41 AM, Mario Giesel wrote:

 Hello, list,

  I'm new to R and I'm trying to produce a chart with currency values
 on the y axis.
 Values should be e.g. 1,00, 1,50, 2,00, etc.
 In fact they are 1,0, 1,5, 2,0, etc.
 How do I get R to show two digits after the comma on that axis?

?sprintf
?format

On the left (geographic) side of the  Atlantic, it might be:

  sprintf(%1.2f, 1)
[1] 1.00

I assume that your system is set up with different options() and that
your punkts are going to be handle to your liking by sprintf.

--

David Winsemius, MD
West Hartford, CT

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On the left (geographic) side of the  Atlantic, it might be:

 sprintf(%1.2f, 1)
[1] 1.00

I assume that your system is set up with different options() and that your 
punkts are going to be handle to your liking by sprintf.

--

David Winsemius, MD
West Hartford, CT
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[R] Numerical Format on axis

2011-11-16 Thread Mario Giesel
Hello, list,
 
 I'm new to R and I'm trying to produce a chart with currency values on the y 
axis.
Values should be e.g. 1,00, 1,50, 2,00, etc.
In fact they are 1,0, 1,5, 2,0, etc.
How do I get R to show two digits after the comma on that axis?
 
Thanks for any help,
 Mario
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