[R] R: polygon error?
It seems to me that you are using polygon in a wrong way. What you probably need could be something like: polygon(c(rev(t$z), t$z), c(rep(0, nrow(t)), t$ht), col=2, border=NA) Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di LL Inviato: sabato 26 maggio 2007 12.34 A: r-help@stat.math.ethz.ch Oggetto: [R] polygon error? Hi.. I'm not sure why polygon returns an area above the standard normal curve. z - pretty(c(-3,3), 100) ht - dnorm(z) data - data.frame(z=z, ht=ht) zc - 1.645 plot(data, type=l) lines(data) t - subset(data, zzc) polygon(t, col=red) Thanks, Lance [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: Simple programming question
try also this dfr$score-factor(dfr$var3 %in% sort(unique(dfr$var3), decr=T)[1:2] * dfr$var3, labels=c(low, mid, high)) Hope this helps, Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Lauri Nikkinen Inviato: venerdì 18 maggio 2007 15.15 A: r-help@stat.math.ethz.ch Oggetto: [R] Simple programming question Hi R-users, I have a simple question for R heavy users. If I have a data frame like this dfr - data.frame(id=1:16, categ=rep(LETTERS[1:4], 4), var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1)) dfr - dfr[order(dfr$categ),] and I want to score values or points in variable named var3 following this kind of logic: 1. the highest value of var3 within category (variable named categ) - high 2. the second highest value - mid 3. lowest value - low This would be the output of this reasoning: dfr$score - factor(c(high,mid,low,low,high,mid,mid,low,high,mid,low,low,high,mid,low,low)) dfr The question is how I do this programmatically in R (i.e. if I have 2000 rows in my dfr)? I appreciate your help! Cheers, Lauri [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: extract from a data frame
Oats[Oats$Variety %in% c(Victory, Golden Rain),] or subset(Oats, Variety %in% c(Victory, Golden Rain)) Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di elyakhlifi mustapha Inviato: lunedì 23 aprile 2007 9.56 A: R-help@stat.math.ethz.ch Oggetto: [R] extract from a data frame hello, I'd like know how to do to extract data from a frame for example how can I do to extract only the data where variety=victory or variety=golden rain thanks. Oats Block Variety nitro yield 1 I Victory 0.0 111 2 I Victory 0.2 130 3 I Victory 0.4 157 4 I Victory 0.6 174 5 I Golden Rain 0.0 117 6 I Golden Rain 0.2 114 7 I Golden Rain 0.4 161 8 I Golden Rain 0.6 141 9 I Marvellous 0.0 105 10 I Marvellous 0.2 140 11 I Marvellous 0.4 118 12 I Marvellous 0.6 156 13II Victory 0.061 14II Victory 0.291 15II Victory 0.497 16II Victory 0.6 100 17II Golden Rain 0.070 18II Golden Rain 0.2 108 19II Golden Rain 0.4 126 20II Golden Rain 0.6 149 21II Marvellous 0.096 22II Marvellous 0.2 124 23II Marvellous 0.4 121 ___ Découvrez une nouvelle façon d'obtenir des réponses à toutes vos questions ! Profitez des connaissances, des opinions et des expériences des internautes sur Yahoo! Questions/Réponses [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: math-operations
I guess you need %% and %/% try 513 %/% 100 [1] 5 513 %% 100 [1] 13 ?%% Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Schmitt, Corinna Inviato: venerdì 30 marzo 2007 15.34 A: r-help@stat.math.ethz.ch Oggetto: [R] math-operations Hallo R-experts, for a function I need to work with the commands div and mod known from Pascal and Ruby. The only help I know is ceiling() and floor() in R. Do div and mod exist? When yes please send me a little example. In Pascal-Syntax I want: 513 div 100 = 5 513 mod 100 = 13 How can I get this in R-syntax? Thanks, Corinna __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: Searching and deleting elements of list
you could try mapply mydata2-mapply([, mydata, lapply(mydata, function(x) !x %in% A)) mydata2[[1]]-A #to replace the obviously deleted elements of A mydata2 mydata2[[1]] mydata2[[2]] mydata2[[3]] mydata2[[4]] Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di jastar Inviato: giovedì 8 marzo 2007 12.31 A: r-help@stat.math.ethz.ch Oggetto: [R] Searching and deleting elements of list Hi, I have a problem. Please, look at example and try to help me!! A-c(aaa,bbb,ccc,ddd,eee) B-c(vvv,ooo,aaa,eee,zzz,bbb) C-c(sss,jjj,ppp,ddd) D-c(bbb,ccc) mydata=list(A,B,C,D) I want to find and delete from 'mydata' all elements which occur in A (except A). I mean after operation: mydata[[1]] [1] aaa bbb ccc ddd eee mydata[[2]] [1] vvv ooo zzz mydata[[3]] [1] sss,jjj,ppp mydata[[4]] NULL My list have about 1 subelements (each contains several strings) so using loops is senseless. Thank's for all replies and sorry for my English (I hope you understand what I'm talking about) :-) -- View this message in context: http://www.nabble.com/Searching-and-deleting-elements-of-list-tf3368489.html#a9372270 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: Grouping columns in a data frame based on the values of a column
Perhaps using 'ave' and 'cut': df - data.frame(x=runif(100, 0.1, 1), y=rnorm(100, 0.2, 0.6)) df$xcut-cut(df$x, seq(0, 1, 0.1)) df$z-ave(df$y, df$xcut) df[order(df$x),] Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di [EMAIL PROTECTED] Inviato: venerdì 15 settembre 2006 17.45 A: r-help@stat.math.ethz.ch Oggetto: [R] Grouping columns in a data frame based on the values of a column Dear R users, This is a trivial question, there might even be an R function for it, but I have to do it many times and wonder if there is an efficient for it. Suppose we have a data frame like this: d - data.frame(x=sample(seq(0.1:1, by=0.01), size=100, replace=TRUE), y=rnorm(100, 0.2, 0.6)) and want to have the average of y for a given interval of x, for example mean(y)[0x0.1]. Is there a simple way of doing this or I need to improvise? Thank you so much for any help Eleni Rapsomaniki __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: data.frame to shape
assuming that the rows are sorted correctly dat id x y 1 50 1647685 4815259 2 50 1647546 4815196 3 50 1647454 4815294 4 50 1647405 4815347 5 50 1647292 4815552 6 50 1647737 4815410 7 74 1647555 4815201 8 74 1647464 4815023 9 74 1646970 4815129 10 74 1646895 4815264 11 74 1646762 4815513 list.dat-split(dat, dat$id) closed.polygons-lapply(list.dat, function(x) rbind(x, x[1,])) do.call(rbind, closed.polygons) id x y 50.1 50 1647685 4815259 50.2 50 1647546 4815196 50.3 50 1647454 4815294 50.4 50 1647405 4815347 50.5 50 1647292 4815552 50.6 50 1647737 4815410 50.11 50 1647685 4815259 74.7 74 1647555 4815201 74.8 74 1647464 4815023 74.9 74 1646970 4815129 74.10 74 1646895 4815264 74.11 74 1646762 4815513 74.71 74 1647555 4815201 but maybe there are better ways to do what you want -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Leonardo Lami Inviato: 09 August, 2006 15:33 A: r-help@stat.math.ethz.ch Oggetto: [R] data.frame to shape Hi all, I have a simple question: I have a data.frame like this: id x y 1 50 1647685 4815259 2 50 1647546 4815196 3 50 1647454 4815294 4 50 1647405 4815347 5 50 1647292 4815552 6 50 1647737 4815410 7 74 1647555 4815201 8 74 1647464 4815023 9 74 1646970 4815129 10 74 1646895 4815264 11 74 1646762 4815513 and I'd like to trasform it with the convert.to.shapefile function (shapefiles package) but to make this I must have a data.frame like this: id x y 1 50 1647685 4815259 2 50 1647546 4815196 3 50 1647454 4815294 4 50 1647405 4815347 5 50 1647292 4815552 6 50 1647737 4815410 7 50 1647685 4815259 8 74 1647555 4815201 9 74 1647464 4815023 10 74 1646970 4815129 11 74 1646895 4815264 12 74 1646762 4815513 13 74 1646762 4815513 with the first point of every id repeated to close the polygon. There is a function to make this indipendently by the number of the id Best regards Leonardo -- Leonardo Lami email + jabber: [EMAIL PROTECTED] www.faunalia.it Cell: (+39)349-1310164 Tel+Fax: (+39) 0587-213742 Piazza Garibaldi 5 - 56025 Pontedera (PI), Italy __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: the first and last case
could it be dat[unlist(tapply(1:nrow(dat), ind, range)),] ? stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Mauricio Cardeal Inviato: 26 July, 2006 14:22 A: r-help@stat.math.ethz.ch Oggetto: [R] the first and last case Hi all Sometime ago I asked for a solution about how to aggregate data and the help was wonderful. Now, I´d like to know how to extract for each individual case below the first and the last observation to obtain this: ind y 18 19 27 2 11 39 3 10 4 8 4 5 # Below the example: ind - c(1,1,1,2,2,3,3,3,4,4,4,4) y - c(8,10,9,7,11,9,9,10,8,7,6,5) dat - as.data.frame(cbind(ind,y)) dat attach(dat) mean.ind - aggregate(dat$y, by=list(dat$ind), mean) mean.ind Thanks Mauricio __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: determination of number of entries in list elements
You need lapply or sapply for example: sapply(yourlist, length) then you can do subset(yourlist, sapply(yourlist, length) yourlength) Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Benjamin Otto Inviato: 19 May, 2006 12:10 A: R-Help Oggetto: [R] determination of number of entries in list elements Hi, is there some elegant way to determine the number of components stored in each list element? Example: The list: - list $Elem1 [1] A B C $Elem1 [1] D $Elem1 [1] E F Then normal command length(list) would return 3. But I would like some command return the array of the single element lengths like [1] 3 1 2 so I can afterwards get my list subset with only entries which have a certain amount of components bigger or lower than a certain threshold. regards Benjamin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: R: [Re:] function to replace missing values with median value?]]
oops!, I pressed the 'send' key too soon, ...
see ?replace
replace(x, is.na(x), median(x, na.rm=T))
take also a look at the function itself
replace
function (x, list, values)
{
x[list] - values
x
}
environment: namespace:base
Stefano
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di
Guazzetti Stefano
Inviato: Thursday, May 04, 2006 07:55 AM
A: [EMAIL PROTECTED]; [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Oggetto: [R] R: [Re:] function to replace missing values with median
value?]]
there is also a replace function
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di
[EMAIL PROTECTED]
Inviato: Thursday, May 04, 2006 07:31 AM
A: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Oggetto: [R] [Re:] function to replace missing values
with median
value?]]
The following should work
sz - function(x) { ifelse(is.na(x) == F, x, median(x,
na.rm=TRUE)) }
best, isaia.
Original Message
Subject: [R] function to replace missing values with
median value?
Date: Wed, 3 May 2006 10:06:40 -0700 (PDT)
From: r user [EMAIL PROTECTED]
To: rhelp r-help@stat.math.ethz.ch
I have a data set with ~10 variables (i.e. columns).
I wrote this little function to replace missing values
with zero.
“ sz - function(x) { ifelse(is.na(x)==F,x,0) } “
Can anyone help with a function that replaces missing
values with the median of the non-missing values?
__
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~~~
~ Ennio D. Isaia
~ Dep. of Statistics Applied Mathematics, University of Torino
~ Piazza Arbarello, 8 - 10122 Torino (Italy)
~ Phone: +39 011 670 57 29 ~~ Fax: +39 011 670 57 83
~~~
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[R] R: [Re:] function to replace missing values with median value?]]
there is also a replace function
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di
[EMAIL PROTECTED]
Inviato: Thursday, May 04, 2006 07:31 AM
A: [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Oggetto: [R] [Re:] function to replace missing values with median
value?]]
The following should work
sz - function(x) { ifelse(is.na(x) == F, x, median(x,
na.rm=TRUE)) }
best, isaia.
Original Message
Subject: [R] function to replace missing values with median value?
Date: Wed, 3 May 2006 10:06:40 -0700 (PDT)
From: r user [EMAIL PROTECTED]
To: rhelp r-help@stat.math.ethz.ch
I have a data set with ~10 variables (i.e. columns).
I wrote this little function to replace missing values
with zero.
“ sz - function(x) { ifelse(is.na(x)==F,x,0) } “
Can anyone help with a function that replaces missing
values with the median of the non-missing values?
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
--
~~~
~ Ennio D. Isaia
~ Dep. of Statistics Applied Mathematics, University of Torino
~ Piazza Arbarello, 8 - 10122 Torino (Italy)
~ Phone: +39 011 670 57 29 ~~ Fax: +39 011 670 57 83
~~~
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide!
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[R] R: Dotplot x-axis
Take a look at the scales argument in dotpolot.
Maybe you need something like:
position-exp(-5:0)/(1+exp(-5:0))
dotplot(type~freq/(3027-freq),
scales=list(x=list(log=T,
at=position, lab=round(position, 3))) )
Stefano
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di Michael Kubovy
Inviato: 14 April 2006 15:29
A: r-help@stat.math.ethz.ch
Oggetto: [R] Dotplot x-axis
Here's a small dataset:
type - c('hierarchical','partial','single','complete','single
+hierarchical','single+partial','partial+hierarchical','sing
le+partial
+hierarchical')
freq - c(1455,729,688,65,29,28,16,17)
lodds - log(freq/(3027 - freq))
dotplot(type~lodds)
I would like to have the x-axis have ticks at nice, close
to equally-
spaced values, of the proportions rather than at round log-odds
values, which appear as -5:0. For example: perhaps have ticks at c
(0.007, 0.02, 0.05, 0.12, 0.27, 0.5), which are approximate
values of
exp(-5:0)/(1+exp(-5:0))
Advice?
_
Professor Michael Kubovy
University of Virginia
Department of Psychology
USPS: P.O.Box 400400Charlottesville, VA 22904-4400
Parcels:Room 102Gilmer Hall
McCormick RoadCharlottesville, VA 22903
Office:B011+1-434-982-4729
Lab:B019+1-434-982-4751
Fax:+1-434-982-4766
WWW:http://www.people.virginia.edu/~mk9y/
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[R] R: Fw: Controling the x or y limit
perhaps you are looking at something like plot(0, xlim=c(0,20), xaxt=n) axis(1, at=pretty(0:20, 10)) Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Subhabrata Inviato: venerdì 10 febbraio 2006 7.05 A: r-help Cc: [EMAIL PROTECTED] Oggetto: [R] Fw: Controling the x or y limit Hello - R-experts, This may sound simple to many --- We can specify the x limit by saying xlim = c(0, 20) for example. Then the graph will show the range of x -axis between 0 - 20. But the coordinate gap will be automatic like 0 then 5 then 15 and 20. Is there any way by which we can set it in a gap of 2 or any number. Thank you for any help. With Regards Subhabrata __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: dataframe subset
Dear Bernhard, if I understand correctly your question may be you want something like df-data.frame(x=sample(1:10, 100, repl=T), y=sample(1:5, 100, repl=T)) subset(df, x%in%y) Regards, Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Bernhard Baumgartner Inviato: 08 February 2006 15:22 A: r-help@stat.math.ethz.ch Oggetto: [R] dataframe subset I have a dataframe with a column, say x consisting of values, each value appearing different times, e.g. x: 1,1,1,1,2,2,4,4,4,9,10,10,10,10,10 ... and a vector, including e.g.: y: 2,9,10,... I need a subset of the dataframe: all rows where x is equal to one of the values in y. Currently I use a loop for this, but because x and y are large this is very slow. Is there any idea how to solve this problem faster? Thank you, Bernhard __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] R: Graph with values of coordinates of points in x axis
try: plot(x, y, type=l, xlab=Months, xaxt=n, ylab=Y values) axis(1, at=0:5*6) Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di Massimiliano Tripoli Inviato: mercoledì 15 giugno 2005 10.35 A: r-help@stat.math.ethz.ch Oggetto: [R] Graph with values of coordinates of points in x axis Hi all, I want to draw a line with the values of x marked in X axis. I tried with x - c(0,6,12,18,24,30) #coordinates of points x y - c(2,5,7,5,7,16) #coordinates of points y plot(x,type=n,xlab=Months,main=main,ylim=c(0,16),xlim=c(0,30)) lines(x,y) The graph shows by default an increment of the sequence in x axis that I'm not able to change. I'would like to have 0,6,12,18,24,30 and not 0,10,15,20,25,30. Any hint is appreciated. Thanks in advance. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] weighted.mean and tapply (again)
what about using mapply?
splitted.value-with(x.1, split(VALUE, GROUP))
splitted.freq-with(x.1, split(FREQUENCY, GROUP))
mapply(weighted.mean, splitted.value, w=splitted.freq)
Stefano
-Messaggio originale-
Da: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] conto di
[EMAIL PROTECTED]
Inviato: mercoledì 25 maggio 2005 17.57
A: Dan Bolser
Cc: R mailing list; [EMAIL PROTECTED]
Oggetto: Re: [R] weighted.mean and tapply (again)
x.1 - read.table('clipboard',header=T)
x.1
GROUP VALUE FREQUENCY
1 2 278
2 2 340
3 2 416
4 2 5 3
5 2 6 1
6 2 8 1
7 3 319
8 3 410
9 3 519
10 3 6 4
by(x.1, x.1$GROUP, function(x) weighted.mean(x$VALUE,
x$FREQUENCY))
x.1$GROUP: 2
[1] 2.654676
---
x.1$GROUP: 3
[1] 4.153846
Jim
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James HoltmanWhat is the problem you are trying to solve?
Executive Technical Consultant -- Office of Technology, Convergys
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Dan Bolser
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Sent by: Subject:
[R] weighted.mean and tapply (again)
[EMAIL PROTECTED]
ath.ethz.ch
05/25/2005 11:33
I read answers to questions including the words tapply and
weighted.mean, but I didn't understand either the problem
(data) or the
solution provided.
Here is my question ...
dat[1:10,]
GROUP VALUE FREQUENCY
1 2 278
2 2 340
3 2 416
4 2 5 3
5 2 6 1
6 2 8 1
7 3 319
8 3 410
9 3 519
1 3 6 4
For each GROUP, I would like to calculate the weighted.mean
of VALUE using
the FREQUENCY as the weight, so for the snippet of data
shown that would
be...
group.2 - weighted.mean(c(2,3,4,5,6,8),c(78,40,16,3,1,1))
group.3 - weighted.mean(c(3,4,5,6),c(19,10,19,4))
cbind(rbind(2,3),rbind(group.2,group.3))
[,1] [,2]
group.22 2.654676
group.33 4.153846
I would like to use tapply to automatically do this across the whole
dataset (dat) - which includes lots of other distinct
grouping factors,
however, like I said, I couldn't understand (and therefore
apply to my
data) any of the other solutions I found, so any help here would be
greatly appreciated!
All the best,
Dan.
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[R] unexpected behaviour of 'curve' function
Dear all, curve(x^3*(1-x)^7, from = 0, to = 1) works as expected but, omitting the xlim or the to and from arguments and calling curve more than once: par(mfrow = c(2,2)) for (i in 1:4) curve(x^3*(1-x)^7) gives an expected (al least to me) result. Note also that a pu object is returned by curve pu [1] -0.1802445 1.1802445 The behaviour is reproducible with both R 2.0.0 and R 2.0.1 I can see that a promise of evaluation of pu is made within curve but I cannot understand completely what happens. Thanks in advance, Stefano Guazzetti platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major2 minor0.0 year 2004 month10 day 04 language R __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] How to plot this
Hi, looking at ?plot.density you will find a zero.line argument: set it to FALSE and no gray lines will appear in the plot. plot(density(y), zero.line = F, main= , ann = F, xlim = c(0, 4), ylim = c(0, 1), lty = 2, col = 4, axes = F) #and the add mtext(side = 1, line = 0, text = Environmental gradient ) mtext(side = 2, line = 0, text = Abundance of species ) arrows(0, 0, 4, 0, angle = 15, length = 0.1, lwd=2) arrows(0, 0, 0, 1, angle = 15, length = 0.1, lwd=2) Hope this helps Stefano -Messaggio originale- Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] conto di [EMAIL PROTECTED] Inviato: mercoledì 17 novembre 2004 7.39 A: [EMAIL PROTECTED] Oggetto: [R] How to plot this Hi there, I produced a plot using the following codes: y-rnorm(1000, 2, 0) x0-c(0, 0) y0-c(0, 0) y1-c(0, 1) x1-c(0, 4) plot(density(y), ylab=Abundance of species, xlab=Environmental gradient, main= , xlim=c(0, 4), ylim=c(0, 1), lty=2, col=4, xaxt=n, yaxt=n, frame.plot=F) lines(x0, y1) # add an axis lines(x1, y0) # add an axis arrows(3.95, 0, 4, 0, angle = 15, length = 0.1) arrows(0, 0.98, 0, 1, angle = 15, length = 0.1) Please help me to remove the grey horizontal line and put the axis labels closer to the axes. And also appreciate any suggestions on how to make those arrows look nicer, e.g. a filled small arrow for each axis, like what from points(0, 1, pch=17), but a slightly narrowed one. Thanks. Regards, Jin Li Jin Li, PhD Climate Impacts Modeller CSIRO Sustainable Ecosystems Atherton, QLD 4883, Australia [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] retrieve rows from frame assuming criterion
Assuming your data.frame is called data data[data$PUNTAR==c(IX49,IX48),] is probably what you want Stefano -Messaggio originale- Da: Luis Rideau Cruz [mailto:[EMAIL PROTECTED] Inviato: venerdì 23 luglio 2004 15.37 A: [EMAIL PROTECTED] Oggetto: [R] retrieve rows from frame assuming criterion Hi all, I have a data frame in which one column(PUNTAR) is of character type. What I want is to retrieve is the frame but only with those rows matching elements of PUNTAR with a list characters (e.g c(IX49,IX48) ) YearTUR STODNR PUNTAR 1994 9412 94020061 IX49 1994 9412 94020062 IX48 1994 9412 94020063 X32 1994 9412 94020065 X23 1994 9412 94020066 X27 1994 9412 94020067 XI19 1994 9412 94020068 XI16 1994 9412 94020069 XI14 1994 9412 94020070 XI8 1994 9412 94020071 X25 1994 9412 94020072 X18 1994 9412 94020073 II23 1994 9412 94020074XII33 1994 9412 94020075XII31 my.function(frame) should be then equal to Year TURNR STODNR M_PUNTAR 1994 9412 94020061 IX49 1994 9412 94020062 IX48 Thank you in advance Luis Ridao Cruz Fiskirannsóknarstovan Nóatún 1 P.O. Box 3051 FR-110 Tórshavn Faroe Islands Phone: +298 353900 Phone(direct): +298 353912 Mobile: +298 580800 Fax: +298 353901 E-mail: [EMAIL PROTECTED] Web:www.frs.fo __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] retrieve rows from frame assuming criterion [corrected]
sorry for my previus (WRONG) answer, as someone already pointed out a solution could be subset(data, PUNTAR==c(IX49,IX48)) -Messaggio originale- Da: Guazzetti Stefano Inviato: venerdì 23 luglio 2004 15.55 A: 'Luis Rideau Cruz'; [EMAIL PROTECTED] Oggetto: R: [R] retrieve rows from frame assuming criterion Assuming your data.frame is called data data[data$PUNTAR==c(IX49,IX48),] is probably what you want Stefano -Messaggio originale- Da: Luis Rideau Cruz [mailto:[EMAIL PROTECTED] Inviato: venerdì 23 luglio 2004 15.37 A: [EMAIL PROTECTED] Oggetto: [R] retrieve rows from frame assuming criterion Hi all, I have a data frame in which one column(PUNTAR) is of character type. What I want is to retrieve is the frame but only with those rows matching elements of PUNTAR with a list characters (e.g c(IX49,IX48) ) YearTUR STODNR PUNTAR 1994 9412 94020061 IX49 1994 9412 94020062 IX48 1994 9412 94020063 X32 1994 9412 94020065 X23 1994 9412 94020066 X27 1994 9412 94020067 XI19 1994 9412 94020068 XI16 1994 9412 94020069 XI14 1994 9412 94020070 XI8 1994 9412 94020071 X25 1994 9412 94020072 X18 1994 9412 94020073 II23 1994 9412 94020074XII33 1994 9412 94020075XII31 my.function(frame) should be then equal to Year TURNR STODNR M_PUNTAR 1994 9412 94020061 IX49 1994 9412 94020062 IX48 Thank you in advance Luis Ridao Cruz Fiskirannsóknarstovan Nóatún 1 P.O. Box 3051 FR-110 Tórshavn Faroe Islands Phone: +298 353900 Phone(direct): +298 353912 Mobile: +298 580800 Fax: +298 353901 E-mail: [EMAIL PROTECTED] Web:www.frs.fo __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: R: [R] retrieve rows from frame assuming criterion [corrected]
Yes, paraphrasing Murphy I can say of myself: Nothing seems to be able to stop a stupid thought in its pathway from the brain to the keyboard. :-) Sorry once again and thank for your patience. Stefano -Messaggio originale- Da: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Inviato: venerdì 23 luglio 2004 16.30 A: Guazzetti Stefano Cc: Luis Rideau Cruz; [EMAIL PROTECTED] Oggetto: Re: R: [R] retrieve rows from frame assuming criterion [corrected] On Fri, 23 Jul 2004, Guazzetti Stefano wrote: sorry for my previus (WRONG) answer, as someone already pointed out a solution could be subset(data, PUNTAR==c(IX49,IX48)) That's still wrong. You want PUNTAR %in% c(IX49,IX48). Using == recycles entries, so it tests the first element against IX49, the second against IX48, the third against IX49 -Messaggio originale- Da: Guazzetti Stefano Inviato: venerdì 23 luglio 2004 15.55 A: 'Luis Rideau Cruz'; [EMAIL PROTECTED] Oggetto: R: [R] retrieve rows from frame assuming criterion Assuming your data.frame is called data data[data$PUNTAR==c(IX49,IX48),] is probably what you want Stefano -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] proportions confidence intervals
You should consider prop.test or binom.test. The problem you will find is that these functions are not intended to do what you want but to give you one confidence interval at a time. However a starting point could be : I-sample(1:50) #the numerator N-sample(50:200, 50) #the denominator conf.intervals-t( sapply( mapply(binom.test, SIMPLIFY=F, x=I, n=N), [[, conf.int) ) cbind(I, N, P=I/N, conf.int) best whishes, Stefano -Messaggio originale- Da: Darren Shaw [mailto:[EMAIL PROTECTED] Inviato: lunedì 12 luglio 2004 17.45 A: [EMAIL PROTECTED] Oggetto: [R] proportions confidence intervals Dear R users this may be a simple question - but i would appreciate any thoughts does anyone know how you would get one lower and one upper confidence interval for a set of data that consists of proportions. i.e. taking a usual confidence interval for normal data would result in the lower confidence interval being negative - which is not possible given the data (which is constrained between 0 and 1) i can see how you calculate a upper and lower confidence interval for a single proportion, but not for a set of proportions many thanks Darren Shaw - Dr Darren J Shaw Centre for Tropical Veterinary Medicine (CTVM) The University of Edinburgh Scotland, EH25 9RG, UK __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] Creating Binary Outcomes from a continuous variable
consider a cutpoint of 20 x-runif(100, min=1, max=50) as.integer(x 20) Stefano -Messaggio originale- Da: Doran, Harold [mailto:[EMAIL PROTECTED] Inviato: mercoledì 7 luglio 2004 14.57 A: [EMAIL PROTECTED] Oggetto: [R] Creating Binary Outcomes from a continuous variable Dear List: I have searched the archives and my R books and cannot find a method to transform a continuous variable into a binary variable. For example, I have test score data along a continuous scale. I want to create a new variable in my dataset that is 1=above a cutpoint (or passed the test) and 0=otherwise. My instinct tells me that this will require a combination of the transform command along with a conditional selection. Any help is much appreciated. Thanks, Harold [[alternative HTML version deleted]] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] help with histogram
Maybe you want something like: x-rnorm(1000) hist(x, breaks=100, col=ifelse(abs((hist(x, breaks=100, main=))$breaks) 1.669, 4,2)) see also the density argument in ?hist Stefano -Messaggio originale- Da: Martin Olivier [mailto:[EMAIL PROTECTED] Inviato: venerdì 7 maggio 2004 14.40 A: r-help Oggetto: [R] help with histogram Hi all, I need some help with the function histogram. Let x be a vector. The command hist(x,col=blue) gives an histogram for the vector x. I would like to add some colors in the graphics such that the histogram is red if abs(x)1.669 and blue otherwise...and what is the solution if I want to change the filling instead of the color Thanks for your help, Olivier __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] lines and glm
You probably mean something like: ti - 1:1000 e1 - rnorm(1000) e2 - rnorm(1000) x - 0.0001*ti+e1 y2 - -2+x+e2 y - ifelse(y20,1,0) plot(x, y, pch = 16, col = darkblue, main = expression(paste(Scatter diagram of , italic(y[t]), against , italic(x[t]))), xlab = expression(italic(x[t])), ylab = expression(italic(y[t]))) model-glm(y ~ x, family = binomial) predProbs-predict(model,data.frame(x=seq(min(x), max(x), length.out=100)), type=response) lines(seq(min(x), max(x), length.out=100), predProbs, col=2, lwd=2) Note also that it is not a good idea to name t a R object, since the name is reserved for a special function. Stefano -Messaggio originale- Da: Jan P. Smit [mailto:[EMAIL PROTECTED] Inviato: giovedì 8 aprile 2004 13.38 A: [EMAIL PROTECTED] Oggetto: [R] lines and glm Dear R-helpers, I'm a beginner using R 1.8.1 on Windows 2000. I'm trying to replicate some examples in Franses' Paap's Quantitative Models in Marketing Research. t - 1:1000 e1 - rnorm(1000) e2 - rnorm(1000) x - 0.0001*t+e1 y2 - -2+x+e2 y - ifelse(y20,1,0) plot(x, y, pch = 16, col = darkblue, + main = expression(paste(Scatter diagram of , italic(y[t]), against , + italic(x[t]))), + xlab = expression(italic(x[t])), + ylab = expression(italic(y[t]))) lines(glm(y ~ x, family = binomial)) Error in xy.coords(x, y) : x and y lengths differ However, length(x) [1] 1000 length(y) [1] 1000 as it should be. I'm sure I must be missing something obvious, but it is not clear to me (after reading the Introduction, FAQ and the relevant functions help pages) what. Sincerely, Jan Smit __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
R: [R] Changing background in splom et al.
trellis.device(bg=white, color=F)
before your call to splom could make what you want but
take also a look at
?trellis.par.set
Stefano
-Messaggio originale-
Da: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Inviato: mercoledì 3 marzo 2004 12.10
A: r-help
Oggetto: [R] Changing background in splom et al.
Context: Windows XP, R 1.8.1
I'm studying Venables-Ripley MASS book and having a go at
the many examples
in library MASS. The code I'm checking (from script ch04.R) now is
..
data(swiss)
splom(~ swiss, aspect = fill,
panel = function(x, y, ...) {
panel.xyplot(x, y, ...); panel.loess(x, y, ...)
}
)
which produces an agreable plot with a gray background and
cyan points,
but.
Copying the plot as a metafile into Word and having a laser
printer I'd
better stick to a b/w plot.
How can I turn the background to white and the cyan points to black?
Ciao - Vittorio
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