Re: [R] replace values in a distance matrix
M4.5[M4.5 < 4.5] <- 18
On 7/27/06, Mari Carmen Garcia Esteban <[EMAIL PROTECTED]> wrote:
>
> Hi to everybody!
> I´m just a beginner in R, and I´m trying to replace values in a distance
> matrix with a concret condition: replace all values (elements) lower than
> 4.5 with value=18.
>
> I´ve tried this, but it doesn´t work...
>
> Dxy would be my 117 x 117 euclidean distance matrix
>
>M18 and M4.5 would be 117 x 117 matrices:
>
> M18<-matrix(rep(18,13689),nrow=117)
> M4.5<-matrix(rep(4.5,13689),nrow=117)
>
> REPLACE ALL VALUES >4.5 WITH 18.0 in Dxy
>
> for(i in 1:117) {
> for(j in 1:117) {
> Dxy[i,j]<-ifelse(Dxy[i,j] }
> }
>
>
> Thanks in advance!
>
> Mari Carmen Garcia
> Ud. Anatomía, Fisiologia y Génetica.
> Escuela Técnica Superior de Ingenieros de Montes de Madrid
> Spain.
>
>
>
>
>
> -
>
>
>
>
>[[alternative HTML version deleted]]
>
>
>
> __
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
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Re: [R] replace values in data frame
Wade,
Given that you appear to have multiple search and replace items to deal
with, here is a possible loop based "Global Search and Replace"
solution:
gsr <- function(Source, Search, Replace)
{
if (length(Search) != length(Replace))
stop("Search and Replace Must Have Equal Number of Items\n")
Changed <- as.character(Source)
for (i in 1:length(Search))
{
cat("Replacing: ", Search[i], " With: ", Replace[i], "\n")
Changed <- replace(Changed, Changed == Search[i], Replace[i])
}
cat("\n")
Changed
}
Source: The source vector, which will be coerced to character
Search: The Search values to be matched in "Source" as a character
vector
Replace: The values that will replace those found in 'Search' on a
one-for-one basis.
This function returns a character vector. As with replace(), the result
must be assigned. The change is not done 'in place'. The function will
also output the search and replace values to the console during the
loop.
Again, using the iris dataset as an example:
> iris$Species
[1] setosa setosa setosa setosa setosa setosa
[7] setosa setosa setosa setosa setosa setosa
[13] setosa setosa setosa setosa setosa setosa
[19] setosa setosa setosa setosa setosa setosa
[25] setosa setosa setosa setosa setosa setosa
[31] setosa setosa setosa setosa setosa setosa
[37] setosa setosa setosa setosa setosa setosa
[43] setosa setosa setosa setosa setosa setosa
[49] setosa setosa versicolor versicolor versicolor versicolor
[55] versicolor versicolor versicolor versicolor versicolor versicolor
[61] versicolor versicolor versicolor versicolor versicolor versicolor
[67] versicolor versicolor versicolor versicolor versicolor versicolor
[73] versicolor versicolor versicolor versicolor versicolor versicolor
[79] versicolor versicolor versicolor versicolor versicolor versicolor
[85] versicolor versicolor versicolor versicolor versicolor versicolor
[91] versicolor versicolor versicolor versicolor versicolor versicolor
[97] versicolor versicolor versicolor versicolor virginica virginica
[103] virginica virginica virginica virginica virginica virginica
[109] virginica virginica virginica virginica virginica virginica
[115] virginica virginica virginica virginica virginica virginica
[121] virginica virginica virginica virginica virginica virginica
[127] virginica virginica virginica virginica virginica virginica
[133] virginica virginica virginica virginica virginica virginica
[139] virginica virginica virginica virginica virginica virginica
[145] virginica virginica virginica virginica virginica virginica
Levels: setosa versicolor virginica
Note that iris$Species is a factor with specific levels. The gsr()
function above will coerce the Source argument to a character vector
internally and return a character vector:
> iris$Species <- gsr(iris$Species,
c("setosa", "versicolor", "virginica"),
c("s1", "v1", "v2"))
Replacing: setosa With: s1
Replacing: versicolor With: v1
Replacing: virginica With: v2
> iris$Species
[1] "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1"
[14] "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1"
[27] "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1"
[40] "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "s1" "v1" "v1"
[53] "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1"
[66] "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1"
[79] "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1"
[92] "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v1" "v2" "v2" "v2" "v2"
[105] "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2"
[118] "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2"
[131] "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2" "v2"
[144] "v2" "v2" "v2" "v2" "v2" "v2" "v2"
HTH,
Marc Schwartz
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Re: [R] replace values in data frame
Thanks for re-posting onlist. My offlist reply: The 'list' argument, as per ?replace, needs to be a vector of one or more indices into another vector, generally the source vector argument 'x', not the actual values that you want to replace. 'list' can either be explicit integers as the indices, or the result of a logical operation as I used in my example. In addition, the objects 'list' and 'replace' in your code are not vectors, but they are data frames. That is the default type of object created when using read.table(). You can use str(list) and str(replace) to check this. The str() function returns the structure of an object. It is likely that you really want list$V1 and replace$V1 as the vectors of interest here. V1 thru Vx are the default column names created when using read.table() if no header row exists in the incoming file. In addition, note that replace() is not vectorized. It will not handle multiple search and replace arguments in a single call. See my example using the iris dataset. You would need to do each one separately. You can create a loop of one type or another to cycle through multiple arguments if you wish, where each cycle through the loop is a single call to replace() with a new set of values as appropriate for the arguments. Finally, you should generally not use R object or function names for user created objects. Both 'list' and 'replace' are such objects. R will generally be able to differentiate, but there is no guarantee and it will help you avoid code debugging headaches. HTH, Marc Schwartz On Fri, 2006-07-07 at 12:58 -0400, Wade Wall wrote: > The format is like this. > > Plot species Value > > P1 ACERRUB 3 > P2 MAGNVIR2 > P3 ARONARB 2 > etc. > > imported using x<-read.table(file="filename.txt") > > I want to replace a list of values in the 2nd column with another list. For > example, I want to replace ARONARB with PHOTPYR. > list<-read.table(file="originalnames.txt") > replace<-read.table(file="replacementnames.txt") > I tried to use replace in this manner: > > newx<-replace(x,list,replace) > however, I get the error message: error in replace, invalid subscript type. > > I have tried transforming the above list and modifying the column names > (column 2), but to no avail. I hope this clarifies a little. Sorry about > that. > > > > > On Fri, 2006-07-07 at 11:20 -0400, Wade Wall wrote: > > > Hi all, > > > > > > I have a three columned list that I have imported into R. The first > > column > > > is a plot (ex. Plot1), the second is a species name (ex ACERRUB) and the > > > third a numeric value. I want to replace some of the second column > > names > > > with other names (for example replace ACERRUB with ACERDRU). The > > original > > > and replacement values are in separate lists (not vectors), but I can't > > seem > > > to find the right function to perform this. The replace function seems > > to > > > only want to work with numbers. > > > > > > Any clues? > > > > > > Wade > > __ 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
Re: [R] replace values in data frame
The format is like this. Plot species Value P1 ACERRUB 3 P2 MAGNVIR2 P3 ARONARB 2 etc. imported using x<-read.table(file="filename.txt") I want to replace a list of values in the 2nd column with another list. For example, I want to replace ARONARB with PHOTPYR. list<-read.table(file="originalnames.txt") replace<-read.table(file="replacementnames.txt") I tried to use replace in this manner: newx<-replace(x,list,replace) however, I get the error message: error in replace, invalid subscript type. I have tried transforming the above list and modifying the column names (column 2), but to no avail. I hope this clarifies a little. Sorry about that. On 7/7/06, Marc Schwartz (via MN) <[EMAIL PROTECTED]> wrote: > > On Fri, 2006-07-07 at 11:20 -0400, Wade Wall wrote: > > Hi all, > > > > I have a three columned list that I have imported into R. The first > column > > is a plot (ex. Plot1), the second is a species name (ex ACERRUB) and the > > third a numeric value. I want to replace some of the second column > names > > with other names (for example replace ACERRUB with ACERDRU). The > original > > and replacement values are in separate lists (not vectors), but I can't > seem > > to find the right function to perform this. The replace function seems > to > > only want to work with numbers. > > > > Any clues? > > > > Wade > > Without seeing the code you are using, we can only guess a syntax error > of some sort. It works fine using the iris dataset: > > > head(iris) > Sepal.Length Sepal.Width Petal.Length Petal.Width Species > 1 5.1 3.5 1.4 0.2 setosa > 2 4.9 3.0 1.4 0.2 setosa > 3 4.7 3.2 1.3 0.2 setosa > 4 4.6 3.1 1.5 0.2 setosa > 5 5.0 3.6 1.4 0.2 setosa > 6 5.4 3.9 1.7 0.4 setosa > 7 4.6 3.4 1.4 0.3 setosa > 8 5.0 3.4 1.5 0.2 setosa > 9 4.4 2.9 1.4 0.2 setosa > 10 4.9 3.1 1.5 0.1 setosa > > > iris$Species <- replace(iris$Species, iris$Species == "setosa", > "NewValue") > > > head(iris) > Sepal.Length Sepal.Width Petal.Length Petal.Width Species > 1 5.1 3.5 1.4 0.2 NewValue > 2 4.9 3.0 1.4 0.2 NewValue > 3 4.7 3.2 1.3 0.2 NewValue > 4 4.6 3.1 1.5 0.2 NewValue > 5 5.0 3.6 1.4 0.2 NewValue > 6 5.4 3.9 1.7 0.4 NewValue > 7 4.6 3.4 1.4 0.3 NewValue > 8 5.0 3.4 1.5 0.2 NewValue > 9 4.4 2.9 1.4 0.2 NewValue > 10 4.9 3.1 1.5 0.1 NewValue > > > HTH, > > Marc Schwartz > > > [[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
Re: [R] replace values in data frame
On Fri, 2006-07-07 at 11:20 -0400, Wade Wall wrote: > Hi all, > > I have a three columned list that I have imported into R. The first column > is a plot (ex. Plot1), the second is a species name (ex ACERRUB) and the > third a numeric value. I want to replace some of the second column names > with other names (for example replace ACERRUB with ACERDRU). The original > and replacement values are in separate lists (not vectors), but I can't seem > to find the right function to perform this. The replace function seems to > only want to work with numbers. > > Any clues? > > Wade Without seeing the code you are using, we can only guess a syntax error of some sort. It works fine using the iris dataset: > head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1 1.5 0.2 setosa 5 5.0 3.6 1.4 0.2 setosa 6 5.4 3.9 1.7 0.4 setosa 7 4.6 3.4 1.4 0.3 setosa 8 5.0 3.4 1.5 0.2 setosa 9 4.4 2.9 1.4 0.2 setosa 10 4.9 3.1 1.5 0.1 setosa > iris$Species <- replace(iris$Species, iris$Species == "setosa", "NewValue") > head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 NewValue 2 4.9 3.0 1.4 0.2 NewValue 3 4.7 3.2 1.3 0.2 NewValue 4 4.6 3.1 1.5 0.2 NewValue 5 5.0 3.6 1.4 0.2 NewValue 6 5.4 3.9 1.7 0.4 NewValue 7 4.6 3.4 1.4 0.3 NewValue 8 5.0 3.4 1.5 0.2 NewValue 9 4.4 2.9 1.4 0.2 NewValue 10 4.9 3.1 1.5 0.1 NewValue HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] replace values?
# reproducing your example
xx<-"x y z
+ 1 2 3
+ 2 3 1
+ 3 2 1
+ 1 1 3
+ 2 1 2
+ 3 2 3
+ 2 1 1"
# you did not tell us the class of your data, assuming data.frame
df<-read.table(textConnection(xx),header=T,colClasses="factor")
# a clean way to do what you want is using factors with ?levels
# (note that data has already been read as factor)
levels(df$x)<-c("a","b","c","d")
levels(df$y)<-c("b","a","c","d")
levels(df$z)<-c("d","c","b","a")
subset(df,x=="a")
x y z
1 a a b
4 a b b
subset(df,x=="a"&y=="a")
x y z
1 a a b
HTH, m
zhijie zhang wrote:
> Dear friends,
> i have a dataset like this:
> x y z
> 1 2 3
> 2 3 1
> 3 2 1
> 1 1 3
> 2 1 2
> 3 2 3
> 2 1 1
> I want to replace x with the following values:1<-a,2<-b,3<-c,4<-d;
> replace y with the following values:1<-b,2<-a,3<-c,4<-d;
> replace z with the following values:1<-d,2<-c,3<-b,4<-a;
> Finally,select two subsets:
> 1. if x='a';
> 2.x='a' and y='a';
> thanks very much!
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Re: [R] replace values?
On 07/02/06 12:39, zhijie zhang wrote:
> Dear friends,
> i have a dataset like this:
> x y z
> 1 2 3
> 2 3 1
> 3 2 1
> 1 1 3
> 2 1 2
> 3 2 3
> 2 1 1
> I want to replace x with the following values:1<-a,2<-b,3<-c,4<-d;
> replace y with the following values:1<-b,2<-a,3<-c,4<-d;
> replace z with the following values:1<-d,2<-c,3<-b,4<-a;
Here's one way. Call your dataset M, and assume it is a
data.frame. This method of replacement works best when you are
replacing consecutive integers, as you are. Note that X[1] is
"a", X[2] is "b" and so on.
X <- c("a","b","c","d")
Y <- c("b","a","c","d")
Z <- c("d","c","b","a")
M$x <- X[M$x]
M$y <- Y[M$y]
M$z <- Z[M$z]
> Finally,select two subsets:
> 1. if x='a';
> 2.x='a' and y='a';
M[M$x=="a",]
M[M$x=="a" & M$y=="a",]
The subsets will be rows. I'm not sure that's what you mean.
Jon
--
Jonathan Baron, Professor of Psychology, University of Pennsylvania
Home page: http://www.sas.upenn.edu/~baron
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Re: [R] replace values in a matrix subject to boolean condition
Uwe Ligges wrote:
> Werner Wernersen wrote:
>
>> Hi everybody!
>>
>> I am sorry to bother you with a question so simple
but
>> I think there might be a better solution:
>> I have a matrix of size 360x501 where I want to
check
>> the value of each 5th column of each row and
replace it (and the 6th,
>> 7th,
>> 8th column) by zero if the value is less than 1000.
I have written a
>> double loop
>> to do that but that requires a lot of time.
>>
>> Is there a faster way to achieve this?
>
>
>
> Two ways to interpret your question:
>
> 1) if col5 < 100 replace col5 & col6 & col7 & col8
by 0:
>
>X[X[,5] < 1000, 5:8] <- 0
>
>
> 2) if col5 < 100 replace col5 by 0, if col6 < 100
replace col6 by 0, ...:
>
>
> for(i in 5:8)
>X[X[,i] < 1000, i] <- 0
>
> Uwe Ligges
>
>
>
>
>
>> Thanks,
>>Werner
>>
>> __
>> 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
>
>
>
>
Thanks for all your answers!
Now I did it with
for (j in seq(3,length(d[1,]),by=5)) {
d[d[,j]<1000,j:(j+3)] <- c(0,0,0,0)
}
which is so much faster than the double loop.
Thanks again,
Werner
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Re: [R] replace values in a matrix subject to boolean condition
Werner Wernersen wrote: Hi everybody! I am sorry to bother you with a question so simple but I think there might be a better solution: I have a matrix of size 360x501 where I want to check the value of each 5th column of each row and replace it (and the 6th, 7th, 8th column) by zero if the value is less than 1000. I have written a double loop to do that but that requires a lot of time. Is there a faster way to achieve this? Two ways to interpret your question: 1) if col5 < 100 replace col5 & col6 & col7 & col8 by 0: X[X[,5] < 1000, 5:8] <- 0 2) if col5 < 100 replace col5 by 0, if col6 < 100 replace col6 by 0, ...: for(i in 5:8) X[X[,i] < 1000, i] <- 0 Uwe Ligges Thanks, Werner __ 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
Re: [R] replace values but different replacement length
What are the lengths of match02$BL02DATE[is.na(match02$BL02DATE)] and match02$BLOCKED? In particular, have you considered the following: match02$BL02DATE[is.na(match02$BL02DATE)] <- match02$BLOCKED[is.na(match02$BL02DATE)] hope this helps. spencer graves Christian Schulz wrote: Hi, have anybody a hint/help how it's possible replace i.e. the NA values from BL02DATE with non-Missing values from BLOCKED and vice versa. Both variables are with an id in a 188.249 rows data.frame and my fr function didn't count the NA's. Many thanks Christian fr(BL02DATE) Count Prcnt 1 140660 84.6 2 25589 15.4 Total 166249 100.0 fr(BLOCKED) Count Prcnt 1 151982 85.4 2 25976 14.6 Total 177958 100.0 match02$BL02DATE[is.na(match02$BL02DATE)] <- match02$BLOCKED Warning message: number of items to replace is not a multiple of replacement length __ [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
