similar to: Use or ??

Hi, A column of my df looks like A 10/20/30 40/20 60/10/10/5 80/10 I want to split it such that the last column has the last composition and if there are not enough the middle columns get the 0s. That way my df would look like A1 A2 A3 A4 10 20 0 30 40 0 0 20 60 10 10 5 80 0 0 10 How can I do that ?? [[alternative HTML version deleted]]

I have 5 (A,B,C,D,E) columns in my dataframe. I want to replace all "x" with "y" and all "a" with "b" within these 5 columns. Can I do it in one step? Thanks [[alternative HTML version deleted]]

Hi, I have a table in which one column has the name of the objects as shown below. Name Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Budlamp-Woodcutter Complex - 15 to 60% slope (60/25/15) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) Terrarossa-Blacktail-Pyeatt Complex - 1 to 40% slope (40/35/15/10) How can I split the single column into three columns

Hi, I have 6 columns in my dataframe (test) A B C D E F a b c d 40 30 a f a b 20 10 x m y m 50 30 If character value in column A matches with that in column C, it gets the percentage in column E, similarly if value in column B matches the value in D, it gets the percentage in F column. If it doesn't match, it gets 0 percentage. How can I do that ? thanks [[alternative HTML version

Hi, My data.frame "A" has FID like this FID a a b b b c c d d d d Now my second data.frame "B" has age value for a, b, c, d like FID Age a      5 b      7 c      9 d      3 How can search for the Age column in "B" and replace the values in "A" so that my new "A" looks like this FID Age a      5 a      5 b      7 b      7 b      7

Hi, My dataframe has two columns one with area and other with percent. How can i add the areas that are within a range of percentage?? My dataframe looks like Area Percent 456       0 3400      10 79          25 56           18 467         0 67          67 839        85 1120      0 3482      85 I want to add the area for values whose percent is <0, 0-25, 25-50, 50-75, >75. How

Hi, my data frame is x<-data.frame(ID=c("abc/def","abc/def/ghi","abc","mno/pqr/st/ab")) I want to split my column ID using "/" as the place to split. How can I do that without telling the code how many sub-columns. I could use nchar(gsub("[^/]","",x$ID)) to get how many "/" are in each row of the column, but could

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Hi all, It seems like I cannot use normal 'if' for data frames. What would be the best way to do the following. if data$col1='high' data$col2='H' else if data$col1='Neutral' data$col2='N' else if data$col='low' data$col2='L' else #chuch a warning? Note that col2 was not an existing column and was newly assigned for this

Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to "" Vec2 : transform 1 to B and 0 to "" Vec3 : transform 1 to C and 0 to "" to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD

Hello R experts, I have go this data frame: 'data.frame': 1 obs. of 20 variables: $ Anno : chr "PREVISIONI VS TARGET" $ OreTot: num 41 $ GioTot: logi NA $ OrGTot: logi NA $ OreCli: num 99 $ GioCli: logi NA $ OrGCli: logi NA $ OreFor: num -27 $ GioFor: logi NA $ OrGFor: logi NA $ OreOrt: num -18 $ GioOrt: logi NA $ OrGOrt: logi NA $ OreSpo: num -6 $ GioSpo: logi

Hi, Your question is still not clear. May be this helps: dat2<- read.table(text=" patient_id????? t???????? scores 1????????????????????? 0??????????????? 1.6 1????????????????????? 1??????????????? 2.6 1????????????????????? 2???????????????? 2.2 1????????????????????? 3???????????????? 1.8 2????????????????????? 0????????????????? 2.3 2?????????????????????? 2???????????????? 2.5

Hello, I'm having trouble performing a certain function within R and I was hoping someone might be able to help. I have a matrix (1000x21) that contains whole-number values ranging from 1-7 and I want to replace all entries within this matrix that have a value of 1 with a random number contained within a different dataframe that has 21 rows,1000 columns. I've tried using the if/else

HI, I met this problem: I have the feature data frame: Feature OS 4 2 4 1 4 3 1 2 4 1 what I want to do is to autimatically create one more column called "group": Feature OS Group 4 2 1 4 1 2 4 3

I want to write code that says "If you find an element equal to 4 in this vector for each person in the data set tested separately, then put in 1 for 2 and 2 for 4, else leave the variable as is" u.ppl <- (unique(init.dat1$grid)) l.ppl <- length(u.ppl) for (i in 1:l.ppl) { if (grep("4",init.dat1$Slide1_RESP)) {2 == 1, 4 == 2}; else

Hi, here i have a Max and Min values Min <-3 Max <-6 and also a matrix like this, ABC XYZ PQR ------ ------- ------- 2 4 3 5 4 8 7 1 3 In this i need to check each particular column values are between Max and Min value. If the coulmn value not coming between Max and

Dear Listers, I try to change the structure of my data. i have an indicator-matrix and want to end up with a factor. i have v1 v2 v3 1 0 0 0 1 0 0 0 1 and want v1 v2 v3 v4 1 0 0 1 0 1 0 2 0 0 1 3 amongst other things i tried the following d <- data.frame(d1=c(1,0,0), d2=c(0,1,0), d3=c(0,0,1)) d$nr <- NA sapply(1:3, function(x)

Hi guys, I was wondering if any one is able to help me on a problem that I was stuck with for a long time. It involves the replacement of character strings with numbers. The character string can take on only 3 possible values, for instance: AA AT TT I would want R to replace AT with 0. Between AA and TT, I want to compare the frequency of either value, and then for the one which occurs more, I

Hi, i would like transfrom characters from a data.frame to factors automatic. > tofac <- function(df){ + i=0 + repeat{ + i <- i+1 + if(!is.character(df[,i])) + next + df[,i] <- as.factor(df[,i]) + print(i) + if(i == length(df)) + break } + } > > tofac(abrdat) [1] 7 [1] 8 [1] 9 [1] 11 [1] 13 [1] 15 Error in "[.data.frame"(df, , i) : undefined columns selected This

Data processing? I have a large number of csv files from animal tracks that look like this: Date_ Time_ Speed Course Type_ Distance 30/03/2012 11:15:05 108 121 -2