search for: col1

Hola: Si codificas "col1" como un factor, lo puedes cambiar todo de una vez renombrando los niveles correspondientes: > df$col1<-as.factor(df$col1) > df$col1  [1] uno   dos   3     4     cinco 6     siete 8     9     diez Levels: 3 4 6 8 9 cinco diez dos siete uno > levels(df$col1)[1:5] <- c(&qu...

Dear all, I have a dataframe which looks like this (dummy): date<-c("jan", "feb", "mar", "apr", "may", "june", "july", "aug","sep","oct","nov","dec") col1<-c(8.2,5.4,4.3,4.1,3.1,2.5,1.1,4.5,3.2,1.9,7.8,6.5) col2<-c(3.1,2.3,4.7,6.9,7.5,1.1,3.6,8.5,7.5,2.5,4.1,2.3) dum<-data.frame(cbind(date,col1,col2)) dum date col1 col2 1 jan 8.2 3.1 2 feb 5.4 2.3 3 mar 4.3 4.7 4 apr 4.1 6.9 5 may 3.1 7.5 6 june 2.5 1.1 7 july 1.1...

Creo que el problema se podría simplificar y aún obtener un resultado aceptable si simplemente haces un ajuste lineal en un entorno del punto. Como ya tienes los valores de los puntos de cada curva como dices Carlos, creo que con esa información podrías identificar un entorno adecuado en el cual ajustar linealmente. Cuéntanos como te va, Saludos !! Eric. On 24-08-20 12:52, Carlos Ortega

Hello! I am merging two datasets and I have encountered a problem with sort. Can someone please point me to my error. Here is the example. ## I have dataframes, first one with factor and second one with factor ## and integer > tmp1 <- data.frame(col1 = factor(c("A", "A", "C", "C", "0", "0"))) > tmp2 <- data.frame(col1 = factor(c("C", "D", "E", "F")), col2 = 1:4) > tmp1 col1 1 A 2 A 3 C 4 C 5 0 6 0 > tmp2 col1 col...

Hello I have 2 data frames df1 and df2. I would like to create a new data frame new_df which will contain only the common rows based on the first 2 columns (chrN and start). The column score in the new data frame should be replaced with a column containing the average score (average_score) from df1 and df2. df1= data.frame(chrN= c(“chr1”, “chr1”, “chr1”, “chr1”, “chr2”, “chr2”, “chr2”),

Hola, me gustar?a hacer algo como en el siguiente ejemplo A un df a?adirle una columna que es la transformaci?n de otra, en plan a todo lo que sea x1, x2, x3 lo llamo prueba 1 todo lo que sea x4,x5,x6 lo llamo prueba 2 el resto de x las dejo como est?n. Ser?a algo as? col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5', 'x35', 'x1','x2', 'x4') df1<-data.frame(col1) attach(df1) df1$transformacion<-ifelse(col1 == "x1"|col1 == "x2"| col1 ==...

Patrick, ## Run the following script an notice the different values of the dataframe "data" in each instance. # I understand you have done something like the following: data <- data.frame(COL1 = c("6/1/14", "7/1/14"), COL2 = c("5/1/15", "5/1/15"), stringsAsFactors = FALSE) data$Date_Flag <- ifelse(data$COL2 > data$COL1, 0,1) data data$COL2 <- as.Date(as.character(data$COL2, format = "%y/%m/%d")) data$COL1 <-...

Take this as an example: > a=data.frame(col1=c(1,2,3,4,5), col2=c ("my","beloved","daughter","son","wife")) > b=data.frame(col1=c(1,2,4), col2=c("my","beloved","son")) > a col1 col2 1 1 my 2 2 beloved 3 3 daughter 4 4 son 5...

Dear R fellows, I created a new column Date_flag to compare the dates of COL1 and COL2 using the code below. But it showed that 5/1/15 is greater than 6/1/2014 and 5/1/2015 greater than 7/1/2014 despite the year is greater. How do I fix that? I did try to format as %y/%m/%d but it does not fix that. data$Date_Flag <- ifelse(data$COL2 > data$COL1, 0,1) COL1...

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 task. Thanks, Sachin [[alternative...

Thank you Jim, I read the data as you suggested but I could not find K1 in col1. rbind(preval,mydat) Col1 Col2 col3 1 <NA> <NA> <NA> 2 X1 <NA> <NA> 3 Y1 <NA> <NA> 4 K2 <NA> <NA> 5 W1 <NA> <NA> 6 Z1 K1 K2 7 Z2 <NA> <NA> 8 Z3 X1 <NA> 9 Z4 Y1 W1 On Sat, Feb 24, 2018...

HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X...

Hi Val, My fault - I assumed that the NA would be first in the result produced by "unique": mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) val23<-unique(unlist(mydat[,c("Col2","col3")])) napos<-which(is.na(val23)) preval<-data.frame(Col1=val23[-napos], Col2=NA,col3=NA) mydat<-rbind(preval,mydat) mydat[is.na(mydat...

Hi all, I put up an exhaustive model to use R's "step" function: ------------------------ mygam=gam(col1 ~ 1 + col2 + col3 + col4 + col2 ^ 2 + col3 ^ 2 + col4 ^ 2 + col2 ^ 3 + col3 ^ 3 + col4 ^ 3 + s(col2, 1) + s(col3, 1) + s(col4, 1) + s(col2, 2) + s(col3, 2) + s(col4, 2) + s(col2, 3) + s(col3, 3) + s(col4, 3) + s(col2, 4) + s(col3, 4) + s(col4, 4) + s(col2, 5) + s(col3, 5) + s(col4, 5) + s(c...

Hola: Como dice Carlos, algo así, por ejemplo: transforma <- function(df) sapply(df, function(x) ifelse(x%in%c("x1","x2","x3"), "prueba1",ifelse(x%in%c("x4","x5","x6"),"prueba2",x))) > transforma(df1)       col1  [1,] "prueba1"  [2,] "prueba1"  [3,] "x11"  [4,] "prueba1"  [5,] "x33"  [6,] "prueba1"  [7,] "prueba2"  [8,] "prueba2"  [9,] "x35" [10,] "prueba1" [11,] "prueba1" [12,] "prue...

Hello list I have been given some Excel sheets with data laid like this: Col1 Col2 A 3 2 3 B 4 5 4 C 1 4 3 I was hoping to import this into R as a csv and then get the mean and SD for each letter in column 1. Could someone give me some guidance on best to approach this? Thanks...

...precheza.cz> Sent: Wednesday, August 23, 2017 11:20:00 AM To: Patrick Casimir; r-help at r-project.org Subject: RE: Comparing 2 dale columns Hi your code is wrong. I get > test<-read.table("clipboard", header=T) > str(test) 'data.frame': 2 obs. of 2 variables: $ COL1: Factor w/ 2 levels "6/1/14","7/1/14": 1 2 $ COL2: Factor w/ 1 level "5/1/15": 1 1 > test$COL2<- as.Date(as.character(test$COL2, format="%y/%m/%d")) > test$COL1<- as.Date(as.character(test$COL1, format="%y/%m/%d"))...

...on your specific weaknesses and become self-sufficient? -- Sent from my phone. Please excuse my brevity. On February 25, 2018 7:55:55 AM PST, Val <valkremk at gmail.com> wrote: >HI Jim and all, > >I want to put one more condition. Include col2 and col3 if they are >not >in col1. > >Here is the data >mydat <- read.table(textConnection("Col1 Col2 col3 >K2 X1 NA >Z1 K1 K2 >Z2 NA NA >Z3 X1 NA >Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) > >The desired out put would be > > Col1 Col2 col3 >1 X1 0 0 >2 K1...

Sorry , I hit the send key accidentally here is my complete message. Thank you Jim and all, I got it. I have one more question on the original question What does this "[-1] " do? preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3")]))[-1], Col2=NA,col3=NA) mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE) preval<-data.frame(Col1=unique(unlist(mydat[,c("Col2","col3...

...ctamente. Se me ocurre que pueda deberse a la versión de R ¿cuál usas? El 10/09/2020 a las 17:51, Samura . escribió: > Gracias por las respuestas. > > Probé lo de hacer la función y no me salía. Pensaba que hacía algo mal. > Ahora con el código de Marcelino tampoco me sale. > > col1 <- c('x1', 'x2', 'x11', 'x1','x33', 'x1','x4', 'x5', 'x35', > 'x1','x2', 'x4') > col2 <- c('x12', 'x4', 'x6', 'x771','x4', 'x2') > col3 &lt...