I have got a data frame like this:
ID A B
1 x5.11 2,34
2 x5.57 5,36
3 x6,13 0,45
I would like to remove the 'x' of all values of the column A. How might I best accomplish this in R.
Thanks!
I have found a very easy way:
data.frama$A <- gsub("x", "", data.frame$A)
Related
I have a large data set in Stata.
There are several item batteries in this data set.
One item battery consists of 8 items (v1 - v8), each scaled from 1 to 7.
I want to code all items that take the value 1 in all items as missing values.
If v1 to v8 have the value "1", all rows to which this applies are to be replaced with missings.
I know how to code missing values with the if qualifier, but the selection with the complex condition causes me difficulties.
The code for R would probably solve this via rowSums, but I need the solution for Stata.
(I assume in R it would work like this:
df[rowSums(df[,c("v1", ... "v8")]!=1)==0, c("v1", .... "v8")] <- NA
But I need a solution for Stata.
If I understood this correctly, you want
egen rowall = concat(v1-v8)
mvdecode v1-v8 if rowall == 8 * "1", mv(1)
That is, all instances in v1-v8 of 1 are recoded as missing if and only if the values of those variables are all 1 in any observation.
I have x as data.frame. with multiple column such as Q1, Q2,....
I have tried to get 1 and 0 for Right and wrong answer. but my code is not workig.
any sugugtion which I can Write all of them and get them in new table. with 1 and 0.
x<-read.csv("data.csv")
unlist(x)
summary(x)
x<-ifelse(x$Q1=="120",1,0)
I got stuck with a specific question in R around concatenating columns of a data frame by using a wildcard. Perhaps I am searching wrongly. However I could not find a matching answer yet.
Here is my question:
I have a data frame df where each column represents a user (U1, U2, U3), e.g.:
> df <-data.frame(U1=1:3, U2=4:6, U3=7:9)
> df
> U1 U2 U3
1 1 4 7
2 2 5 8
3 3 6 9
I would like to concatenate the values from all users into a single vector as one would do using the c() function, e.g.:
> c(df$U1, df$U2, df$U3)
[1] 1 2 3 4 5 6 7 8 9
However, my number of users is large and varies over time. So, I look for an elegant dynamic way of concatenating the columns such as
> c(df$U*)
Unfortunately this does not seem to work. I played around with grep and regular expressions but could not get it to work. For sure, I could use a for-loop and program my own cat function but I assume there is a better way. I just don't find it. Maybe I am just blind. Hope you can help.
sub_df <- df[, grep(pattern ='^U.*', names(df))]
stack(df)$values
Hope this works for you. You could first subset some columns according to your need.
Coerce the data frame to a matrix first:
as.vector(as.matrix(df))
Use the bracket [ to select columns whose names match a certain expression:
df[, grep("U.*", colnames(df)), drop = FALSE]
I am trying to change the name of a variable to be called in r on the fly.
For example, dataframe trades_long_final has many columns "prob_choice1" and "prob_choice2", ... "prob_choiceN" and "col1", "col2", ... "colN".
I want to change the value of each on the fly.
For example,
trades_long_final$"prob_choice1"[1] = 10 and
trades_long_final$"prob_choice2"[1] = 10
works
but not
trades_long_final$gsub("1","2","prob_choice1")[1] = 10
as a way to call trades_long_final$"prob_choice2"[1] by substituting the 1 in prob_choice1 with a 2 because I get the error
Error: attempt to apply non-function
I need this to work because I need to loop over the columns using something like trades_long_final$gsub("i","2","prob_choicei")[1] in a loop for all i.
Thank you so much for your help. It must be a command I don't know how to use...
Instead of using $, you can use [ to change the variable name and assign the value in one line.
trades_long_final[,gsub("1","2","prob_choice1")][1] <- 10
But, it is not clear why you need to do this. Simply
trades_long_final[1, "prob_choice2"] <- 10
would be easier. From the description, "prob_choice2" is already a column in the dataset. So, it is confusing.
data
set.seed(24)
trades_long_final <- data.frame(prob_choice1 =runif(10),
prob_choice2=rnorm(10), col1=rnorm(10,10), col2=rnorm(10,30))
as akrun said, the way to do it was to use [ so that the way I did it was:
for (k in 1:numtrades){
trades_long_final[[paste("prob_choice", k, sep="")]] =
{some complex procedure...}
}
Suppose I have two dataframes a and b,
a has one column called 'detail':
pure water
wood fire
mineral water
water
fire work
and b has one column called 'type':
water
fire
Many R functions require input text to get match, grep('fire',a), but my question is if there is a way to match a using b? I tried loop but failed. Following SQLDF got all false result for match.
ab <- sqldf(select *,case when detail in (select distinct types from b) then 1 else 0 end as match) from a)
Ideally, one can using something like c <- grep(a$detail,b$types). not sure if it is allowed in R though.
Thanks in advance!
Create a type column in a and then merge on it:
merge(transform(a, type = sub(".* ", "", a$detail)), b, all = TRUE)