Replace Given Value in Vector

Replace given value in vector

Perhaps replace is what you are looking for:

> x = c(3, 2, 1, 0, 4, 0)
> replace(x, x==0, 1)
[1] 3 2 1 1 4 1

Or, if you don't have x (any specific reason why not?):

replace(c(3, 2, 1, 0, 4, 0), c(3, 2, 1, 0, 4, 0)==0, 1)

Many people are familiar with gsub, so you can also try either of the following:

as.numeric(gsub(0, 1, x))
as.numeric(gsub(0, 1, c(3, 2, 1, 0, 4, 0)))

Update

After reading the comments, perhaps with is an option:

with(data.frame(x = c(3, 2, 1, 0, 4, 0)), replace(x, x == 0, 1))

How to replace specific vector values?

Try this replacing scheme. The vector xori has the original values whereas x has been changed with different conditions. Additionally, the shortcut 1:3 is equivalent to c(1,2,3). Here the code:

#Code
set.seed(1)
x=sample(c(1,2,3,4,5,6,7,8,9),size=15,replace=TRUE)
xori <- x
#Replace
x[x %in% 1:3]<-1
x[x %in% 6:8]<-2
x[x %in% c(4,5,9)]<-3
#Print
xori
x

Output:

xori
 [1] 9 4 7 1 2 7 2 3 1 5 5 6 7 9 5
x
 [1] 3 3 2 1 1 2 1 1 1 3 3 2 2 3 3

If one way is required:

#One way
x <- ifelse(x %in% 1:3,1,
            ifelse(x %in% 6:8,2,
                   ifelse(x %in% c(4,5,9),3,NA)))

Output:

x
 [1] 3 3 2 1 1 2 1 1 1 3 3 2 2 3 3

Replace values from a column for vector value

x1<-merge(data.frame(A=x),data.frame(A=c('A','B','C'),B=c("Red,"Blue","Green")),by='A')[,2]

did that and do the trick

Is there a way to replace a character in a vector with a NULL value in R?

It's worth remembering the difference between NULL and NA. NA values are a dodgy value, NULL is no value whatsoever. In order to get the second output to be the same as the first output, you would have something the same as the following

column <- c("None", "Some", "NULL", "Many", "All")
column <- column[column != "NULL"]

This creates a shorter vector, which is why str_replace doesn't like it.

R - Given a List of Vectors replace values in Vectors for each Vector with LOOP

It seems to me that this is what you are looking for

vec1 <- list(c(1, 2, 3, 4, 5)) # as per your requirement
list_knot <- lapply(vec1[[1]], function(v, x) x - v^3, x = 1:800)

, i.e., subtract each element of vec1 from the same x (1:800) and return a list of vectors. As you would like to replace every single element for each h in list_knot, you might just drop that list completely and construct a new one.

Search for and replace a value in a vector of structs

To change the elements in the vector, you need to take a reference in the range-for loop:

for (auto &e : data)

Otherwise you are making a copy, which doesn't change the original elements.

The 4th parameter of replace_if needs to take an info object, not just the new_name value.

Instead of replace_if, the appropriate algorithm to use here would be for_each, which modifies the info objects as needed:

std::for_each(data.begin(), data.end(), 
               [old_name, new_name](info &i) { 
                 if (i.name == old_name)
                    i.name = new_name;
              });

However, in this case, the range-for loop is probably less code, and easier to read and write.

Also, please avoid using namespace std;, it's bad practice.

Replace values in a vector in R

If the outlier detection is based on values greater than 2000, then

data$students <- with(data, replace(students, students > 2000, mean(students)))

Regarding the mean part, it is not clear whether the mean takes the outlier values too. If it is not

i1 <- data$students >2000
data$students <- with(data, replace(students, i1, mean(students[i1])))

---

Related Topics