CODEWITHSUNDEEP
rreplacena
Hugo
Hugo

Reputation: 25

Replace blank cell with "no" in R

I like to replace blank cells (" ") in a column with "no". The missing entries do have a meaning for me (no score determined yet) and I like to use the factor variable in a regression tree later.

I found a similar question here (Replace blank cells with character) and tried the following, but then the blank cells are converted to NA and not as text:

> Test$SCORE[Test$SCORE==" "]<- "no"

Warning message:
In `[<-.factor`(`*tmp*`, Test$SCORE == " ", value = c(NA, NA, 8L,  :
  invalid factor level, NA generated

Is there a way to avoid NA and use my own text?

Please see example data "Test": 

ID  Score  
 1. A
 2. " "
 3. B
 4. " "
 5. C

Is there a way to avoid NA and use my own text? This is the result I like to achieve: 

ID  Score
1   A
2   "no"
3   B 
4   "no"
5   C

The dataset is very large therefore a manual solution via indexing specific rows is quite time consuming. I appreciate your help because R is quite new for me.

Thank you very much in advance.

Additional info:

str(Test$SCORE) Factor w/ 13 levels " ","A","B","C",..

Please excuse the format of the example table, but this is my first question.

Upvotes: 2

Views: 12057

Answers (2)

Sowmya S. Manian
Sowmya S. Manian

Reputation: 3833

> df <- data.frame(Test=1:5,Score=c("A"," ","B"," "," "))
> df
   Test Score
 1    1     A
 2    2      
 3    3     B
 4    4      
 5    5      

> df[,2] <- as.character(df$Score)
> is.character(df[,2])
[1] TRUE

> df$Score[df$Score==" "] <- "No"
> df
  Test Score
1    1     A
2    2    No
3    3     B
4    4    No
5    5    No

Upvotes: 0

Roland
Roland

Reputation: 132706

Work on the factor levels:

DF <- read.table(text = 'ID  Score  
                 1. A
                 2. " "
                 3. B
                 4. " "
                 5. C', header = TRUE)
levels(DF$Score)[levels(DF$Score) == " "] <- "no"
#  ID Score
#1  1     A
#2  2    no
#3  3     B
#4  4    no
#5  5     C

This is very efficient since there are usually far less factor levels than elements in your vector.

Upvotes: 7

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