for loop & if function in R

Question

I was writing a loop with if function in R. The table is like below:

ID  category
1   a
1   b
1   c
2   a
2   b
3   a
3   b
4   a
5   a

I want to use the for loop with if function to add another column to count each grouped ID, like below count column: ID category Count

My code is (output is the table name):

for (i in 2:nrow(output1)){
  if(output1[i,1] == output[i-1,1]){
    output1[i,"rn"]<- output1[i-1,"rn"]+1
  } 

  else{
     output1[i,"rn"]<-1
   }

}

But the result returns as all count column values are all "1".

ID  category    Count
1   a   1
1   b   1
1   c   1
2   a   1
2   b   1
3   a   1
3   b   1
4   a   1
5   a   1

Please help me out... Thanks

There are functions that can do this operation quickly, but it is always good to practice logical control flows with the loop. try adding output1$rn <- 1 before the loop — Pierre L
– Pierre L, Commented Sep 17, 2015 at 17:23
try grouping by id and counting rows, library(dplyr); dat %>% group_by(ID) %>% mutate(Count = 1:n()) — Rorschach
– Rorschach, Commented Sep 17, 2015 at 17:24

Pierre L · Accepted Answer · 2015-09-17 17:32:19Z

3

There are packages and vectorized ways to do this task, but if you are practicing with loops try:

output1$rn <- 1
for (i in 2:nrow(output1)){
  if(output1[i,1] == output1[i-1,1]){
    output1[i,"rn"]<- output1[i-1,"rn"]+1
  } 

  else{
     output1[i,"rn"]<-1
   } 
}

With your original code, when you made this call output1[i-1,"rn"]+1 in the third line of your loop, you were referencing a row that didn't exist on the first pass. By first creating the row and filling it with the value 1, you give the loop something explicit to refer to.

output1
#   ID category rn
# 1  1        a  1
# 2  1        b  2
# 3  1        c  3
# 4  2        a  1
# 5  2        b  2
# 6  3        a  1
# 7  3        b  2
# 8  4        a  1
# 9  5        a  1

With the package dplyr you can accomplish it quickly with:

library(dplyr)
output1 %>% group_by(ID) %>% mutate(rn = 1:n())

Or with data.table:

library(data.table)
setDT(output1)[,rn := 1:.N, by=ID]

With base R you can also use:

output1$rn <- with(output1, ave(as.character(category), ID, FUN=seq))

There are vignettes and tutorials on the two packages mentioned, and by searching ?ave in the R console for the last approach.

edited Sep 17, 2015 at 17:32

answered Sep 17, 2015 at 17:24

Pierre L

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1 Comment

Arun Over a year ago

From the next version on (v1.9.8), we'd be able to do this simply as: dt[, rn := rowid(ID)]

abhiieor · Accepted Answer · 2015-09-17 17:31:53Z

1

looping solution will be painfully slow for bigger data. Here is one line solution using data.table:

require(data.table)
a<-data.table(ID=c(1,1,1,2,2,3,3,4,5),category=c('a','b','c','a','b','a','b','a','a'))
a[,':='(category_count = 1:.N),by=.(ID)]

answered Sep 17, 2015 at 17:31

abhiieor

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Comments

Dhawal Kapil · Accepted Answer · 2015-09-17 17:40:59Z

1

what you want is actually a column of factor level. do this

df$count=as.numeric(df$category)

this will give out put as

  ID category count
1  1        a     1
2  1        b     2
3  1        c     3
4  2        a     1
5  2        b     2
6  3        a     1
7  3        b     2
8  4        a     1
9  5        a     1

provided your category is already a factor. if not first convert to factor

df$category=as.factor(df$category)
df$count=as.numeric(df$category)

answered Sep 17, 2015 at 17:40

Dhawal Kapil

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5 Comments

David Arenburg Over a year ago

This will only work for this specific example. Provided that category is some real name this could mess up, for example as.numeric(factor(c("shoe","bag","tie")))

Dhawal Kapil Over a year ago

Cant understand why. It will work for any factor. Even if factor levels are of any length .

David Arenburg Over a year ago

Did you try the code I posted in my previous comment?

Dhawal Kapil Over a year ago

yes..by mess up you meant it will generate factor levels in order of alphabets and will assign shoe=2,bag=1,tie=3?

David Arenburg Over a year ago

Yes, so I'm guessing that OP could have different categories which could produce wrong result.

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