Creating a new column using for/nested loop in r

Question

Just getting started using R and I need some help in understanding the application of for/nested loop.

StudyID<-c(1:5)
SubjectID<-c(1:5)

df<-data.frame(StudyID=rep(StudyID, each=5), SubjectID=rep(SubjectID, each=1))

How can I create a new column called as ID, which would use the combination of studyID and subjectID to create a unique ID ?

So for this data, unique ID should be from 1:25.

So the final data looks like this:

UniqueID<- c(1:25)

df<-cbind(df,UniqueID)

View(df)

Is there any other way which is faster and more efficient that looping ?

M-- · Accepted Answer · 2017-05-23 20:58:12Z

2

Using the dplyr package, you could do:

library(dplyr)
df$Id = group_indices(df,StudyID,SubjectID)

This returns:

#StudyID   SubjectID   Id
#   1         1        1
#   1         2        2
#   1         3        3
#   1         4        4
#   1         5        5
#   2         1        6
#   2         2        7
#   2         3        8
#   2         4        9
#   2         5       10
#   3         1       11
#   3         3       13
#   3         4       14
#   3         5       15
#   4         1       16
#   4         2       17
#   4         3       18
#   4         4       19
#   4         5       20
#   5         1       21
#   5         2       22
#   5         3       23
#   5         4       24
#   5         5       25

edited May 23, 2017 at 20:58

M--

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answered May 23, 2017 at 20:44

Lamia

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Comments

M-- · Accepted Answer · 2017-05-30 20:31:19Z

2

Another method to achieve that without loading any library (base R) would be this (assuming data frame is sorted based on the two columns):

StudyID<-c(1:5)
SubjectID<-c(1:5)
df<-data.frame(StudyID=rep(StudyID, each=5), SubjectID=rep(SubjectID, each=1))

df$uniqueID <- cumsum(!duplicated(df[1:2]))

or you can use this solution, mentioned in the comments (I prefer this over the first solution):

df$uniqueID <- as.numeric(factor(do.call(paste, df)))

The output would be:

> print(df, row.names = FALSE)
#StudyID  SubjectID  uniqueID
#   1         1          1
#   1         2          2
#   1         3          3
#   1         4          4
#   1         5          5
#   2         1          6
#   2         2          7
#   2         3          8
#   2         4          9
#   2         5         10
#   3         1         11
#   3         2         12
#   3         3         13
#   3         4         14
#   3         5         15
#   4         1         16
#   4         2         17
#   4         3         18
#   4         4         19
#   4         5         20
#   5         1         21
#   5         2         22
#   5         3         23
#   5         4         24
#   5         5         25

edited May 30, 2017 at 20:31

answered May 23, 2017 at 20:48

M--

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3 Comments

And_R Over a year ago

Just imagine the last observation were not (5, 5) but (1, 1). Your code will not recognize that it's about the first observation. It will just stop counting. Compare with the above solution to see the difference!

user20650 Over a year ago

another base R interation: as.numeric(factor(do.call(paste, df)))

M-- Over a year ago

@And that can be solved by sorting. Notice that I am not changing the dataframe by sorting. Just apply it on the sorted df and transform back to the original order. user20650 answer would be much better and easier in that case and that's why I included it. Thank you both.

989 · Accepted Answer · 2017-05-23 21:46:43Z

1

You could go for interaction in base R:

df$uniqueID <- with(df, as.integer(interaction(StudyID,SubjectID)))

For example (this example expresses better what you are after):

set.seed(10)
df <- data.frame(StudyID=sample(5,10,replace = T), SubjectID=rep(1:5,times=2))
df$uniqueID <- with(df, as.integer(interaction(StudyID,SubjectID)))

     # StudyID SubjectID uniqueID
# 1        3         1        3
# 2        2         2        6
# 3        3         3       11
# 4        4         4       16
# 5        1         5       17
# 6        2         1        2
# 7        2         2        6
# 8        2         3       10
# 9        4         4       16
# 10       3         5       19

answered May 23, 2017 at 21:46

989

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1 Comment

Ravula Abhigyan Over a year ago

This give the solution, however sorting has to be done in case you desire the ID's in an order.

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