CODEWITHSUNDEEP
r
steppermotor
steppermotor

Reputation: 701

Creating a new column based on unique ID with values in r

For each of these IDs, I would like to create a new column called Age populated with the values 0 to 5 for each ID (r=0:5) as shown below.

Data Frame

ID         
1124
1123

Desired Outcome

ID   Age
1124  0
1124  1
1124  2
1124  3
1124  4
1124  5
1123  0
1123  1
1123  2
1123  3
1123  4
1123  5

Upvotes: 0

Views: 285

Answers (3)

steveb
steveb

Reputation: 5532

Here is a base R version:

df = data_frame(ID = c(1124, 1123))
expand.grid(ID = df$ID, Age = 0:5)

##      ID Age
## 1  1124   0
## 2  1123   0
## 3  1124   1
## 4  1123   1
## 5  1124   2
## 6  1123   2
## 7  1124   3
## 8  1123   3
## 9  1124   4
## 10 1123   4
## 11 1124   5
## 12 1123   5

This is sorted differently from the tidyr::expand result.

EDIT

As @thelatemail suggested, you can do the following to avoid renaming df

expand.grid(c(Age=list(0:5), df))

or

merge(df, list(Age=0:5))

EDIT 2

Here is a data.table example:

library(data.table)
setDT(df) # Convert df to a data.table.
df[, do.call(CJ, list(ID = ID, Age = 0:5))]

For large data sets, one might want to benchmark the various methods.

Upvotes: 2

Marius
Marius

Reputation: 60060

This can be done with tidyr::expand:

library(tidyverse)

df = data_frame(ID = c(1124, 1123))

df %>%
    expand(ID, Age = 0:5)

Output:

# A tibble: 12 x 2
      ID   Age
   <dbl> <int>
 1  1123     0
 2  1123     1
 3  1123     2
 4  1123     3
 5  1123     4
 6  1123     5
 7  1124     0
 8  1124     1
 9  1124     2
10  1124     3
11  1124     4
12  1124     5

Upvotes: 1

FKneip
FKneip

Reputation: 378

library(tidyverse)
your_data_frame %>%
    group_by(ID) %>%
    mutate(Age = (1:n()) - 1)

This works also if you have more than 6 Age values per ID.

Upvotes: 0

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