CODEWITHSUNDEEP
rlistdata.tablelapplysapply
Ahmadov
Ahmadov

Reputation: 1607

R data.table Accessing elements from a column of lists

I have been following a tutorial on data.tables here Let's say that I have the following table (I have changed the original table to fit my question)

##    gear cyl gearsL
## 1:    4   6  4,3,5
## 2:    4   6  4,3,5
## 3:    4   4  4,3,5
## 4:    3   6  5,6,7
## 5:    3   8  5,6,7
## 6:    3   6  5,6,7

I now want to create a new column which will "ungroup" the gearsL column, as follows:

##    gear cyl gearsL  gearA
## 1:    4   6  4,3,5  4
## 2:    4   6  4,3,5  3
## 3:    4   4  4,3,5  5
## 4:    3   6  5,6,7  5
## 5:    3   8  5,6,7  6
## 6:    3   6  5,6,7  7

I can use the following snippet of code to extract a static element, e.g. element at index 2. 

dt[,gearL1:=lapply(gearsL, function(x) x[2])]
dt[,gearS1:=sapply(gearsL, function(x) x[2])]

This will result in the following table: 

##    gear cyl gearsL  gearL1 gearS1
## 1:    4   6  4,3,5  3      3
## 2:    4   6  4,3,5  3      3
## 3:    4   4  4,3,5  3      3
## 4:    3   6  5,6,7  6      6
## 5:    3   8  5,6,7  6      6
## 6:    3   6  5,6,7  6      6

However, I want a "dynamic" index. First, I created a new field, called IDX, which acts as a row-number with groups. 

dt[,IDX:=1:.N,by='gear']

which will result in the following table:

##    gear cyl gearsL  gearL1 gearS1  IDX
## 1:    4   6  4,3,5  3      3        1
## 2:    4   6  4,3,5  3      3        2
## 3:    4   4  4,3,5  3      3        3
## 4:    3   6  5,6,7  6      6        1
## 5:    3   8  5,6,7  6      6        2
## 6:    3   6  5,6,7  6      6        3

Using the newly created IDX column, I would like to access the elements of each list as follows:

 dt[,gearA:=sapply(gearsL, function(x) x[IDX])]
 dt[,gearA:=lapply(gearsL, function(x) x[IDX])]

However, the above snippet doesn't work as expected. How can I access the elements of a list based on the value of another column?

Upvotes: 1

Views: 2716

Answers (1)

Nathan Werth
Nathan Werth

Reputation: 5263

dt[, gearA := mapply('[[', gearsL, IDX, SIMPLIFY = TRUE)]

This runs along both gearsL and IDX, giving them as arguments to the [[ function. I.e., gears[[i]][[IDX[[i]]]].

Upvotes: 2

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