# Mapping

Suppose we load the data which is represented by the following table:

Name Age Interests Expertise
Ken 24 reading, music, movies R:2, C#:4, Python:3
James 25 sports, music R:3, Java:2, C++:5
Penny 24 movies, reading R:1, C++:4, Python:2

list.load() is designed for loading data from given data source. The data source can be either local or remote and the function by default uses the file extension to decide the way to read it.

library(rlist)
people <- list.load("https://renkun-ken.github.io/rlist-tutorial/data/sample.json")
str(people)

# List of 3
#  $:List of 4 # ..$ Name     : chr "Ken"
#   ..$Age : int 24 # ..$ Interests: chr [1:3] "reading" "music" "movies"
#   ..$Expertise:List of 3 # .. ..$ R     : int 2
#   .. ..$CSharp: int 4 # .. ..$ Python: int 3
#  $:List of 4 # ..$ Name     : chr "James"
#   ..$Age : int 25 # ..$ Interests: chr [1:2] "sports" "music"
#   ..$Expertise:List of 3 # .. ..$ R   : int 3
#   .. ..$Java: int 2 # .. ..$ Cpp : int 5
#  $:List of 4 # ..$ Name     : chr "Penny"
#   ..$Age : int 24 # ..$ Interests: chr [1:2] "movies" "reading"
#   ..$Expertise:List of 3 # .. ..$ R     : int 1
#   .. ..$Cpp : int 4 # .. ..$ Python: int 2


NOTE: str() previews the structure of an object. We may use this function more often to avoid verbose representation of list objects.

To extract the name of each people (list element), traditionally we can call lapply() like the following:

lapply(people, function(x) {
x$Name })  # [[1]] # [1] "Ken" # # [[2]] # [1] "James" # # [[3]] # [1] "Penny"  Using rlist's list.map() the task is made extremely easy: list.map(people, Name)  # [[1]] # [1] "Ken" # # [[2]] # [1] "James" # # [[3]] # [1] "Penny"  List mapping is to evaluate an expression for each list member. It is the fundamental operation in rlist functionality. Almost all functions in this package that work with expressions are using mapping but in different ways. The following examples demonstrate several types of mapping in more details. ## list.map The simplest way of mapping is provided by list.map() as we have just demonstrated. Basically, it evaluates an expression for each list element. The function makes it easier to query a list by putting all fields of the list member in mapping to the environment where the expression is evaluated. In other words, the expression is evaluated in the context of one list member each time. For example, the following code maps each list member in people by expression age. Therefore, it results in a list where each item becomes the value of that expression for each member of people. list.map(people, Age)  # [[1]] # [1] 24 # # [[2]] # [1] 25 # # [[3]] # [1] 24  Since the expression does not have to be a field name of the list member, we can evaluate whatever we want in the context of a list member. The following code maps each list member to the sum of years of the programming languages they use. list.map(people, sum(as.numeric(Expertise)))  # [[1]] # [1] 9 # # [[2]] # [1] 10 # # [[3]] # [1] 7  If we need more than one values for each member, we can evaluate a vector or list expression. The following code maps each list member to a new list of the age and range of number of years using programming languages. list.map(people, list(age=Age, range=range(as.numeric(Expertise))))  # [[1]] # [[1]]$age
# [1] 24
#
# [[1]]$range # [1] 2 4 # # # [[2]] # [[2]]$age
# [1] 25
#
# [[2]]$range # [1] 2 5 # # # [[3]] # [[3]]$age
# [1] 24
#
# [[3]]$range # [1] 1 4  In some cases we need to refer to the item itself, or its index in the list, or even its name. In the expression, . represents the item itself, .i represents its index, and .name represents its name. For example, nums <- c(a=3, b=2, c=1) list.map(nums, . + 1)  #$a
# [1] 4
#
# $b # [1] 3 # #$c
# [1] 2

list.map(nums, .i)

# $a # [1] 1 # #$b
# [1] 2
#
# $c # [1] 3  list.map(nums, paste0("name: ", .name))  #$a
# [1] "name: a"
#
# $b # [1] "name: b" # #$c
# [1] "name: c"


If the default symbols clash with the data, we can use lambda expression to specify other symbols. We will cover this later.

NOTE: rlist functions are general enough to work smoothly with vectors. list.map() works very much like lapply() so that the input will be finally transformed to list.

## list.mapv

If we want to get the mapping results as a vector rather than a list, we can use list.mapv(), which basically calls unlist() to the list resulted from list.map().

list.mapv(people, Age)

# [1] 24 25 24

list.mapv(people, sum(as.numeric(Expertise)))

# [1]  9 10  7


## list.select

In contrast to list.map(), list.select() provides an easier way to map each list member to a new list. This functions basically evaluates all given expressions and put the results into a list.

If a field name a list member is selected, its name will automatically preserved. If a list item evaluated from other expression is selected, we may better give it a name, or otherwise it will only have an index.

list.select(people, Name, Age)

# [[1]]
# [[1]]$Name # [1] "Ken" # # [[1]]$Age
# [1] 24
#
#
# [[2]]
# [[2]]$Name # [1] "James" # # [[2]]$Age
# [1] 25
#
#
# [[3]]
# [[3]]$Name # [1] "Penny" # # [[3]]$Age
# [1] 24

list.select(people, Name, Age, nlang=length(Expertise))

# [[1]]
# [[1]]$Name # [1] "Ken" # # [[1]]$Age
# [1] 24
#
# [[1]]$nlang # [1] 3 # # # [[2]] # [[2]]$Name
# [1] "James"
#
# [[2]]$Age # [1] 25 # # [[2]]$nlang
# [1] 3
#
#
# [[3]]
# [[3]]$Name # [1] "Penny" # # [[3]]$Age
# [1] 24
#
# [[3]]$nlang # [1] 3  ## list.iter Sometimes we don't really need the result of a mapping but its side effects. For example, if we only need to print out something about each list member, we don't need to carry on the output of mapping. list.iter() performs iterations over a list and returns the input data invisibly for further data transformation. list.iter(people, cat(Name, ":", Age, "\n"))  # Ken : 24 # James : 25 # Penny : 24  ## list.maps All the previous functions work with a single list. However, there are scenarios where mapping multiple lists is needed. list.maps() evaluates an expression with multiple lists each of which is represented by a user-defined symbol at the function call. l1 <- list(p1=list(x=1,y=2), p2=list(x=3,y=4), p3=list(x=1,y=3)) l2 <- list(2, 3, 5) list.maps(a$x*b+a$y, a=l1, b=l2)  #$p1
# [1] 4
#
# $p2 # [1] 13 # #$p3
# [1] 8


list.maps() does not follow the conventions of many other functions like list.map() and list.iter() where the data comes first and expression comes the second. Since list.maps() supports multi-mapping with a group of lists, only implicit lambda expression is supported to avoid ambiguity. After that the function still allows users to define the symbol that represents each list being mapped in ....

In the example above, ... means a = l1, b = l2, so that a and b are meaningful in the first expression a$x*b+a$y where a and b mean the iterating element of each list, respectively.

The function does not require named be supplied with the lists as arguments. In this case, we can use ..1, ..2, etc. to refer to the first, second or other lists.

list.maps(..1$x*..2 + ..1$y, l1, l2)

# $p1 # [1] 4 # #$p2
# [1] 13
#
# \$p3
# [1] 8