将“双精度”列表的名称创建为“命名向量列表”

Question

What I try to do:

In aphid package there is a function deriveHMM() which needs to be fed with a list like:

x <- list(c("c"="10.0", "b"="5.0","c"="10.0", "a"="1.0", "a"="2.0",...))

wich needs to be created of a very large input vector like

iv <- c(10, 5, 10, 1, 2,...)

It is important, that the order of my original input vector remains unchanged.

I need to automatically create this list by a large input of doubles from a .csv file (import of doubles to R worked fine). Each double has to get a name depending on its closest distance to a predefined value, for example:

all doubles rangig from 0 to 2.5 should be named "a"

all doubles ranging from 2.5 to 7.5 should be named "b"

all doubles greater than 7.5 should be named "c"

and after that all doubles be converted to a character (or string (?)) so the method deriveHMM() accepts the input.

I would be very happy to have suggestions. I am new to R and this is my first post on Stackoverflow.com. I am not an experienced programmer, but I try my best to understand your help.

EDIT: Updated the question, because what i need is a "List of named vectors of characters", exactly like in my example above without changing the order.

Answer 1

This solution uses findInterval to get an index into a tags vector, the vector of names.

set.seed(1234)    # Make the results reproducible
x <- runif(10, 0, 20)

tags <- letters[1:3]
breaks <- c(0, 2.5, 7.5, Inf)

names(x) <- tags[findInterval(x, breaks)]

x
#         a          c          c          c          c 
# 2.2740682 12.4459881 12.1854947 12.4675888 17.2183077 
#         c          a          b          c          c 
#12.8062121  0.1899151  4.6510101 13.3216752 10.2850228

Edit.

If you need x to be of class "character" , get the index into tags first, then coerce x to character and only then assign the names attribute.

i <- findInterval(x, breaks)
x <- as.character(x)
names(x) <- tags[i]
x
#                  a                   c                   c 
# "2.27406822610646"  "12.4459880962968"  "12.1854946576059" 
#                  c                   c                   c 
# "12.4675888335332"  "17.2183076711372"  "12.8062121057883" 
#                  a                   b                   c 
#"0.189915127120912"  "4.65101012028754"   "13.321675164625" 
#                  c 
# "10.2850228268653" 

Answer 2

Here is an example, where x represents your input vector.

x <- seq(1, 10, 0.5)

The first step is to give your elements names depending on their values.

names(x) <- ifelse(x <= 2.5, "a", ifelse(x > 2.5 & x <= 7.5, "b", "c"))

Next, split your vector and a apply as.character . We can use by here.

lst <- by(x, names(x), as.character, simplify = TRUE)
is.list(lst)
# [1] TRUE

Result

lst
#names(x): a
#[1] "1"   "1.5" "2"   "2.5"
#----------------------------------------------------------------------------------------------------------------------- 
#names(x): b
# [1] "3"   "3.5" "4"   "4.5" "5"   "5.5" "6"   "6.5" "7"   "7.5"
#----------------------------------------------------------------------------------------------------------------------- 
#names(x): c
#[1] "8"   "8.5" "9"   "9.5" "10" 

---

You could also use split and lapply as shown below, by is shorthand of such an approach.

lapply(split(x, names(x)), as.character)