How to Map a Vector of Values to Another Vector with My Own Custom Map in R

How do I map a vector of values to another vector with my own custom map in R

A couple of options, all using:

myVector<-c(1,2,3,2,3,3,1)

Factor

newvals <- c(.2,.4,.5)
newvals[as.factor(myVector)]
#[1] 0.2 0.4 0.5 0.4 0.5 0.5 0.2

Named vector

newvals <- c(`1`=.2,`2`=.4,`3`=.5)
newvals
#  1   2   3 
#0.2 0.4 0.5 

newvals[as.character(myVector)]
#  1   2   3   2   3   3   1 
#0.2 0.4 0.5 0.4 0.5 0.5 0.2 

Lookup table

mapdf <- data.frame(old=c(1,2,3),new=c(.2,.4,.5))
mapdf$new[match(myVector,mapdf$old)]
#[1] 0.2 0.4 0.5 0.4 0.5 0.5 0.2

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Benchmarks to quantify @Joe 's comment below and address @Ananda's comment as well.

myVector <- c(1,2,3,2,3,3,1)
# setup for the benchmarking
test <- sample(myVector,1e6,replace=TRUE)
newvals <- c(.2,.4,.5)
newvalsvec <- c(`1`=.2,`2`=.4,`3`=.5)
mapdf <- data.frame(old=c(1,2,3),new=c(.2,.4,.5))

microbenchmark(
  newvals[as.factor(test)],
  newvalsvec[as.character(test)],
  mapdf$new[match(test,mapdf$old)],
  newvals[test],
  times=10L
)

#Unit: milliseconds
#         expr        min         lq     median         uq        max
#factor        1863.40146 1876.04197 1890.99147 1913.13046 2014.23609
#namedvector   1809.26883 1812.76272 1837.18852 1851.42954 1858.44996
#lookup          38.48697   38.83405   39.90146   69.65140   71.75051
#newvals[test]   34.07380   34.55885   50.61287   65.69495   66.08699

Pass a vector of arguments to map function

Skip the group_by() step and just use nest() - otherwise your data will remain grouped after nesting and need to be ungrouped. To get your function to work, just pass the parameters as a list.

library(tidyverse)

mtcars %>%
  nest(data = -cyl) %>%
  mutate(
    newold = map2_df(data, list(c(5, 10)), myf)
  ) %>%
  unpack(newold)

# A tibble: 3 x 4
    cyl data                 old   new
  <dbl> <list>             <dbl> <dbl>
1     6 <tibble [7 x 10]>   19.7  30.7
2     4 <tibble [11 x 10]>  26.7  31.1
3     8 <tibble [14 x 10]>  15.1  17.0

Map a function over a vector

With purrr, we use map

library(purrr)
map(myknots,  ~ myfoo(x, y, nknots = .x))

Applying purrr::map over each of a vector of characters

Your syntax is a little bit off, you'd either use map(animals, make_df) or map(animals, ~ make_df(.)), the second argument of map needs to be a function, which is the same as lapply:

data.frame(animals) %>% mutate(ldf = map(animals, make_df)) %>% as.tibble()
# A tibble: 3 x 2
#  animals                  ldf
#   <fctr>               <list>
#1   sheep <data.frame [5 x 3]>
#2     cow <data.frame [5 x 3]>
#3   horse <data.frame [5 x 3]>

data.frame(animals) %>% mutate(ldf = map(animals, ~ make_df(.))) %>% as.tibble()
# A tibble: 3 x 2
#  animals                  ldf
#   <fctr>               <list>
#1   sheep <data.frame [5 x 3]>
#2     cow <data.frame [5 x 3]>
#3   horse <data.frame [5 x 3]>

Or if using the data.frame constructor, you need to use I to create a list type column:

data.frame(animals, ldf = I(lapply(animals, make_df)))
#                         ^

How to replace the elements of a large character vector

You could create a lookup dataframe with vec1 and vec2 as columns. Note that, there were some whitespaces in the data that your shared which I have removed.

dat <- data.frame(a = c("Zone 1A", "Zone 1C","Zone 2B","Zone 3C"), b = 1:4)
vec1 <- c("Zone 1A","Zone 1B","Zone 1C","Zone 2A","Zone 2B","Zone 2C") 
vec2 <- c("Zone 1","Zone 1","Zone 1","Zone 2","Zone 2","Zone 2")
lookup <- data.frame(vec1, vec2)

You can use match to replace values.

dat$a <- lookup$vec2[match(dat$a, lookup$vec1)]

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Based on the data shared we can remove the last character in a column which returns us the proper "Zone" value. You can use sub to do that.

dat$a <- sub('[A-Z]$', '', dat$a)

Replace values in a vector based on another vector

Working with factors might be faster:

xf <- as.factor(x)
y[xf]

Note, that levels(xf) gives you a character vector similar to your x.lvl. Thus, for this method to work, elements of y should correspond to appropriate elements of levels(xf).

How to sort a vector by alternating its values

We may use rowid

library(data.table)
x[order(rowid(x))]
[1] 4 5 4 5

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