问题
I have trivial question: I couldn't find a dictionary data structure in R, so I used list instead (like "word"->number) So, right now I have problem how to get the list of keys. Anybody knows?
回答1:
Yes, the list type is a good approximation. You can use names() on your list to set and retrieve the 'keys':
> foo <- vector(mode="list", length=3)
> names(foo) <- c("tic", "tac", "toe")
> foo[[1]] <- 12; foo[[2]] <- 22; foo[[3]] <- 33
> foo
$tic
[1] 12
$tac
[1] 22
$toe
[1] 33
> names(foo)
[1] "tic" "tac" "toe"
>
回答2:
You do not even need lists if your "number" values are all of the same mode. If I take Dirk Eddelbuettel's example:
> foo <- c(12, 22, 33)
> names(foo) <- c("tic", "tac", "toe")
> foo
tic tac toe
12 22 33
> names(foo)
[1] "tic" "tac" "toe"
Lists are only required if your values are either of mixed mode (for example characters and numbers) or vectors.
For both lists and vectors, an individual element can be subsetted by name:
> foo["tac"]
tac
22
Or for a list:
> foo[["tac"]]
[1] 22
回答3:
To extend a little bit answer of Calimo I present few more things you may find useful while creating this quasi dictionaries in R:
a) how to return all the VALUES of the dictionary:
>as.numeric(foo)
[1] 12 22 33
b) check whether dictionary CONTAINS KEY:
>'tic' %in% names(foo)
[1] TRUE
c) how to ADD NEW key, value pair to dictionary:
c(foo,tic2=44)
results:
tic tac toe tic2
12 22 33 44
d) how to fulfill the requirement of REAL DICTIONARY - that keys CANNOT repeat(UNIQUE KEYS)? You need to combine b) and c) and build function which validates whether there is such key, and do what you want: e.g don't allow insertion, update value if the new differs from the old one, or rebuild somehow key(e.g adds some number to it so it is unique)
e) how to DELETE pair BY KEY from dictionary:
foo<-foo[which(foo!=foo[["tac"]])]
回答4:
The reason for using dictionaries in the first place is performance. Although it is correct that you can use named vectors and lists for the task the issue is that they are becoming quite slow and memory hungry with more data.
Yet what many people don't know is that R has indeed an inbuilt dictionary data structure: environments with the option hash = TRUE
See the following example for how to make it work:
# vectorize assign, get and exists for convenience
assign_hash <- Vectorize(assign, vectorize.args = c("x", "value"))
get_hash <- Vectorize(get, vectorize.args = "x")
exists_hash <- Vectorize(exists, vectorize.args = "x")
# keys and values
key<- c("tic", "tac", "toe")
value <- c(1, 22, 333)
# initialize hash
hash = new.env(hash = TRUE, parent = emptyenv(), size = 100L)
# assign values to keys
assign_hash(key, value, hash)
## tic tac toe
## 1 22 333
# get values for keys
get_hash(c("toe", "tic"), hash)
## toe tic
## 333 1
# alternatively:
mget(c("toe", "tic"), hash)
## $toe
## [1] 333
##
## $tic
## [1] 1
# show all keys
ls(hash)
## [1] "tac" "tic" "toe"
# show all keys with values
get_hash(ls(hash), hash)
## tac tic toe
## 22 1 333
# remove key-value pairs
rm(list = c("toe", "tic"), envir = hash)
get_hash(ls(hash), hash)
## tac
## 22
# check if keys are in hash
exists_hash(c("tac", "nothere"), hash)
## tac nothere
## TRUE FALSE
# for single keys this is also possible:
# show value for single key
hash[["tac"]]
## [1] 22
# create new key-value pair
hash[["test"]] <- 1234
get_hash(ls(hash), hash)
## tac test
## 22 1234
# update single value
hash[["test"]] <- 54321
get_hash(ls(hash), hash)
## tac test
## 22 54321
Edit: On the basis of this answer I wrote a blog post with some more context: http://blog.ephorie.de/hash-me-if-you-can
回答5:
The package hash is now available: https://cran.r-project.org/web/packages/hash/hash.pdf
Examples
h <- hash( keys=letters, values=1:26 )
h <- hash( letters, 1:26 )
h$a
# [1] 1
h$foo <- "bar"
h[ "foo" ]
# <hash> containing 1 key-value pair(s).
# foo : bar
h[[ "foo" ]]
# [1] "bar"
回答6:
You may want to look at the hash package on CRAN.
回答7:
Shorter variation of Dirk's answer:
# Create a Color Palette Dictionary
> color <- c('navy.blue', 'gold', 'dark.gray')
> hex <- c('#336A91', '#F3C117', '#7F7F7F')
> # Create List
> color_palette <- as.list(hex)
> # Name List Items
> names(color_palette) <- color
>
> color_palette
$navy.blue
[1] "#336A91"
$gold
[1] "#F3C117"
$dark.gray
[1] "#7F7F7F"
回答8:
I'll just comment you can get a lot of mileage out of table when trying to "fake" a dictionary also, e.g.
> x <- c("a","a","b","b","b","c")
> (t <- table(x))
x
a b c
2 3 1
> names(t)
[1] "a" "b" "c"
> o <- order(as.numeric(t))
> names(t[o])
[1] "c" "a" "b"
etc.
来源：https://stackoverflow.com/questions/2858014/working-with-dictionaries-lists-in-r