CODEWITHSUNDEEP
rdataframe
user3841581
user3841581

Reputation: 2747

Building hash map from an R dataFrame

I have a pandas DataFrame df with the following content:

Serial N       voltage           
   B              12            
   B              10            
   C              12            
   C              40            
   .               .             
   AB             10             
   AB             13

I would like to build a hashmap from this dataFrame where the key is the serial number (unique) and the value is a list corresponding the voltage of that serial number. For example, I would like to have something like

df[[B]]=[12,10] or df[[c]]=[12,40].

How can I do in R?

Upvotes: 0

Views: 2312

Answers (3)

Florian
Florian

Reputation: 663

df <- data.frame(s=sample(LETTERS[1:2], 5, T), v=sample(10:40, 5), 
                 stringsAsFactors=FALSE)
h <- aggregate(v~s, data=df, FUN=c)
setNames(h[,2], h[,1])

> df
  s  v
1 A 29
2 B 22
3 A 33
4 B 26
5 A 37

> setNames(h[,2], h[,1])
$A
[1] 29 33 37

$B
[1] 22 26

Upvotes: 0

nrussell
nrussell

Reputation: 18612

Here's another base R approach: 

Hash <- sapply(unique(Df$Key), function(x) {
    Df[Df$Key == x, 2]
}, simplify = FALSE)

Df[Df$Key %in% c("A", "B"),]
#   Key Value
# 1   A    61
# 2   A    52
# 3   B    46
# 4   B    53

Hash[["A"]]
#[1] 61 52]

Hash[["B"]]
#[1] 46 53

Data: 

Df <- data.frame(
    Key = sort(rep(LETTERS, 2)),
    Value = rpois(52, 50)
)

Upvotes: 2

hrbrmstr
hrbrmstr

Reputation: 78832

You just need to do some basic transformations (all in base R):

set.seed(1492)
dat <- data.frame(serial_n=sample(c("A", "B", "C", "AB", "AC", "BC"), 100, replace=TRUE),
                  voltage=sample(10:30, 100, replace=TRUE),
                  stringsAsFactors=FALSE)

head(dat)
##   serial_n voltage
## 1        B      20
## 2        B      21
## 3        B      18
## 4        A      19
## 5        A      11
## 6        A      16

unclass(by(dat, dat$serial_n, function(x) {
  tmp <- x$voltage
  setNames(tmp, x$serial_n[1])
  tmp
})) -> df

df[["A"]]
##  [1] 19 11 16 23 28 13 20 23 27 27 28 11 27 12 25 26 19

df[["AB"]]
##  [1] 20 28 29 16 10 28 18 19 27 20 27 26 15 20 28 21 28 21 24 19 12

Upvotes: 2

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