Optimizing preprocessing data frame in R

Question

I have the following data frame with the name dataValues :

   dates         hours
1  2015-10-12    1
5  2015-10-12    5
9  2015-10-12    9
11 2015-10-12    11
14 2015-10-12    14
15 2015-10-12    15
17 2015-10-12    17
19 2015-10-12    19
22 2015-10-12    22
23 2015-10-12    23
24 2015-10-12    24
27 2015-10-13    3
29 2015-10-13    5
33 2015-10-13    9
36 2015-10-13    12
37 2015-10-13    13
38 2015-10-13    14
40 2015-10-13    16
42 2015-10-13    18
44 2015-10-13    20
45 2015-10-13    21
46 2015-10-13    22
47 2015-10-13    23
49 2015-10-14    1
54 2015-10-14    6
56 2015-10-14    8
59 2015-10-14    11
60 2015-10-14    12
61 2015-10-14    13
63 2015-10-14    15
64 2015-10-14    16
66 2015-10-14    18
69 2015-10-14    21
71 2015-10-14    23
72 2015-10-14    24

I have preprocessed this data frame to get all hours on a certain day, which is variable totallist and has output:

[[1]]

[1] 1 5 9 11 14 15 17 19 22 23 24

[[2]]

[1] 3 5 9 12 13 14 16 18 20 21 22 23

[[3]]

[1] 1 6 8 11 12 13 15 16 18 21 23 24

The code I used for this is the following:

uniqueDates <- unique(dataValues$dates)
totallist <- {}
for(date in uniqueDates){
  templist <- {}
  for(i in 1:length(dataValues$dates)){
    if(dataValues$dates[i]==date){
      newlist <- append(templist,dataValues$hours[i])
    }
  }
  totallist <- append(totallist,list(templist))
}

For the example in this question (with 3 days) it works fine and the result is what I want, but if I use this on a large dataset (which has about 260 days), it takes about 6 to 7 minutes to finish.

My question is if there is an optimized way to do what I want?

Answer 1

Try any of these:

# 1
with(unique(dataValues), split(hours, dates))

# 1a - variation of last solution
with(dataValues, lapply(split(hours, dates), unique))

# 2
unstack(unique(dataValues), hours ~ dates)

# 2a - variation of last solution
lapply(unstack(dataValues, hours ~ dates), unique)

Note that if the data values are known to be unique already, as is the case in the sample data shown in the question, then unique(dataValues) in #1 and #2 could be replaced with just dataValues .

Answer 2

I believe you would be better by using the tapply function. I've created a simpler dataframe just to show what it is doing:

df <- data.frame(dates=rep(c("2015-01-02","2015-01-03","2015-01-04"),10),hours=trunc(runif(30,1,10)))

tapply(df$hours,df$dates,unique)

Output:

$`2015-01-02`
[1] 2 8 6 1 5

$`2015-01-03`
[1] 7 5 2 3

$`2015-01-04`
[1] 1 2 6 5 8 4 9