# random generation with arrival rate for fixed time

how i can generate 5 random numbers for 3 hrs with input of arrival rate, one way is runif() function and giving delay through sleep.sys( ) but it will only add delay, however i want to set the arrival rate please guide is there any way to do it through any distribution function where arrival rate can be set.

Hello,

The question is not very clear to me. Could you write out a few examples of input and output so we know what you mean. Especially this arrival rate is unclear

PJ

1 Like

x=runif(5,1,10)
like above example is there any function or can be created to generate unkown numbers between 1 and 10 after 2 hours for simulation purposes

Hi,

I still don't understand what it is you like. Do you like R to generate numbers "live" by spitting our a random number to to console for a 2 hour period (if so why?). Or do you want to generate data that has random timestamps in a span of 2h with random numbers?

Meanwhile, you might want to take a at look at the sample() function. Here, you can set a specific set of values (can be text, numbers, ...) you sample randomly from.

``````sample(c(1,5,20), size = 10, replace = T)
[1]  5  5  1 20 20  5 20  1  5  1
``````

PJ

1 Like

yes to generate data that has random timestamps in a span of 2h with random numbers because

1. i want to match two set of data frames (a &b, each with lat and long)after generation at specific time(not instantly generated min after 2 hrs)
2. match two data frames(a,b) of location (each with lat and long) based on the distance of 5 kms
3. unmatched locations(having distance more than 5 kms) in data frame (a) shud be matched with new periodic generation data frame (b) based on the distance of 5kms , similarly after two periodic generation of data frame (b) if all locs of data frame (a) is not matched then all those lat and long of data frame (a) must be saved and displayed.

Hi,

I'm sorry but I don't think I can help you because I still don't understand it. I present below a final idea of what I think you want, but if this is not what you are looking for, I suggest you create a reprex and hopefully someone else can figure it out. A reprex consists of the minimal code and data needed to recreate the issue/question you're having. You can find instructions how to build and share one here:

``````library(lubridate)

#Pick the starting conditions
sDate = "2020-11-20"
sTime = "14:30:00"
intervalMin = 120
nTimestamps = 10

#Generate the random timestamps
randomTimeStamps = as.POSIXct(paste(sDate, sTime)) +
seconds(runif(nTimestamps)*120*60)

sort(randomTimeStamps)
#>  [1] "2020-11-20 14:31:35 EST" "2020-11-20 14:35:11 EST"
#>  [3] "2020-11-20 14:48:33 EST" "2020-11-20 14:49:36 EST"
#>  [5] "2020-11-20 14:57:45 EST" "2020-11-20 15:08:26 EST"
#>  [7] "2020-11-20 15:23:44 EST" "2020-11-20 15:47:57 EST"
#>  [9] "2020-11-20 15:48:02 EST" "2020-11-20 16:26:12 EST"
``````

Created on 2020-11-20 by the reprex package (v0.3.0)

Good luck,
PJ

1 Like

here is the reprex code, what it is doing is matching two data frames based on the nearest distances now what requires is it should match only nearest locations within distance of 5000m/ 5kms only if not found it should run again and again (data frame df is fixed once lat and long generated in first run it should match with other data frame ref , data frame ref lat and long may be regenerated unless found within distance of 5 kms of data frame df) till match found within 5 kms. hope its understood now.

``````library(data.table)
df<-structure(list(ID=c(1:5), lat = c(runif(5,33.45,33.75))
, lon = c(runif(5,72.83,73.17)))
, col.Names = c("ID","lat", "lon"), row.names = c(NA, -10L), class = c("data.table","data.frame"))
df
#>    ID      lat      lon
#> 1:  1 33.54173 72.88869
#> 2:  2 33.62414 73.03707
#> 3:  3 33.55003 73.03129
#> 4:  4 33.73126 73.14129
#> 5:  5 33.66748 72.91560
ref<-structure(list(ID=letters[1:5], lat = c(runif(5,33.45,33.75)), lon = c(runif(5,72.83,73.17)))
, col.Names = c("ID","lat", "lon"),row.names = c(NA, -10L), class = c("data.table","data.frame"))
ref
#>    ID      lat      lon
#> 1:  a 33.67757 72.83289
#> 2:  b 33.52380 73.00690
#> 3:  c 33.63362 72.83984
#> 4:  d 33.63149 72.90808
#> 5:  e 33.47954 73.05804
#Setting to data.table format
setDT(df)
setDT(ref)
#creating a table with cartesian join
df1<-setkey(df[,c(k=1,.SD)],k)[ref[,c(k=1,.SD)],allow.cartesian=TRUE][,k:=NULL]
df1
#calculating the Euclidean distance and giving a rank in ascending order of distance
df1[,EuDist:=sqrt((lat-i.lat)^2+(lon-i.lon)^2)][,distRank:=rank(EuDist,ties="random"),by=.(ID)]
df1
#>     ID      lat      lon i.ID    i.lat    i.lon     EuDist distRank
#>  1:  1 33.54173 72.88869    a 33.67757 72.83289 0.14684973        4
#>  2:  2 33.62414 73.03707    a 33.67757 72.83289 0.21105457        5
#>  3:  3 33.55003 73.03129    a 33.67757 72.83289 0.23584679        5
#>  4:  4 33.73126 73.14129    a 33.67757 72.83289 0.31303955        4
#>  5:  5 33.66748 72.91560    a 33.67757 72.83289 0.08332276        3
#>  6:  1 33.54173 72.88869    b 33.52380 73.00690 0.11956520        3
#>  7:  2 33.62414 73.03707    b 33.52380 73.00690 0.10477654        1
#>  8:  3 33.55003 73.03129    b 33.52380 73.00690 0.03581683        1
#>  9:  4 33.73126 73.14129    b 33.52380 73.00690 0.24718507        1
#> 10:  5 33.66748 72.91560    b 33.52380 73.00690 0.17022953        4
#> 11:  1 33.54173 72.88869    c 33.63362 72.83984 0.10407033        2
#> 12:  2 33.62414 73.03707    c 33.63362 72.83984 0.19745978        4
#> 13:  3 33.55003 73.03129    c 33.63362 72.83984 0.20889675        4
#> 14:  4 33.73126 73.14129    c 33.63362 72.83984 0.31686899        5
#> 15:  5 33.66748 72.91560    c 33.63362 72.83984 0.08298131        2
#> 16:  1 33.54173 72.88869    d 33.63149 72.90808 0.09183140        1
#> 17:  2 33.62414 73.03707    d 33.63149 72.90808 0.12919979        2
#> 18:  3 33.55003 73.03129    d 33.63149 72.90808 0.14769367        3
#> 19:  4 33.73126 73.14129    d 33.63149 72.90808 0.25365657        2
#> 20:  5 33.66748 72.91560    d 33.63149 72.90808 0.03676788        1
#> 21:  1 33.54173 72.88869    e 33.47954 73.05804 0.18040661        5
#> 22:  2 33.62414 73.03707    e 33.47954 73.05804 0.14610514        3
#> 23:  3 33.55003 73.03129    e 33.47954 73.05804 0.07539468        2
#> 24:  4 33.73126 73.14129    e 33.47954 73.05804 0.26512483        3
#> 25:  5 33.66748 72.91560    e 33.47954 73.05804 0.23580952        5
#>     ID      lat      lon i.ID    i.lat    i.lon     EuDist distRank
#selecting the shortest distance
df1<-df1[distRank==1]
df1
#>    ID      lat      lon i.ID    i.lat    i.lon     EuDist distRank
#> 1:  2 33.62414 73.03707    b 33.52380 73.00690 0.10477654        1
#> 2:  3 33.55003 73.03129    b 33.52380 73.00690 0.03581683        1
#> 3:  4 33.73126 73.14129    b 33.52380 73.00690 0.24718507        1
#> 4:  1 33.54173 72.88869    d 33.63149 72.90808 0.09183140        1
#> 5:  5 33.66748 72.91560    d 33.63149 72.90808 0.03676788        1
``````

Created on 2020-11-22 by the reprex package (v0.3.0)

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