I have a dataset which contains the timing of events in football. A game exceeds 60 minutes, and I'm trying to calculate intervals. This is the data I have:
data <- c("11:14", "17:27", "25:34", "39:17", "39:59", "42:32", "50:15", "50:53", "64:22", "67:39")
My issue arises from the fact that minutes exceed 60 (they can be between 0 and 90 minutes). So essentially, I would like code that would print out the intervals between the events:
"6:13", "8:07", "13:43",..., "3:17"
Would it be better to convert the data into hours and then go from there? Just an idea I had to make it easier. I've had a look at other questions, but I couldn't find any that had been asked for R. If it has and I missed it, feel free to criticize but please link me the duplicate.
Thanks in advance!
Look into lubridate for this kind of thing.
There's probably an easier way to do it, but this works:
library(lubridate)
data <- c("11:14", "17:27", "25:34", "39:17", "39:59", "42:32", "50:15", "50:53", "64:22", "67:39")
out <- seconds_to_period(diff(as.numeric(ms(data)))
If you want the output as a formatted string instead of a period, use sprintf:
sprintf('%02d:%02d', minute(out), second(out))
This might work too
data1 <- c("11:14", "17:27", "25:34", "39:17", "39:59", "42:32", "50:15", "50:53", "64:22", "67:39")
data2 = sapply(strsplit(data1,":"), # from http://stackoverflow.com/a/5187350/7128934
function(x) {
x <- as.numeric(x)
x[1]+x[2]/60
}
)
difference = list()
for (i in 1: (length(data1) - 1)) {
difference[i] = data2[i+1] - data2[i]
}
Another base R attempt using as.difftime to specify the units explicitly:
out <- diff(sapply(strsplit(data, ":"), function(x)
Reduce(`+`, Map(as.difftime, as.numeric(x), units=c("mins","secs"))))
)
# time difference in seconds
out
#[1] 373 487 823 42 153 463 38 809 197
# formatted string
sprintf("%d:%02d", out %/% 60, out %% 60)
#[1] "6:13" "8:07" "13:43" "0:42" "2:33" "7:43" "0:38" "13:29" "3:17"
Related
Although there seem to be numerous posts concerned with this issue or related issues, I could not find a post providing a solution in R.
The problem should be easy to solve once you know how to do it: I have vectors with milliseconds. I'd like to mutate them into hours, minutes, seconds (with three decimals). For example:
x <- c(29, 300, 1000, 213451)
The expected result is this:
# "00:00:00.029" "00:00:00.300" "00:00:01.000" "00:03:33.451"
What I've tried so far is this convoluted series of operations:
hrs = x / (60 * 60 * 1000)
mins = (hrs %% 1) * 60
secs <- sprintf(((mins %% 1) * 60), fmt = '%#.3f')
paste(trunc(hrs), trunc(mins), secs, sep = ":")
[1] "0:0:0.029" "0:0:0.300" "0:0:1.000" "0:3:33.451"
The result is better than nothing but still at a remove from the expected result and, what's more, the code to get there is anything but straightforward or elegant.
What's a quicker and more elegant way to convert milliseconds into the timestamp format?
EDIT:
Alternatively, what I've tried is this:
library(chron)
times(x / (24 * 60 * 60 * 1000))
But this fails to print the decimals.
You can set "%OSn" to give the seconds truncated to n decimal places, where n is between 0 and 6.
format(as.POSIXct(x / 1000, "UTC", origin = "1970-01-01"), "%H:%M:%OS3")
# [1] "00:00:00.029" "00:00:00.300" "00:00:01.000" "00:03:33.450"
I have been using the following code to convert decimals into binaries, however I cannot seem to convert a decimal greater than 1069 into binary.
run_id = 1:1600
run_ids = as.data.frame(run_id)
run_ids$bin = 0
for (i in 1:length(x)){
run_ids[i,2] = as.numeric(paste(rev(as.numeric(intToBits(as.numeric(run_ids[i,1])))), collapse=""))
}
Unfortunately the range of numbers that I want to convert goes up to 1,600.
I have tried to use as.double, as.numeric to fix the issue however these do not work.
What am I missing?
Thanks in advance for any help you can give me.
Your code works after a minor correction.
The desired ouput can be obtained by replacing for(i in 1:length(x)){ with for(i in run_id){:
run_id <- 1:1600
run_ids <- as.data.frame(run_id)
run_ids$bin <- 0
for (i in run_id){
run_ids[i,2] <- as.numeric(paste(rev(as.numeric(intToBits(as.numeric(run_ids[i,1])))), collapse=""))
}
> tail(run_ids)
# run_id bin
#1595 1595 11000111011
#1596 1596 11000111100
#1597 1597 11000111101
#1598 1598 11000111110
#1599 1599 11000111111
#1600 1600 11001000000
I am new to R and struggling with the fact that functions are able to operate on whole vectors without having to explicitly specify this.
My goal
I have a data frame calls with multiple columns, one of which is a “date” column. Now I want to add a new column, “daytime”, that labels the daytime the particular entry’s date falls into:
> calls
call_id length date direction daytime
1 258 531 1400594572974 outgoing afternoon
2 259 0 1375555528144 unanswered evening
3 260 778 1385922648396 incoming evening
What I have done so far
I have already implemented methods that return a vector of booleans like that:
# Operates on POSIXlt timestamps
is.earlymorning <- function(date) {
hour(floor_date(date, "hour")) >= 5 & hour(floor_date(date, "hour")) < 9
}
The call is.earlymorning(“2014-05-20 16:02:52”, “2013-08-03 20:45:28”, “2013-12-01 19:30:48”) would thus return (“FALSE”, “FALSE”, “FALSE”). What I am currently struggling with is to implement a function that actually returns labels. What I would like the function to do is the following:
# rawDate is a long value of the date as ms since 1970
Daytime <- function(rawDate) {
date <- as.POSIXlt(as.numeric(rawDate) / 1000, origin = "1970-01-01")
if (is.earlymorning(date)) {
"earlymorning"
} else if (is.morning(date)) {
"morning"
} else if (is.afternoon(date)) {
"afternoon"
} else if (is.evening(date)) {
"evening"
} else if (is.earlynight(date)) {
"earlynight"
} else if (is.latenight(date)) {
"latenight"
}
}
The problem
Obviously, my above approach does not work since the if-conditions would operate on whole vectors in my example. Is there an elegant way to solve this problem? I am sure I am confusing or missing some important points, but as I mentioned I am pretty new to R.
In short, what I want to implement is a function that returns a vector of labels according to a vector of date values:
# Insert new column with daytime labels
calls$daytime <- Daytime(df$date)
# or something like that:
calls$daytime <- sapply(df$date, Daytime)
# Daytime(1400594572974, 1375555528144, 1385922648396) => (“afternoon”, “evening”, “evening”)
One approach would be to use cut rather than ifelse. I am not entirely sure how you want to label hours, but this will give you the idea. foo is your data (i.e., calls).
library(dplyr)
# Following your idea
ana <- transform(foo, date = as.POSIXlt(as.numeric(date) / 1000, origin = "1970-01-01"))
ana %>%
mutate(hour = cut(as.numeric(format(date, "%H")),
breaks = c(00,04,08,12,16,20,24),
label = c("late night", "early morning",
"morning", "afternoon",
"evening", "early night")
)
)
# call_id length date direction daytime hour
#1 258 531 2014-05-20 23:02:52 outgoing afternoon early night
#2 259 0 2013-08-04 03:45:28 unanswered evening late night
#3 260 778 2013-12-02 03:30:48 incoming evening late night
There is no need to have 6 different functions to establish which period of the day a given date is. It suffices to define a vector which matches the hour with the daytime. For instance:
Daytime<-function(rawDate) {
#change the vector according to your definition of the daytime.
#the first value corresponds to hour 0 and the last to hour 23
hours<-c(rep("latenight",5),rep("earlymorning",4),rep("morning",4),rep("afternoon",4),rep("evening",4),rep("earlynight",3))
hours[as.POSIXlt(as.numeric(rawDate) / 1000, origin = "1970-01-01")$hour+1]
}
Given Thomas' hint, I solved my problem in the following (addmittedly unelegant) way:
Daytime <- function(rawDates) {
dates <- as.POSIXlt(as.numeric(rawDates) / 1000, origin = "1970-01-01")
ifelse(is.earlymorning(dates), "earlymorning",
ifelse(is.morning(dates), "morning",
ifelse(is.afternoon(dates), "afternoon",
ifelse(is.evening(dates), "evening",
ifelse(is.earlynight(dates), "earlynight",
ifelse(is.latenight(dates), "latenight",
"N/A")
)
)
)
)
)
}
Considering a case with more labels this approach will get unmaintainable soon. Right now it serves my purposes and I will leave it at that since I must focus on analysing the data as soon as possible. But I will let you know if I had time left and found a less complicated solution! Thank you for your quick response, Thomas.
I'm still working on a question from couple of days ago and would like to receive feedback/support on how I could create a function. Your expertise is highly appreciated.
I have created the following:
##### 1)
> raceIDs
[1] "GER" "SUI" "NZ2" "US1" "US2" "POR" "FRA" "AUS" "NZ1" "SWE"
##### 2)
#For each "raceIDs", there is a csv file which I have made a loop to read and created a list of data frames (assigned to the symbol "boatList")
#For example, if I select "NZ1" the output is:
> head(boatList[[9]]) #Only selected the first six lines as there is more than 30000 rows
Boat Date Secs LocalTime SOG
1 NZ1 01:09:2013 38150.0 10:35:49.997 22.17
2 NZ1 01:09:2013 38150.2 10:35:50.197 22.19
3 NZ1 01:09:2013 38150.4 10:35:50.397 22.02
4 NZ1 01:09:2013 38150.6 10:35:50.597 21.90
5 NZ1 01:09:2013 38150.8 10:35:50.797 21.84
6 NZ1 01:09:2013 38151.0 10:35:50.997 21.95
##### 3)
# A matrix showing the race times for each raceIDs
> raceTimes
start finish
GER "11:10:02" "11:35:05"
SUI "11:10:02" "11:35:22"
NZ2 "11:10:02" "11:34:12"
US1 "11:10:01" "11:33:29"
US2 "11:10:01" "11:36:05"
POR "11:10:02" "11:34:31"
FRA "11:10:02" "11:34:45"
AUS "11:10:03" "11:36:48"
NZ1 "11:10:01" "11:35:16"
SWE "11:10:03" "11:35:08"
What I need to do is I need to calculate the average speed (SOG) of a boat "while it was racing" (between start and finish times) by creating a function called meanRaceSpeed and having three arguments:
What I have tried so far is to create a function with 3 arguments (with a bit of help from experts here):
meanRaceSpeed <- function(raceIDs, boatList, raceTimes)
{
#Probably need to compare times, and thought it might be useful to convert character values into `DateTime` values but not to sure how to use it
#DateTime <- as.POSIXct(paste(boatList$Date, boatList$Time), format="%Y%m%d %H%M%S")
#To get the times for each boat
start_time <- raceTimes$start[rownames(raceTimes) = raceIDs]
finish_time <- raceTimes$finish[rownames(raceTimes) = raceIDs]
start_LocalTime <- min(grep(start_time, boatList$LocalTime))
finish_LocalTime <- max(grep(finish_time, boatList$LocalTime))
#which `SOG`s contain all the `LocalTimes` between start and finish
#take their `mean`
mean(boatList$SOG[start_LocalTime : finish_LocalTime])
}
### Obviously, my code does not work :( and I don't know where.
So basically, I need to create a function with three arguments and the expected result is:
#e.g For NZ1
> meanRaceSpeed("NZ1", boatList, raceTimes)
[1] 18.32 #Mean speed for NZ1 between 11:10:01 - 11:35:16
#e.g for US1
> meanRaceSpeed("US1", boatList, raceTimes)
[1] 17.23 #Mean speed for US1 between 11:10:01 - 11:33:29
Any helps where I could have gone wrong? Highly appreciate your help please.
I'm going to give some general advice for R, but I will also help you with your specific question. Whenever I have a problem in R, I usually find that it helps to make things more explicit.
If the function isn't working with these methods (is that a data frame or a matrix in your function?) then you should try another method. If those table manipulation methods aren't working, try a different one. How?
Here's a few different things you can do to test your function, and a few suggestions that may move you along a bit. (I don't want to fix the whole thing for you, since it's your homework, but rather get you on your way.)
1) Why not try using a loop instead of brackets?
start_time <- raceTimes$start[rownames(raceTimes) = raceIDs]
Make that into a for loop. It's not too hard to do.
2) Debug your functions. There are a lot of tools to do this built into R, and in packages you can add. Since you, likely, don't have time for that with your homework. I'd suggest doing this. Take apart the function and apply each part of it with a variable you want. Are they of the right length? Are they the right data type? Are they getting the right answer before you put them all together? Make sure of that.
3) If all else fails, don't be afraid if the function and code is not elegant. R is not always an elegant language. (Actually, it's rarely an elegant language.) Especially when you're a beginner, your code will likely be ugly. Just make sure it works.
Since I, already, had experience with your data, I sat to make a complete example.
First, data that look like yours:
raceIDs <- c("GER", "SUI", "NZ2", "US1", "US2", "POR", "FRA", "AUS", "NZ1", "SWE")
raceTimes <- as.matrix(read.table(text = ' start finish
GER "11:10:02" "11:35:05"
SUI "11:10:02" "11:35:22"
NZ2 "11:10:02" "11:34:12"
US1 "11:10:01" "11:33:29"
US2 "11:10:01" "11:36:05"
POR "11:10:02" "11:34:31"
FRA "11:10:02" "11:34:45"
AUS "11:10:03" "11:36:48"
NZ1 "11:10:01" "11:35:16"
SWE "11:10:03" "11:35:08"', header = T))
#turn matrix to data.frame or, else, `$` won't work
raceTimes <- as.data.frame(raceTimes, stringsAsFactors = F)
blDF <- data.frame(Boat = rep(raceIDs, 3),
LocalTime = c(raceTimes$start, rep("11:20:25", length(raceIDs)), raceTimes$finish),
SOG = runif(3 * length(raceIDs), 15, 25), stringsAsFactors = F)
boatList <- split(blDF, blDF$Boat)
#remove `names` to create them from scratch
names(boatList) <- NULL
Then:
#create `names` by searching each element of
#`boatList` of what `boat` it contains
names(boatList) <- unlist(lapply(boatList, function(x) unique(x$Boat)))
#the function
meanRaceSpeed <- function(ID, boatList, raceTimes)
{ #named the first argument `ID` instead of `raceIDs`
start_time <- raceTimes$start[rownames(raceTimes) == ID]
finish_time <- raceTimes$finish[rownames(raceTimes) == ID]
start_LocalTime <- min(grep(start_time, boatList[[ID]]$LocalTime))
finish_LocalTime <- max(grep(finish_time, boatList[[ID]]$LocalTime))
mean(boatList[[ID]]$SOG[start_LocalTime : finish_LocalTime])
}
Test:
meanRaceSpeed("US1", boatList, raceTimes)
#[1] 19.7063
meanRaceSpeed("NZ1", boatList, raceTimes)
#[1] 21.74729
mean(boatList$NZ1$SOG) #to test function
#[1] 21.74729
mean(boatList$US1$SOG) #to test function
#[1] 19.7063
I want to convert my geographic coordinates from degrees to decimals, my data are as follows:
lat long
105252 30°25.264 9°01.331
105253 30°39.237 8°10.811
105255 31°37.760 8°06.040
105258 31°41.190 8°06.557
105259 31°41.229 8°06.622
105260 31°38.891 8°06.281
I have this code but I can not see why it is does not work:
convert<-function(coord){
tmp1=strsplit(coord,"°")
tmp2=strsplit(tmp1[[1]][2],"\\.")
dec=c(as.numeric(tmp1[[1]][1]),as.numeric(tmp2[[1]]))
return(dec[1]+dec[2]/60+dec[3]/3600)
}
don_convert=don1
for(i in 1:nrow(don1)){don_convert[i,2]=convert(as.character(don1[i,2])); don_convert[i,3]=convert(as.character(don1[i,3]))}
The convert function works but the code where I am asking the loop to do the job for me does not work.
Any suggestion is apperciated.
Use the measurements package from CRAN which has a unit conversion function already so you don't need to make your own:
x = read.table(text = "
lat long
105252 30°25.264 9°01.331
105253 30°39.237 8°10.811
105255 31°37.760 8°06.040
105258 31°41.190 8°06.557
105259 31°41.229 8°06.622
105260 31°38.891 8°06.281",
header = TRUE, stringsAsFactors = FALSE)
Once your data.frame is set up then:
# change the degree symbol to a space
x$lat = gsub('°', ' ', x$lat)
x$long = gsub('°', ' ', x$long)
# convert from decimal minutes to decimal degrees
x$lat = measurements::conv_unit(x$lat, from = 'deg_dec_min', to = 'dec_deg')
x$long = measurements::conv_unit(x$long, from = 'deg_dec_min', to = 'dec_deg')
Resulting in the end product:
lat long
105252 30.4210666666667 9.02218333333333
105253 30.65395 8.18018333333333
105255 31.6293333333333 8.10066666666667
105258 31.6865 8.10928333333333
105259 31.68715 8.11036666666667
105260 31.6481833333333 8.10468333333333
Try using the char2dms function in the sp library. It has other functions that will additionally do decimal conversion.
library("sp")
?char2dms
A bit of vectorization and matrix manipulation will make your function much simpler:
x <- read.table(text="
lat long
105252 30°25.264 9°01.331
105253 30°39.237 8°10.811
105255 31°37.760 8°06.040
105258 31°41.190 8°06.557
105259 31°41.229 8°06.622
105260 31°38.891 8°06.281",
header=TRUE, stringsAsFactors=FALSE)
x
The function itself makes use of:
strsplit() with the regex pattern "[°\\.]" - this does the string split in one step
sapply to loop over the vector
Try this:
convert<-function(x){
z <- sapply((strsplit(x, "[°\\.]")), as.numeric)
z[1, ] + z[2, ]/60 + z[3, ]/3600
}
Try it:
convert(x$long)
[1] 9.108611 8.391944 8.111111 8.254722 8.272778 8.178056
Disclaimer: I didn't check your math. Use at your own discretion.
Thanks for answers by #Gord Stephen and #CephBirk. Sure helped me out.
I thought I'd just mention that I also found that measurements::conv_unit doesn't deal with "E/W" "N/S" entries, it requires positive/negative degrees.
My coordinates comes as character strings "1 1 1W" and needs to first be converted to "-1 1 1".
I thought I'd share my solution for that.
df <- c("1 1 1E", "1 1 1W", "2 2 2N","2 2 2S")
measurements::conv_unit(df, from = 'deg_min_sec', to = 'dec_deg')
[1] "1.01694444444444" NA NA NA
Warning message:
In split(as.numeric(unlist(strsplit(x, " "))) * c(3600, 60, 1), :
NAs introduced by coercion
ewns <- ifelse( str_extract(df,"\\(?[EWNS,.]+\\)?") %in% c("E","N"),"+","-")
dms <- str_sub(df,1,str_length(df)-1)
df2 <- paste0(ewns,dms)
df_dec <- measurements::conv_unit(df2,
from = 'deg_min_sec',
to = 'dec_deg'))
df_dec
[1] "1.01694444444444" "-1.01694444444444" "2.03388888888889" "-2.03388888888889"
as.numeric(df_dec)
[1] 1.016944 -1.016944 2.033889 -2.033889
Have a look at the command degree in the package OSMscale.
As Jim Lewis commented before it seems your are using floating point minutes. Then you only concatenate two elements on
dec=c(as.numeric(tmp1[[1]][1]),as.numeric(tmp2[[1]]))
Having degrees, minutes and seconds in the form 43°21'8.02 which as.character() returns "43°21'8.02\"", I updated your function to
convert<-function(coord){
tmp1=strsplit(coord,"°")
tmp2=strsplit(tmp1[[1]][2],"'")
tmp3=strsplit(tmp2[[1]][2],"\"")
dec=c(as.numeric(tmp1[[1]][1]),as.numeric(tmp2[[1]][1]),as.numeric(tmp3[[1]]))
c<-abs(dec[1])+dec[2]/60+dec[3]/3600
c<-ifelse(dec[1]<0,-c,c)
return(c)
}
adding the alternative for negative coordinates, and works great for me . I still don't get why char2dms function in the sp library didn't work for me.
Thanks
Another less elegant option using substring instead of strsplit. This will only work if all your positions have the same number of digits. For negative co-ordinates just multiply by -1 for the correct decimal degree.
x$LatDD<-(as.numeric(substring(x$lat, 1,2))
+ (as.numeric(substring(x$lat, 4,9))/60))
x$LongDD<-(as.numeric(substring(x$long, 1,1))
+ (as.numeric(substring(x$long, 3,8))/60))