I have participant data during an exercise test, which includes participant ID, the condition (either Environmental or Control) and the total time taken to complete the test. A small example of my data:
RawData <- data.frame(
ParticipantID = c (1:6),
Condition = c("Control","Experimental","Experimental","Control","Experimental","Control"),
Time = c("04:34:22","02:48:47","04:22:06","02:57:11","02:07:11","05:34:22"))
I then used the lubridate package so I have time in hms via:
RawData <- RawData %>%
mutate(TotalTime = hms::as_hms(Time))
Now I wish to create a new column, that bins each RawData$TotalTime result into a category including: Sub2, Sub230, Sub3, Sub330, Sub4, Sub430, Sub5, Sub530 and Sub6. I could probably do this via a long case_when statement but is there an easy way to do this in lubridate given I am after 30 minute intervals?
My desired output would be:
RawData <- data.frame(
ParticipantID = c (1:6),
Condition = c("Control","Experimental","Experimental","Control","Experimental","Control"),
Time = c("04:34:22","02:48:47","04:22:06","02:57:11","02:07:11","05:34:22"),
Category = c("Sub5","Sub3","Sub430","Sub3","Sub230","Sub6"))
Thank you!
You can use ceiling_date function with units as "30 mins".
library(dplyr)
library(lubridate)
RawData %>%
mutate(TotalTime = as.POSIXct(Time, format = '%T'),
Category = format(ceiling_date(TotalTime, '30 mins'), "%H%M")) %>%
select(-TotalTime)
# ParticipantID Condition Time Category
#1 1 Control 04:34:22 0500
#2 2 Experimental 02:48:47 0300
#3 3 Experimental 04:22:06 0430
#4 4 Control 02:57:11 0300
#5 5 Experimental 02:07:11 0230
#6 6 Control 05:34:22 0600
Related
I have made measurements of temperature in a high time resolution of 10 minutes on different urban Tree species, whose reactions should be compared. Therefore I am researching especially periods of heat. The Task that I fail to do on my Dataset is to choose complete days from a maximum value. E.G. Days where there is one measurement above 30 °C should be subsetted from my Dataframe completely.
Below you find a reproducible example that should illustrate my problem:
In my Measurings Dataframe I have calculated a column indicating wether the individual Measurement is above or below 30°C. I wanted to use that column to tell other functions wether they should pick a day or not to produce a New Dataframe. When anytime of the day the value is above 30 ° C i want to include it by Date from 00:00 to 23:59 in that New Dataframe for further analyses.
start <- as.POSIXct("2018-05-18 00:00", tz = "CET")
tseq <- seq(from = start, length.out = 1000, by = "hours")
Measurings <- data.frame(
Time = tseq,
Temp = sample(20:35,1000, replace = TRUE),
Variable1 = sample(1:200,1000, replace = TRUE),
Variable2 = sample(300:800,1000, replace = TRUE)
)
Measurings$heat30 <- ifelse(Measurings$Temp > 30,"heat", "normal")
Measurings$otheroption30 <- ifelse(Measurings$Temp > 30,"1", "0")
The example is yielding a Dataframe analog to the structure of my Data:
head(Measurings)
Time Temp Variable1 Variable2 heat30 otheroption30
1 2018-05-18 00:00:00 28 56 377 normal 0
2 2018-05-18 01:00:00 23 65 408 normal 0
3 2018-05-18 02:00:00 29 78 324 normal 0
4 2018-05-18 03:00:00 24 157 432 normal 0
5 2018-05-18 04:00:00 32 129 794 heat 1
6 2018-05-18 05:00:00 25 27 574 normal 0
So how do I subset to get a New Dataframe where all the days are taken where at least one entry is indicated as "heat"?
I know that for example dplyr:filter could filter the individual entries (row 5 in the head of the example). But how could I tell to take all the day 2018-05-18?
I am quite new to analyzing Data with R so I would appreciate any suggestions on a working solution to my problem. dplyris what I have been using for quite some tasks, but I am open to whatever works.
Thanks a lot, Konrad
Create variable which specify which day (droping hours, minutes etc.). Iterate over unique dates and take only such subsets which in heat30 contains "heat" at least once:
Measurings <- Measurings %>% mutate(Time2 = format(Time, "%Y-%m-%d"))
res <- NULL
newdf <- lapply(unique(Measurings$Time2), function(x){
ss <- Measurings %>% filter(Time2 == x) %>% select(heat30) %>% pull(heat30) # take heat30 vector
rr <- Measurings %>% filter(Time2 == x) # select date x
# check if heat30 vector contains heat value at least once, if so bind that subset
if(any(ss == "heat")){
res <- rbind(res, rr)
}
return(res)
}) %>% bind_rows()
Below is one possible solution using the dataset provided in the question. Please note that this is not a great example as all days will probably include at least one observation marked as over 30 °C (i.e. there will be no days to filter out in this dataset but the code should do the job with the actual one).
# import packages
library(dplyr)
library(stringr)
# break the time stamp into Day and Hour
time_df <- as_data_frame(str_split(Measurings$Time, " ", simplify = T))
# name the columns
names(time_df) <- c("Day", "Hour")
# create a new measurement data frame with separate Day and Hour columns
new_measurings_df <- bind_cols(time_df, Measurings[-1])
# form the new data frame by filtering the days marked as heat
new_df <- new_measurings_df %>%
filter(Day %in% new_measurings_df$Day[new_measurings_df$heat30 == "heat"])
To be more precise, you are creating a random sample of 1000 observations varying between 20 to 35 for temperature across 40 days. As a result, it is very likely that every single day will have at least one observation marked as over 30 °C in your example. Additionally, it is always a good practice to set seed to ensure reproducibility.
I am trying to create two new column of people's ages and their age groups(5 year intervals) given their date of birth from a data frame.
The current data frame For example:
Person Date of Birth
A 1/2/2000
B 3/2/1998
C 4/5/2008
The expected outcome is :
Person Date of Birth Age Age-Group
A 1/2/2000 18 15-20
B 3/2/1990 28 25-30
C 4/5/2008 10 5-10
What is the best way to do this on the most efficient way for a large data set? Thanks
Something like this? BTW, I slightly adjusted the age groups you used in your example since using 5-10 and 15-20 would mean you will use an age group of 11-14 as well, which would seem weird to me.
df <- read.table(text = "
Person DateofBirth
A 1/2/2000
B 3/2/1998
C 4/5/2008", header = T)
library(lubridate)
df$age <- interval(as.Date(df$DateofBirth, "%d/%m/%Y"), Sys.Date()) %/% years(1)
df$agegroup <- cut(df$age, seq(5,30,5), c("5-10", "11-15", "16-20", "21-25", "25-30"))
df
Person DateofBirth age agegroup
1 A 1/2/2000 18 16-20
2 B 3/2/1998 20 16-20
3 C 4/5/2008 10 5-10
If you many more agegroups, you could as well consider to generalize the last cut argument like this:
df1 <- data.frame(age = 1:100)
df1$agegroup <- cut(df1$age, seq(0,100,5), paste0(seq(1,96, 5), "-", seq(5,100,5)))
I have two columns of dates. Two example dates are:
Date1= "2015-07-17"
Date2="2015-07-25"
I am trying to count the number of Saturdays and Sundays between the two dates each of which are in their own column (5 & 7 in this example code). I need to repeat this process for each row of my dataframe. The end results will be one column that represents the number of Saturdays and Sundays within the date range defined by two date columns.
I can get the code to work for one row:
sum(weekdays(seq(Date1[1,5],Date2[1,7],"days")) %in% c("Saturday",'Sunday')*1))
The answer to this will be 3. But, if I take out the "1" in the row position of date1 and date2 I get this error:
Error in seq.Date(Date1[, 5], Date2[, 7], "days") :
'from' must be of length 1
How do I go line by line and have one vector that lists the number of Saturdays and Sundays between the two dates in column 5 and 7 without using a loop? Another issue is that I have 2 million rows and am looking for something with a little more speed than a loop.
Thank you!!
map2* functions from the purrr package will be a good way to go. They take two vector inputs (eg two date columns) and apply a function in parallel. They're pretty fast too (eg previous post)!
Here's an example. Note that the _int requests an integer vector back.
library(purrr)
# Example data
d <- data.frame(
Date1 = as.Date(c("2015-07-17", "2015-07-28", "2015-08-15")),
Date2 = as.Date(c("2015-07-25", "2015-08-14", "2015-08-20"))
)
# Wrapper function to compute number of weekend days between dates
n_weekend_days <- function(date_1, date_2) {
sum(weekdays(seq(date_1, date_2, "days")) %in% c("Saturday",'Sunday'))
}
# Iterate row wise
map2_int(d$Date1, d$Date2, n_weekend_days)
#> [1] 3 4 2
If you want to add the results back to your original data frame, mutate() from the dplyr package can help:
library(dplyr)
d <- mutate(d, end_days = map2_int(Date1, Date2, n_weekend_days))
d
#> Date1 Date2 end_days
#> 1 2015-07-17 2015-07-25 3
#> 2 2015-07-28 2015-08-14 4
#> 3 2015-08-15 2015-08-20 2
Here is a solution that uses dplyr to clean things up. It's not too difficult to use with to assign the columns in the dataframe directly.
Essentially, use a reference date, calculate the number of full weeks (by floor or ceiling). Then take the difference between the two. The code does not include cases in which the start date or end data fall on Saturday or Sunday.
# weekdays(as.Date(0,"1970-01-01")) -> "Friday"
require(dplyr)
startDate = as.Date(0,"1970-01-01") # this is a friday
df <- data.frame(start = "2015-07-17", end = "2015-07-25")
df$start <- as.Date(df$start,"", format = "%Y-%m-%d", origin="1970-01-01")
df$end <- as.Date(df$end, format = "%Y-%m-%d","1970-01-01")
# you can use with to define the columns directly instead of %>%
df <- df %>%
mutate(originDate = startDate) %>%
mutate(startDayDiff = as.numeric(start-originDate), endDayDiff = as.numeric(end-originDate)) %>%
mutate(startWeekDiff = floor(startDayDiff/7),endWeekDiff = floor(endDayDiff/7)) %>%
mutate(NumSatsStart = startWeekDiff + ifelse(startDayDiff %% 7>=1,1,0),
NumSunsStart = startWeekDiff + ifelse(startDayDiff %% 7>=2,1,0),
NumSatsEnd = endWeekDiff + ifelse(endDayDiff %% 7 >= 1,1,0),
NumSunsEnd = endWeekDiff + ifelse(endDayDiff %% 7 >= 2,1,0)
) %>%
mutate(NumSats = NumSatsEnd - NumSatsStart, NumSuns = NumSunsEnd - NumSunsStart)
Dates are number of days since 1970-01-01, a Thursday.
So the following is the number of Saturdays or Sundays since that date
f <- function(d) {d <- as.numeric(d); r <- d %% 7; 2*(d %/% 7) + (r>=2) + (r>=3)}
For the number of Saturdays or Sundays between two dates, just subtract, after decrementing the start date to have an inclusive count.
g <- function(d1, d2) f(d2) - f(d1-1)
These are all vectorized functions so you can just call directly on the columns.
# Example data, as in Simon Jackson's answer
d <- data.frame(
Date1 = as.Date(c("2015-07-17", "2015-07-28", "2015-08-15")),
Date2 = as.Date(c("2015-07-25", "2015-08-14", "2015-08-20"))
)
As follows
within(d, end_days<-g(Date1,Date2))
# Date1 Date2 end_days
# 1 2015-07-17 2015-07-25 3
# 2 2015-07-28 2015-08-14 4
# 3 2015-08-15 2015-08-20 2
I have a large file of time-series data, which looks as follows. The dataset covers years, in increments of 15 minutes. A small subset looks like:
uniqueid time
a 2014-04-30 23:30:00
a 2014-04-30 23:45:00
a 2014-05-01 00:00:00
a 2014-05-01 00:15:00
a 2014-05-12 13:45:00
a 2014-05-12 14:00:00
b 2014-05-12 13:45:00
b 2014-05-12 14:00:00
b 2014-05-12 14:30:00
To reproduce above:
time<-c("2014-04-30 23:30:00","2014-04-30 23:45:00","2014-05-01 00:00:00","2014-05-01 00:15:00",
"2014-05-12 13:45:00","2014-05-12 14:00:00","2014-05-12 13:45:00","2014-05-12 14:00:00",
"2014-05-12 14:30:00")
uniqueid<-c("a","a","a","a","a","a","b","b","b")
mydf<-data.frame(uniqueid,time)
My goal is to count the number of rows per unique id, per consecutive timeflow. A consecutive timespan is when a unique id is stamped for each 15 minutes in a row (such as id A, which is stamped from 30.04.14 23.30 hrs until 01.05.14 00.15 hrs - hence 4 rows), yet when this flow of 15-minute iterations is disrupted (after 01.05.14 00:15, it is not stamped at 01.05.14 00:30 hence it is disrupted), it should count the next timestamp as start of a new consecutive timeflow and again calculate the number of rows until this flow is disrupted again. Time is POSIX.
As you can see in above example; a consecutive timeflow may cover different days, different months, or different years. I have many unique ids (and as said, a very large file), so I'm looking for a way that my computer can handle (loops probably wouldn't work).
I am looking for output something like:
uniqueid flow number_rows
a 1 4
a 2 2
b 3 2
b 4 1
I have looked into some time packages (such as lubridate), but given my limited R knowledge, I don't even know where to begin.
I hope all is clear - if not, I'd be happy to try to clarify it further. Thank you very much in advance!
Another way to do this with data.table also using a time difference would be to make use of the data.table internal values for group number and number of rows in each group:
library(data.table)
res<-setDT(mydf)[, list(number_rows=.N,flow=.GRP),
by=.(uniqueid,cumsum(as.numeric(difftime(time,shift(time,1L,type="lag",fill=0))) - 15))][,cumsum:=NULL]
print(res)
uniqueid number_rows flow
1: a 4 1
2: a 2 2
3: b 2 3
4: b 1 4
Also since the sample data you posted didn't align with the subset you posted, I have included my data below:
Data
time<-as.POSIXct(c("2014-04-30 23:30:00","2014-04-30 23:45:00","2014-05-01 00:00:00","2014-05-01 00:15:00",
"2014-05-12 13:45:00","2014-05-12 14:00:00","2014-05-12 13:45:00","2014-05-12 14:00:00",
"2014-05-12 14:30:00"))
uniqueid<-c("a","a","a","a","a","a","b","b","b")
mydf<-data.frame(uniqueid,time)
You can groupby the uniqueid and the cumulative sum of the difference of time between rows which is not equal to 15 min and that gives the flow id and then a count of rows should give you what you need:
A justification of the logic is whenever the time difference is not equal to 15 within each uniqueid, a new flow process should be generated so we label it as TRUE and combine that with the cumsum, it becomes a new flow id with the following consecutive rows:
library(dplyr)
mydf$time <- as.POSIXct(mydf$time, "%Y-%m-%d %H:%M:%S")
# convert the time column to POSIXct class so that we can apply the diff function correctly
mydf %>% group_by(uniqueid, flow = 1 + cumsum(c(F, diff(time) != 15))) %>%
summarize(num_rows = n())
# Source: local data frame [4 x 3]
# Groups: uniqueid [?]
#
# uniqueid flow num_rows
# <fctr> <dbl> <int>
# 1 a 1 4
# 2 a 2 2
# 3 b 3 2
# 4 b 4 1
Base R is pretty fast. Using crude benchmarking, I found it finished in half the time of DT, and I got tired of waiting for dplyr.
# estimated size of data, years x days x hours x 15mins x uniqueids
5*365*24*4*1000 # = approx 180M
# make data with posixct and characters of 180M rows, mydf is approx 2.5GB in memory
time<-rep(as.POSIXct(c("2014-04-30 23:30:00","2014-04-30 23:45:00","2014-05-01 00:00:00","2014-05-01 00:15:00",
"2014-05-12 13:45:00","2014-05-12 14:00:00","2014-05-12 13:45:00","2014-05-12 14:00:00",
"2014-05-12 14:30:00")),times = 20000000)
uniqueid<-rep(as.character(c("a","a","a","a","a","a","b","b","b")),times = 20000000)
mydf<-data.frame(uniqueid,time = time)
rm(time,uniqueid);gc()
Base R:
# assumes that uniqueid's are in groups and in order, and there won't be a followed by b that have the 15 minute "flow"
starttime <- Sys.time()
# find failed flows
mydf$diff <- c(0,diff(mydf$time))
mydf$flowstop <- mydf$diff != 15
# give each flow an id
mydf$flowid <- cumsum(mydf$flowstop)
# clean up vars
mydf$time <- mydf$diff <- mydf$flowstop <- NULL
# find flow length
mydfrle <- rle(mydf$flowid)
# get uniqueid/flowid pairs (unique() is too slow)
mydf <- mydf[!duplicated(mydf$flowid), ]
# append rle and remove separate var
mydf$number_rows <- mydfrle$lengths
rm(mydfrle)
print(Sys.time()-starttime)
# Time difference of 30.39437 secs
data.table:
library(data.table)
starttime <- Sys.time()
res<-setDT(mydf)[, list(number_rows=.N,flow=.GRP),
by=.(uniqueid,cumsum(as.numeric(difftime(time,shift(time,1L,type="lag",fill=0))) - 15))][,cumsum:=NULL]
print(Sys.time()-starttime)
# Time difference of 57.08156 secs
dplyr:
library(dplyr)
# convert the time column to POSIXct class so that we can apply the diff function correctly
starttime <- Sys.time()
mydf %>% group_by(uniqueid, flow = 1 + cumsum(c(F, diff(time) != 15))) %>%
summarize(num_rows = n())
print(Sys.time()-starttime)
# too long, did not finish after a few minutes
I think the assumption of uniqueid's and times being in order is huge, and the other solutions might be able to take advantage of that better. order() is easy enough to do.
I'm not sure about the impact of memory, or of the impact of different data sets that aren't so simple. It should be easy enough to break it into chunks and process if memory is an issue. It takes more code in Base R for sure.
Having both ordered "id" and "time" columns, we could build a single group to operate on by creating a logical vector of indices wherever either "id" changes or "time" is > 15 minutes.
With:
id = as.character(mydf$uniqueid)
tm = mydf$time
find where "id":
id_gr = c(TRUE, id[-1] != id[-length(id)])
and "time":
tm_gr = c(0, difftime(tm[-1], tm[-length(tm)], unit = "mins")) > 15
change and combine them in:
gr = id_gr | tm_gr
which shows wherever either "id" changed or "time" > 15.
And to get the result:
tab = tabulate(cumsum(gr)) ## basically, the only operation per group -- 'n by group'
data.frame(id = id[gr], flow = seq_along(tab), n = tab)
# id flow n
#1 a 1 4
#2 a 2 2
#3 b 3 2
#4 b 4 1
On a larger scale:
set.seed(1821); nid = 1e4
dat = replicate(nid, as.POSIXct("2016-07-07 12:00:00 EEST") +
cumsum(sample(c(1, 5, 10, 15, 20, 30, 45, 60, 90, 120, 150, 200, 250, 300), sample(5e2:1e3, 1), TRUE)*60),
simplify = FALSE)
names(dat) = make.unique(rep_len(letters, nid))
dat = data.frame(id = rep(names(dat), lengths(dat)), time = do.call(c, dat))
system.time({
id = as.character(dat$id); tm = dat$time
id_gr = c(TRUE, id[-1] != id[-length(id)])
tm_gr = c(0, difftime(tm[-1], tm[-length(tm)], unit = "mins")) > 15
gr = id_gr | tm_gr
tab = tabulate(cumsum(gr))
ans1 = data.frame(id = id[gr], flow = seq_along(tab), n = tab)
})
# user system elapsed
# 1.44 0.19 1.66
For comparison, included MikeyMike's answer:
library(data.table)
dat2 = copy(dat)
system.time({
ans2 = setDT(dat2)[, list(flow = .GRP, n = .N),
by = .(id, cumsum(as.numeric(difftime(time,
shift(time, 1L, type = "lag", fill = 0),
unit = "mins")) > 15))][, cumsum := NULL]
})
# user system elapsed
# 3.95 0.22 4.26
identical(as.data.table(ans1), ans2)
#[1] TRUE
I have the following data frame representing user subscriptions:
User StartDate EndDate
1 2015-09-03 2015-10-17
2 2015-10-27 2015-12-25
...
How can I transform it into a time series that gives me the count of active monthly subscriptions over time (assuming it is active in the month if at least for one day in that month). Something like this (based on the example above, assuming only 2 records):
Month Count
2015-08 0
2015-09 1
2015-10 2
2015-11 1
2015-12 1
2016-01 0
Rem: I took some arbitrary start and end dates for the time series, to make the example clear.
Prepare the data and make sure that the date columns are actually stored as dates:
data <- read.table(text = "User StartDate EndDate
1 2015-09-03 2015-10-17
2 2015-10-27 2015-12-25", header = TRUE)
data$StartDate <- as.Date(StartDate)
data$EndDate <- as.Date(EndDate))
This function returns a vector with all month that are within a subscription:
library(lubridate)
subscr_month <- function(start, end) {
start <- floor_date(start, "month")
seq <- seq(start, end, by = "1 month")
months <- format(seq, format = "%Y-%m")
return(months)
}
It uses the function floor_date() from the lubridate package. It is necessary to round of the start date, because otherwise the last month might be missing. For example, for user 2, if you add two month to the start date, you end up on 2015-12-27, which is after the end date, such that no date from December will be included in seq. The last line converts the Dates to character that only include year and month.
Now, you can apply this function to each start and end date from your data using mapply(). Afterwards, table() creates a table of counts of all dates in the resulting list:
all_month <- mapply(subscr_month, data$StartDate, data$EndDate, SIMPLIFY = FALSE)
table(unlist(all_month))
## 2015-09 2015-10 2015-11 2015-12
## 1 2 1 1
You can also convert the table to a data frame:
as.data.frame(table(unlist(all_month)))
## Var1 Freq
## 1 2015-09 1
## 2 2015-10 2
## 3 2015-11 1
## 4 2015-12 1
Your example output also includes the counts for months that do not appear in the data set. If you want to have this, you can convert the vector of months to a factor and set the levels to all the months you want to include:
month_list <- format(seq(as.Date("2015-08-01"), as.Date("2016-01-01"), by = "1 month"), format = "%Y-%m")
all_month_factor <- factor(unlist(all_month), levels = month_list)
table(all_month_factor)
## all_month_factor
## 2015-08 2015-09 2015-10 2015-11 2015-12 2016-01
## 0 1 2 1 1 0
read the data frame mentioned.
df = structure(list(StartDate = structure(c(16681, 16735), class = "Date"),
EndDate = structure(c(16735, 16794), class = "Date")), class = "data.frame", .Names = c("StartDate",
"EndDate"), row.names = c(NA, -2L))
Could make good use of do in dplyr package and seq
df %>%
rowwise() %>% do({
w <- seq(.$StartDate,.$EndDate,by = "15 days") #for month difference less than 1 complete month
m <- format(w,"%Y-%m") %>% unique
data.frame(Month = m)
}) %>%
group_by(Month) %>%
summarise(Count = length(Month))