I'm trying to make a timeline like you'd make with any of the timevis, vistime, or timeline R packages, but I'm only interested in times and not dates. I don't mind putting a placeholder date in there, but it seems that all of these packages require the start and end times to include dates and include the date in the timeline.
I've been searching for ways to either not include dates in a timeline or only print the time but not the date in any of these package, but haven't been able to find anything. Does anyone have any ideas?
All of those packages use as.POSIXct under the hood, which requires objects to be Date objects and doesn't work with times only. So, if your data is about only one day, you can add the date on the clock times (using paste) and e.g. vistime will display only the time (ok, a date almost completely hidden in the corner):
dat <- data.frame(event = 1:2,
start = c("14:00", "16:00"),
end = c("15:30", "17:00"))
# add a Date
dat[,c("start", "end")] <- sapply(dat[,c("start", "end")], function(x) paste(Sys.Date(), x))
vistime(dat)
I use vistime version 0.7.0.9000 which can be obtained by executing devtools::install_github("shosaco/vistime").
If you want to represent times without any date information, you should try out the package hms. It is part of the tidyverse collection and is described as:
A simple class for storing durations or time-of-day values and displaying them in the hh:mm:ss format.
Example use:
library(hms)
hms(56, 34, 12)
#> 12:34:56
Related
I have data below for work hours which I need to compare - start and stop with date and time. I first extract the time portion of each as start and stop variables, then use the chron package to change them from factor data to something I can compare more easily.
require(chron)
eg_data3 <- data.frame(
id = c('42', '42', '42', '42', '42'),
time_in = as.factor(c('11/5/2017 13:52', '11/4/2017 14:25', '11/5/2017 15:30', '11/5/2017 17:10', '11/6/2017 18:20')),
time_out = as.factor(c('11/5/2017 13:59', '11/4/2017 14:59', '11/5/2017 16:00', '11/5/2017 17:45', '11/6/2017 18:50')))
eg_data3$start_time <- substring(strptime(eg_data3$time_in, format = "%m/%d/%Y %H:%M"),12,19)
eg_data3$end_time <- substring(strptime(eg_data3$time_out, format = "%m/%d/%Y %H:%M"),12,19)
eg_data3$end_time <- chron(times = eg_data3$end_time)
eg_data3$start_time <- chron(times = eg_data3$start_time)
Next, I generate another variable which compares the difference between stop time 1, and start time 2, IE stop time in row 1 with start time in row 2, to see the gap between them.
require(dplyr)
eg_data3 <- eg_data3 %>% group_by(id) %>% mutate(diff_outX0_inX1 = start_time - lag(end_time))
When I do this, the variable is formatted as a decimal. I cannot for the life of me get it to display as hh:mm:ss. I have tried specifying out.format as hh:mm:ss in chron, changing time_in / time_out to numeric and character before and after extraction and applying chron(times), changing the format of the diff_ variable after, etc.
What seems like a very simple question -
How do I get the result comparison (diff_outX0_inX1) variable to display as time, either hh:mm or hh:mm:ss ?? I know the formula to convert fractional days into minutes in Excel, but I'd prefer to not write out a two step function, I assume it's a simple formatting issue.
Any help is appreciated.
EDIT - got flagged as a duplicate...OK. I asked if there was a way to do this that did not involve writing a function. The answer that was linked involves a function. First comment provided a clean simple answer. I can reproduce the answer in the comment, I could not reproduce the function myself, not nearly as helpful. I also added another solution that does not requre dplyr. No where I looked online showed me something as simple as "just format the result with chron."
I am relatively new to R and I have a dataset in which I am trying to convert a date and time into a numeric value. The date and time are in the format 01JUN17:00:00:00 under a variable called pickup_datetime. I have tried using the code
cab_small_sample$pickup_datetime <- as.numeric(as.Date(cab_small_sample$pickup_datetime, format = '%d%b%y'))
but this way doesn't incorporate time, I tried to add the time format to the format section of code but still did not work. Is there an R function that will convert the data into a numeric value>
R has two main time classes: "Date" and "POSIXct". POSIXct is a datetime class and you can get all the gory details at: ? DateTimeClasses. The help page for the formats used at the time of data input, however, are at ?striptime.
cab_small_sample <- data.frame(pickup_datetime = "01JUN17:00:00:00")
cab_small_sample$pickup_dt <- as.numeric(as.POSIXct(cab_small_sample$pickup_datetime,
format = '%d%b%y:%H:%M:%S'))
cab_small_sample
# pickup_datetime pickup_dt
#1 01JUN17:00:00:00 1496300400 # seconds since 1970-01-01
I find that a "destructive reassignment of values" is generally a bad idea so as a "my (best?) practice rule" I don't assign to the same column until I'm sure I have the code working properly. (And I always leave an untouched copy somewhere safe.)
lubridate is an extremely handy package for dealing with dates. It includes a variety of functions which do the date/time parsing for you, as long as you can provide the order of components. In this case, since your data is in day-month-year-hms form, you can use the dmy_hms function.
library(lubridate)
cab_small_sample <- dplyr::tibble(
pickup_datetime = c("01JUN17:00:00:00", "01JUN17:11:00:00"))
cab_small_sample$pickup_POSIX <- dmy_hms(cab_small_sample$pickup_datetime)
I'm trying to plot the created time of twitters. I can extract the created time with the following code:
tweets <- searchTwitter('weather', n=100,lang='en')
t <- twListToDF(tweets)
s <- t[, c("created")]
The time format I get is something like: 2017-02-25 18:52:06 UTC
Trying to plot it with plot(s) provides just a list of dots. I'm not sure if it is due to the date/time format.
I want to create a barchart that each bar represents the count of tweets on an hourly period. The x axis would represent time and the y axis the number of tweets.
Any ideas?
One approach that I've used uses the 'lubridate' package, which is available on CRAN.
library(lubridate)
date(now())
hour(now())
You would replace now() with your vector s. If your s vector has the class POSIXct, I believe that this will work. There might be alternative solutions that involve the lubridate R package, too. I hope that this helps.
I have a dataset with some date time like this "{datetime:2015-07-01 09:10:00" So I wanted to remove the text, and then keep the date & the time as as.Date returns only the date. So I write this code but the only problem I have is that during the second line with strsplit, it only returns me the date time of the first line and so erase the others... I woud love to get ALL my date time not only the first. I thought about sapply maybe, but I can't make it right I have many errors or maybe with a loop for? I am novice to R so I don't really know how to do this the best way.
Could you help me please? Besides If you have another idea for the time & date format or a simple way to do it, it should be very nice of you too.
data$`Date Time`=as.character(data$`Date Time`)
data$`Date Time`=unlist(strsplit(data[,1], split='e:'))[2]
date=substr(data$`Date Time`,0,10)
date=as.Date(date)
time=substr(data$`Date Time`,12,19)
data$Date=date
data$Time=time
Thank you very much for your help!
You could use the format argument to avoid all the strsplit:
times <- as.POSIXct(data$`Date Time`, format='{datetime:%Y-%m-%d %H:%M:%S')
(The reason for the "{datetime:" in the format is because you mentioned this is the format of your strings).
This object has both date and time in it, and then you can just store it in the dataframe as a single column of type POSIXct rather than two columns of type string e.g.
data$datetime <- times
but if you do want to store the date as a Date and the time as a string (as in your example above):
data$Date <- as.Date(times)
data$Time <- strftime(times, format='%H:%M:%S')
See ?as.Date, ?as.POSIXct, ?strptime for more details on that format argument and various conversions between date and string.
I have a set of data that need to be plotted (1M rows) with R. The time column (column 1) is in hh:mm:ss.000 format. I would like to plot the graph in a time range, say from 08:05:00 to 09:00:00. How do I do it? I have searched the web and couldn't find a way to set the xlim properly.
Here's a short example of the data. Column 1 is time, Column 2, 3, 4.. will be on y axis.
07:51:19.553,10.785,0.000,0.392,1.512,1.527,1.553,1.560,2.838
08:05:00.661,-1.555,0.000,0.041,0.310,0.314,0.321,0.327,1.474
08:06:58.250,30.781,0.000,0.093,0.156,0.160,0.168,0.173,1.411
08:30:02.506,-0.002,0.000,0.052,0.120,0.123,0.132,0.137,1.361
09:05:00.997,-1.802,0.000,0.032,0.078,0.080,0.087,0.090,1.258
10:05:00.661,-1.555,0.000,0.041,0.310,0.314,0.321,0.327,1.474
Thanks in advance for your help.
You really want to use a proper time series class such as zoo or xts
Subsetting, plotting, ... then come for free. Start with the excellent zoo documentation before maybe switching to xts for even better performance and subsetting.
Now, one million rows is too many as you end up with more data than pixels -- but at least this will give you a chance to summarize your data.
Here is a quick illustration:
> options(digits.sec=3) ## important: turn on milli-sec via print()
> library(xts)
Loading required package: zoo
> X <- xts(cumsum(rnorm(100)), order.by=Sys.time()+cumsum(runif(100)/10))
> plot(X)
To change character vector to "date & time" object, POSIXlt(ct) object, function strptime() will come handy. Here's a short example how it's done.
dtm <- strptime(c("1.1.2010 11:35"), format = "%d.%m.%Y %H:%M", tz = "CET")