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R tick data : merging date and time into a single object
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Hi Have 2 columns in a dataframe. Column 1 has Dates like 2017-01-01 and column 2 has time stamp like 1:00 PM.
I need to create another column that combines these 2 information and gives me the 2017-01-01 13:00:00
Use as.POSIXct to convert from character to date format.
df$date.time <- as.POSIXct(paste(df$date, df$time), format = "%Y-%m-%d %I:%M %p")
EDIT:
To provide some further context... You paste the date and the time column together to get the string 2017-001-01 1:00 PM.
You then input the format of the string as a POSIXct argument using format =. You can see the relationship between symbols and their meaning here.
Reproducible example
library(lubridate)
A <- data.frame(X1 = ymd("2017-01-01"),
X2 = "1:00 PM", stringsAsFactors=F)
# X1 X2
# 1 2017-01-01 1:00 PM
solution
library(dplyr)
library(lubridate)
temp <- A %>%
mutate(X3 = ymd_hm(paste(X1, X2)))
output
X1 X2 X3
<date> <chr> <dttm>
1 2017-01-01 1:00 PM 2017-01-01 13:00:00
multi-row input
B <- data.frame(X1 = ymd("2017-01-01", "2016-01-01"),
X2 = c("1:00 PM", "2:00 AM"), stringsAsFactors=F)
temp <- B %>%
mutate(X3 = ymd_hm(paste(X1, X2)))
# X1 X2 X3
# <date> <chr> <dttm>
# 1 2017-01-01 1:00 PM 2017-01-01 13:00:00
# 2 2016-01-01 2:00 AM 2016-01-01 02:00:00
Related
I currently have a dataset with multiple different time formats(AM/PM, numeric, 24hr format) and I'm trying to turn them all into 24hr format. Is there a way to standardize mixed format columns?
Current sample data
time
12:30 PM
03:00 PM
0.961469907
0.913622685
0.911423611
09:10 AM
18:00
Desired output
new_time
12:30:00
15:00:00
23:04:31
21:55:37
21:52:27
09:10:00
18:00:00
I know how to do them all individually(an example below), but is there a way to do it all in one go because I have a large amount of data and can't go line by line?
#for numeric time
> library(chron)
> x <- c(0.961469907, 0.913622685, 0.911423611)
> times(x)
[1] 23:04:31 21:55:37 21:52:27
The decimal times are a pain but we can parse them first, feed them back as a character then use lubridate's parse_date_time to do them all at once
library(tidyverse)
library(chron)
# Create reproducible dataframe
df <-
tibble::tibble(
time = c(
"12:30 PM",
"03:00 PM",
0.961469907,
0.913622685,
0.911423611,
"09:10 AM",
"18:00")
)
# Parse times
df <-
df %>%
dplyr::mutate(
time_chron = chron::times(as.numeric(time)),
time_chron = if_else(
is.na(time_chron),
time,
as.character(time_chron)),
time_clean = lubridate::parse_date_time(
x = time_chron,
orders = c(
"%I:%M %p", # HH:MM AM/PM 12 hour format
"%H:%M:%S", # HH:MM:SS 24 hour format
"%H:%M")), # HH:MM 24 hour format
time_clean = hms::as_hms(time_clean)) %>%
select(-time_chron)
Which gives us
> df
# A tibble: 7 × 2
time time_clean
<chr> <time>
1 12:30 PM 12:30:00
2 03:00 PM 15:00:00
3 0.961469907 23:04:31
4 0.913622685 21:55:37
5 0.911423611 21:52:27
6 09:10 AM 09:10:00
7 18:00 18:00:00
I have a dataframe (vlinder) like the following, whereby the date and the timestamp (in UTC) are in separate columns:
date time.utc variable
1/04/2020 0:00:00 12
1/04/2020 0:05:00 54
In a first step, I combined the date and time variables into one column called dateandtime using the following code:
vlinder$dateandtime <- paste(vlinder$date, vlinder$time.utc)
which resulted in an extra column in dataframe vlinder:
date time.utc variable dateandtime
1/04/2020 0:00:00 12 1/04/2020 0:00:00
1/04/2020 0:05:00 54 1/04/2020 0:05:00
I want to convert the time of UTC into local time (which is CEST, so a time difference of 2 hours).
I tried using the following code, but I get something totally different.
vlinder$dateandtime <- as.POSIXct(vlinder$dateandtime, tz = "UTC")
vlinder$dateandtime.cest <- format(vlinder$dateandtime, tz = "Europe/Brussels", usetz = TRUE)
which results in:
date time.utc variable dateandtime dateandtime.cest
1/04/2020 0:00:00 12 0001-04-20 0001-04-20 00:17:30 LMT
1/04/2020 0:05:00 54 0001-04-20 0001-04-20 00:17:30 LMT
How can I solve this?
Many thanks!
Here's a lubridate and tidyverse answer. Some data tidying, data type changes, and then bam. Check lubridate::OlsonNames() for valid time zones (tz). (I'm not positive I chose the correct tz.)
library(tidyverse)
library(lubridate)
df <- read.table(header = TRUE,
text = "date time.utc variable
1/04/2020 00:00:00 12
1/04/2020 00:05:00 54")
df <- df %>%
mutate(date = mdy(date),
datetime_utc = as_datetime(paste(date, time.utc)),
datetime_cest = as_datetime(datetime_utc, tz = 'Europe/Brussels'))
date time.utc variable datetime_utc datetime_cest
1 2020-01-04 00:00:00 12 2020-01-04 00:00:00 2020-01-04 01:00:00
2 2020-01-04 00:05:00 54 2020-01-04 00:05:00 2020-01-04 01:05:00
The default format of as.POSIXct expects an date ordered by Year-Month-Day. Therefore the date 01/04/2020 is translated into the 20th April of Year 1.
You just need to add your timeformat to as.POSIXct:
vlinder$dateandtime <- as.POSIXct(vlinder$dateandtime, tz = "UTC", format = "%d/%m/%Y %H:%M:%S")
format(vlinder$dateandtime, tz = "Europe/Brussels", usetz = TRUE)
I have two columns in a data frame first is water consumption and the second column is for date+hour. for example
Value Time
12.2 1/1/2016 1:00
11.2 1/1/2016 2:00
10.2 1/1/2016 3:00
The data is for 4 years and I want to create separate columns for month date year and hour.
I would appreciate any help
We can convert to Datetime and then extract the components. We assume the format of 'Time' column is 'dd/mm/yyyy H:M' (in case it is different i.e. 'mm/dd/yyyy H:M', change the dmy_hm to mdy_hm)
library(dplyr)
library(lubridate)
df1 %>%
mutate(Time = dmy_hm(Time), month = month(Time),
year = year(Time), hour = hour(Time))
# Value Time month year hour
#1 12.2 2016-01-01 01:00:00 1 2016 1
#2 11.2 2016-01-01 02:00:00 1 2016 2
#3 10.2 2016-01-01 03:00:00 1 2016 3
In base R, we can either use strptime or as.POSIXct and then use either format or extract components
df1$Time <- strptime(df1$Time, "%d/%m/%Y %H:%M")
transform(df1, month = Time$mon+1, year = Time$year + 1900, hour = Time$hour)
# Value Time month year hour
#1 12.2 2016-01-01 01:00:00 1 2016 1
#2 11.2 2016-01-01 02:00:00 1 2016 2
#3 10.2 2016-01-01 03:00:00 1 2016 3
data
df1 <- structure(list(Value = c(12.2, 11.2, 10.2), Time = c("1/1/2016 1:00",
"1/1/2016 2:00", "1/1/2016 3:00")), class = "data.frame", row.names = c(NA,
-3L))
I have a bunch of character date and times that I would like to merge into 1 column of date time.
For example I have:
Date Time
1/1/2018 2:00:00 PM
1/1/2018 9:00:00 AM
I would like the result to end like:
Date time
2018-01-01 14:00:00
2018-01-01 9:00:00
I first tried
paste(Date,Time)
but then I realized it does not take 'PM' into account when combining the two columns.
What should I do to merge the two columns in the correct format?
We can use use as.POSIXct after pasteing the 'Date' and 'Time' columns (assuming that the Date format is month/day/year)
datetime <- with(df1, as.POSIXct(paste(Date, Time),
format = "%m/%d/%Y %I:%M:%S %p"))
data.frame(datetime)
# datetime
#1 2018-01-01 14:00:00
#2 2018-01-01 09:00:00
data
df1 <- structure(list(Date = c("1/1/2018", "1/1/2018"), Time = c("2:00:00 PM",
"9:00:00 AM")), class = "data.frame", row.names = c(NA, -2L))
I have an excel dataset in which there are dates and time points as follows:
record_id date_E1 time_E1 date_E2 time_E2 ...
1 2019/8/24 09:00:00 2019/8/25 18:00:00
I would like to construct a variable which contains the number of hours past the first time and date, (09:00 a.m 2019/8/24). When I read the excel file with
read_excel("C:/visit.xlsx")
the time_E1 .. appears as 0.3750000 0.7736111 0.4131944 0.4131944,
and the date appears as 43640 43640 43641 43642, in R. I use visit_dates<-as.Date(as.numeric(visit_date_L$Day), origin = "1899-12-30")
to convert dates to 2019-8-24 and .. but do not know how to convert time of the day and convert to the hours past the first time point. What I expect is a vector like: 0, 42, ... hours past first time point.
I have used the following code:
as.POSIXct(visit_times, format = " %H-%M", origin = "09:00:00"),
but it returns a NULL vector. After that I could use the following code to transpose and combine date and time data:
visit_time <- subset(MY_visit, select = c(record_id, time_E1, ...)
visit_date <- subset(MY_visit, select = c(record_id, date_E1,...)
visit_time_L <- melt(visit_time, id.vars=c("record_id"))
visit_date_L <- melt(visit_date, id.vars=c("record_id"))
names(visit_time_L)[names(visit_time_L)=="value"] <- "time"
names(visit_date_L)[names(visit_date_L)=="value"] <- "Day"
visit_all <- cbind(visit_time_L, visit_date_L)
Any ideas how can I solve this problem?
Here is an approach that you can try. I have dates/times stored in an Excel file. Read it in and keep the columns as characters. Convert the dates to their proper format, as you did. Convert the fractions of the time of day to numeric and multiply by 24. Paste the dates/times together and convert to date format, then find the difference between the two in hours (the result will be in days, so multiply by 24).
library(dplyr);library(readxl); library(lubridate)
df <- read_excel('Book1.xlsx',col_types = c('text'))
# A tibble: 1 x 4
date1 time1 date2 time2
<chr> <chr> <chr> <chr>
1 43466 0.375 43467 0.41666666666666669
df %>% mutate_at(c('date1','date2'), ~ as.Date(as.numeric(.),origin='1899-12-30')) %>%
mutate_at(c('time1','time2'), ~ as.numeric(.)*24) %>%
mutate(t1=ymd_h(paste(date1,time1)),
t2=ymd_h(paste(date2,time2)),
diff=as.numeric(t2-t1)*24)
# A tibble: 1 x 7
date1 time1 date2 time2 t1 t2 diff
<date> <dbl> <date> <dbl> <dttm> <dttm> <dbl>
1 2019-01-01 9 2019-01-02 10 2019-01-01 09:00:00 2019-01-02 10:00:00 25