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I am trying to work out the difference between dates between two date columns in a dataframe df$started and df$done. The result is put in a third column called df$diff Some of the df$diff has NA and in these rows I want to enter the difference between the current date and df$started. How can I do this?
In the future, please follow the questions guidelines.
library(lubridate)
start = as.Date(c("14.01.2015", "26.03.2015"),format = "%d.%m.%Y")
end = as.Date(c("18.01.2015", NA),format = "%d.%m.%Y")
diff = ifelse(!is.na(end),difftime(end,start,units="days"),difftime(Sys.time(),start,units="days"))
df = data.frame(start,end,diff)
View(df)
start end diff
1 2015-01-14 2015-01-18 4.0000
2 2015-03-26 NA 174.7846
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I have a dataset with time, where the time intervals are 6 hours apart and I have a column of heaterstatus.
The dataset :
I would like to know the percentage of zero occurred in each day for heaterstatus. New to R, any suggestion will be helpful.
Not tested since you only provided data as an image, but this should do what you want:
library(dplyr)
dat %>%
group_by(day = as.Date(Time)) %>%
summarize(pct_0 = mean(HeaterStatus == 0))
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activity$ActivityDate = as.POSIXct(activity$ActivityDate, format="%d/%m/%y", tz=Sys.timezone()) returns NA in ActivityDate and Date columns in R
activity$ActivityDate = as.POSIXct(activity$ActivityDate, format="%d/%m/%y", tz=Sys.timezone()) ends up with NA in ActivityDate column in R
I can only guess what you would like to achieve based on your question title - it would help if you could include a minimal reproducible example and some prose describing your problem.
It seems that lubridate's parsing functions might do what you need:
ActivityDate <- c("07/31/2022", "07-31-2022")
lubridate::mdy(ActivityDate)
#> [1] "2022-07-31" "2022-07-31"
Created on 2022-12-21 with reprex v2.0.2
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I have 16 variables that are numeric, and I need to create an extra column that is YES (otherwise NO) when 3 or more variables out of those 16 have a value above 1015.
How could I do that?
Thanks
You can try with rowSums :
cols <- 1:16
df$res <- ifelse(rowSums(df[cols] > 1015, na.rm = TRUE) >= 3, 'Yes', 'No')
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I have tried to plot sales order against time. graph is as follows:
I want to replace number of days by dates:
which means replace 0 by 22/09/2016
50 by 11/11/2016
100 by 31/12/2016
150 by 19/02/2017
200 by 10/04/2017
I don't know how to proceed.
You will want to make use of the xts package which can handle daily values
library(xts)
v <- 0:200 # your data here
d <- seq.Date(as.Date('2016-09-22'), as.Date('2017-04-10'), length.out = 201)
plot(xts(v, order.by = as.POSIXct(d)))
Here is a great cheat sheet:
https://www.datacamp.com/community/blog/r-xts-cheat-sheet#gs.1XjXRyI
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I'm trying to calculate the number of retained students using R. The two variables I'm working with are 'registration_date' (mm/dd/yr) and 'date_of_last_login' (mm/dd/yr). A student is considered retained if they logged-in in the preceding 30 days.
ID 1 , 2, 3, 4, 5
registration_date 2/1/15, 2/1/15, 3/15/15, 2/10/15, 4/15/15
date_of_last_login 2/3/15, 3/15/15, 4/30/15, 4/25/15, 5/16/15
I imagine the idea is to create a new variable: 'retained students' but I am not sure how to set up the formula in R.
Assuming you mean the 30 days previous to today:
last_login <- c("2/3/15","3/15/15","4/30/15")
login <- as.Date(last_login, format = '%m/%d/%y')
retained_students <- (Sys.Date()-login < 30)
retained_students
retained_students is then a vector with either TRUE or FALSE for each login