Very trick and at time daunting working with dates in R. But you made it so simple and straitforward with the mdy(), ymd() and dmy() functions. Well done!

@danieldbonneau

Жыл бұрын

Thanks for the kind words, Osorio!

Thanks, very helpful!

@danieldbonneau

2 ай бұрын

Thanks! I'm glad it was helpful

Thank you for simplifying this function !

@danieldbonneau

Жыл бұрын

Happy I could help!

Thank you so much, i was so frustrated

@danieldbonneau

Жыл бұрын

Thanks for the positive feedback! If you have any other R problems you're frustrated by let me know and I may be able to make a video to help you out.

Thank you!

@danieldbonneau

6 ай бұрын

Glad I could help!

Thanks

@danieldbonneau

Жыл бұрын

You're welcome, Yaw!

Hi, thanks for this video, it's very helpful. If you don't mind me asking, what would you do in cases where the date and time are given together in a dataset? for example "Dec 04, 2021, 11:30:00AM" all in one column. I'm unsure of the best way to split them out and format them as dates/times. Thanks!

@danieldbonneau

Жыл бұрын

Thanks for the question. To just convert that directly into a date format, you could use the mdy_hms() function within the lubridate package. Then, to get something like just the date without the time, you could use the date() function on that newly created variable. The same works for if you want to get the year, month, day, hour, minute, second. Then, if you wanted to also get a column that was just the time, you could load in the hms package and use the as_hms() function on the formatted datetime object. Here's some simple code showing a few of these things: dates % mutate(date_format = mdy_hms(dates)) %>% mutate(just_date = date(date_format)) %>% mutate(month = month(date_format)) %>% mutate(hour = hour(date_format)) %>% mutate(just_time = as_hms(date_format))

Thank you

@danieldbonneau

Жыл бұрын

You're welcome!

is there a function for month/year or is it just for d/m/y?

@danieldbonneau

Жыл бұрын

There are a couple of ways I'm interpreting your question, so I'll try to answer both here. 1. If you're looking to get month and year as separate variables. You could use separate() in the tidyr package to split it into two columns. Something like: df %>% separate(date, into = c("month", "year"), sep = "/") 2. To convert month/year into a formal date object. Off the top of my head, that's not possible without adding the day in to the date. So you would need to do something like: date

How do you retain the format of month-day-year?

@danieldbonneau

6 ай бұрын

For the column in your data frame? Date variables will always just show as Year-Month-Day. If you wanted to have a variable that displays it as Month-Day-Year, you could just paste it together in that way in a different column. However, you'd still use your actual date column if you wanted to run any calculations with that date variable.

@Svykle

6 ай бұрын

@@danieldbonneau omg. Thanks . God I spent hours on this lol . So how would display my date like in a histogram using year -date? Thanks . R is fun but challenging at first .

My dataset is different. The column names are the dates and not the cells below them. It doesn't seem to work.

@danieldbonneau

Жыл бұрын

If your column names are dates, you want to start by reorganizing your data set. Your data set should (almost always) be in a "long" format, where each column holds one particular piece of information - such as date or sales. So before applying the conversion from this video, you're going to want to use pivot_longer() from the tidyr package. Let's suppose your data set has dates as the column names, and sales as the values within the first row of your data set and that this data set is called "my_data". You would run this code first to reorganize your data into two columns: "date" and "sales" pivot_df % pivot_longer(cols = everything(), names_to = "date", values_to = "sales") With that line, you should be left with a new data set that has two columns, date and sales. Then, you can transform the date column into an actual date data type using the steps in this video. To see the pivot_longer code in action with a different example, I have this video: kzread.info/dash/bejne/d3-hpJSMc9CZmpc.html

how to convert a number such as this '41738.955555555556' into date ?

@danieldbonneau

Жыл бұрын

You can use the as.Date() function to convert that to a date. However, you'll need to also supply an origin argument. I've never seen it with decimals, but it's the same with decimals or the whole number. Running as.Date(41738, origin = "1899-12-30") gives 2014-04-09. Is that the date you're looking for? The way this works is the numeric value (41738 in this case) is the number of days from the origin supplied. In other words, 2014-04-09 is 41,738 days from 1899-12-30. Hope that helps.

@koushikvadali7859

Жыл бұрын

@@danieldbonneau thanks