fetch date using Regex in R

Question

I have the various file name as given below: file<-"/path/path/path/pth.0p25.2015011500.p264.path2.pathpathh254004.nc"

Problem 1: I want datetime part(2015011500)from the above file. I have written the below code:

# Fetching the date vlaue from the filename
a<-substr(gsub("[A-z]|[////]","",file),6,13)
 a
[1] "20150115"
hrs<-substr(gsub("[A-z]|[////]","",file),14,15)
hrs
[1] "00"
#concatenating both date and a as one
chr<-paste(a,hrs, sep=" ")
chr
[1] "20150115 00"

But when I am trying to convert chr to date. I am getting NULL value.

#Converting chr variable to date
datetime<-as.Date(a,format="%Y%m%d %H")

Result:
datetime
[1] NA

Problem 2: I want to fetch 264 from the file name. Code for this is given below:

validhrs<-substr(gsub("[A-z]|[////]","",ncfname),17,19)
Result:
validhrs
[1] "264"

I want to convert this validhr to time(hour).And then add this to datetime

Anyone could help me with this?

Thanks in advance!!

you use `a` instead of `chr` in the `as.Date` function. If you use `chr` the `as.Date.` or `as.POSIXct` will work properly. — D.sen, Sep 12 '17 at 20:15
For your frist problem, use `as.POSIXct` (which is for date-times) instead of `as.Date` (which is only for dates). Dates don't include hours and minutes. — Nathan Werth, Sep 12 '17 at 20:15
FYI: [`[A-z]` matches more than just letters](http://stackoverflow.com/a/29771926/3832970). — Wiktor Stribiżew, Sep 12 '17 at 20:15
@D.sen and @NathanWerth, I used as.POSIXct and I am getting the following value:`datetime datetime [1] NA`. Any Comments? — Kaushik, Sep 12 '17 at 20:18
^ Do not add the `%M%S`, your string does not reflect that format — D.sen, Sep 12 '17 at 20:21
@D.Sen , thanks it worked. I am getting below output: `datetime [1] "2015-01-15 EST"`. Here, I am not getting time component. Is it because hour is 00 in my case? — Kaushik, Sep 12 '17 at 20:22
Correct, switch `hrs` to `1` for testing purposes. You'll see the hour value. — D.sen, Sep 12 '17 at 20:23
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/154268/discussion-between-d-sen-and-kaushik). — D.sen, Sep 12 '17 at 20:26

www · Answer 1 · 2017-09-12T20:44:07.740

1

Answer to Question 1: How to extract date from string, then convert it to date class:

x <- "/path/path/path/pth.0p25.2015011501.p264.path2.pathpathh254004.nc"

#Extract the date:
x <- gsub(".*(\\d{10}).*",'\\1',x)

#Convert to POSIXct:
x <- as.POSIXct(x,format="%Y%m%d %H")

Output:

[1] "2015-01-15 01:00:00 PST"

Answer to Question 2: How to add number of hours, "...p264..." in this case, to x:

hrs <- gsub(".*[p]+(\\d{3}).*","\\1",x)

require(lubridate)

x + hours(hrs)

Output:

[1] "2015-01-26 01:00:00 PST"

Note: 264 hours (i.e. 11 days) were added to what was once "2015-01-15 01:00:00".

edited Sep 12 '17 at 20:44

answered Sep 12 '17 at 20:23

www

4,034
1
9
22

then how to convert it to datetime? Any comments on problem2? – Kaushik Sep 12 '17 at 20:25
Thanks. Could you help me in solving problem2? – Kaushik Sep 12 '17 at 20:30
@Kaushik - Solved both; see above. – www Sep 12 '17 at 20:45
@Kaushik - If this was helpful, please remember to mark this answer as the solution to help the community know it's solved and to help others with similar questions to find their answer even faster. – www Sep 12 '17 at 21:58

fetch date using Regex in R

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