[R] code for year month day hr format

[Rui Barradas]

Às 20:00 de 15/06/2024, Jibrin Alhassan escreveu:
> I have solar-geophysical data e.g as blow:
> YEAR DOY HR   IMF  SW   SSN    Dst f10.7
> 2012 214  0   3.4  403. 132    -9 154.6
> 2012 214  1   3.7  388. 132   -10 154.6
> 2012 214  2   3.7  383. 132   -10 154.6
> 2012 214  3   3.7  391. 132    -9 154.6
> 2012 214  4   4.2  399. 132    -7 154.6
> 2012 214  5   4.1  411. 132    -6 154.6
> 2012 214  6   4.0  407. 132    -6 154.6
> 2012 214  7   4.2  404. 132    -4 154.6
> 2012 214  8   4.3  405. 132    -6 154.6
> 2012 214  9   4.4  409. 132    -6 154.6
> 2012 214 10   4.4  401. 132    -6 154.6
> 2012 214 11   4.5  385. 132    -7 154.6
> 2012 214 12   4.7  377. 132    -8 154.6
> 2012 214 13   4.7  382. 132    -6 154.6
> 2012 214 14   4.3  396. 132    -4 154.6
> 2012 214 15   4.1  384. 132    -2 154.6
> 2012 214 16   4.0  382. 132    -1 154.6
> 2012 214 17   3.9  397. 132     0 154.6
> 2012 214 18   3.8  390. 132     1 154.6
> 2012 214 19   4.2  400. 132     2 154.6
> 2012 214 20   4.6  408. 132     1 154.6
> 2012 214 21   4.8  401. 132    -3 154.6
> 2012 214 22   4.9  395. 132    -5 154.6
> 2012 214 23   5.0  386. 132    -1 154.6
> 2012 215  0   5.0  377. 143    -1 138.6
> 2012 215  1   4.9  384. 143    -2 138.6
> 2012 215  2   4.9  390. 143    -4 138.6
> 2012 215  3   4.9  372. 143    -6 138.6
> 2012 215  4   5.1  371. 143    -4 138.6
> I want to process it to be of the format as shown below
>   y   m  d  hr imf  sws  ssn    Dst f10.7
> 2012-08-01 10 3.4  403. 132    -9 154.6
> 2012-08-01 12 3.7  388. 132   -10 154.6
> 2012-08-01 15 3.7  383. 132   -10 154.6
> 2012-08-01 17 3.7  391. 132    -9 154.6
> I want to request an R code to accomplish this task. Thanks for your time.
> *Jibrin Adejoh Alhassan (Ph.D)*
> Department of Physics and Astronomy,
> University of Nigeria, Nsukka
> 
> 	[[alternative HTML version deleted]]
> 
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> and provide commented, minimal, self-contained, reproducible code.
Hello,

To create a date column, paste the first two columns and coerce to class 
"Date" with conversion specifications %Y for the 4 digit year and %j for 
the day of year. See

help("strptime")



df1 <- read.table(text = "YEAR DOY HR   IMF  SW   SSN    Dst f10.7
2012 214  0   3.4  403. 132    -9 154.6
2012 214  1   3.7  388. 132   -10 154.6
2012 214  2   3.7  383. 132   -10 154.6
2012 214  3   3.7  391. 132    -9 154.6
2012 214  4   4.2  399. 132    -7 154.6
2012 214  5   4.1  411. 132    -6 154.6
2012 214  6   4.0  407. 132    -6 154.6
2012 214  7   4.2  404. 132    -4 154.6
2012 214  8   4.3  405. 132    -6 154.6
2012 214  9   4.4  409. 132    -6 154.6
2012 214 10   4.4  401. 132    -6 154.6
2012 214 11   4.5  385. 132    -7 154.6
2012 214 12   4.7  377. 132    -8 154.6
2012 214 13   4.7  382. 132    -6 154.6
2012 214 14   4.3  396. 132    -4 154.6
2012 214 15   4.1  384. 132    -2 154.6
2012 214 16   4.0  382. 132    -1 154.6
2012 214 17   3.9  397. 132     0 154.6
2012 214 18   3.8  390. 132     1 154.6
2012 214 19   4.2  400. 132     2 154.6
2012 214 20   4.6  408. 132     1 154.6
2012 214 21   4.8  401. 132    -3 154.6
2012 214 22   4.9  395. 132    -5 154.6
2012 214 23   5.0  386. 132    -1 154.6
2012 215  0   5.0  377. 143    -1 138.6
2012 215  1   4.9  384. 143    -2 138.6
2012 215  2   4.9  390. 143    -4 138.6
2012 215  3   4.9  372. 143    -6 138.6
2012 215  4   5.1  371. 143    -4 138.6", header = TRUE)


with(df1, paste(YEAR, DOY)) |> as.Date(format = "%Y %j")
#>  [1] "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01"
#>  [6] "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01"
#> [11] "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01"
#> [16] "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01"
#> [21] "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-01" "2012-08-02"
#> [26] "2012-08-02" "2012-08-02" "2012-08-02" "2012-08-02"

# now create the column
df1$Date <- with(df1, paste(YEAR, DOY)) |> as.Date(format = "%Y %j")
# remove the columns no longer needed
df1 <- df1[-(1:2)]
# relocate the new date column
df1 <- df1[c(ncol(df1), 1:(ncol(df1) - 1L))]
head(df1)
#>         Date HR IMF  SW SSN Dst f10.7
#> 1 2012-08-01  0 3.4 403 132  -9 154.6
#> 2 2012-08-01  1 3.7 388 132 -10 154.6
#> 3 2012-08-01  2 3.7 383 132 -10 154.6
#> 4 2012-08-01  3 3.7 391 132  -9 154.6
#> 5 2012-08-01  4 4.2 399 132  -7 154.6
#> 6 2012-08-01  5 4.1 411 132  -6 154.6


Hope this helps,

Rui Barradas


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