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I have created a dataframe p based on a larger dataset d, meaning that all information in p can be found in d, but not the other way around.

Unfortunately, I just realized that I forgot to extract d$Gender before initiating datamanagement of p. Therefore, I am looking for a solution (preferable in dplyr) to match/extract d$Gender corresponding the same patient in p.

Fortunately, I know the exact date of surgery and age of the patients. So, the extraction from d$Gender to p$Gender should be: if p$DOS = d$DOS & p$Age = d$Age then d$Gender --> p$Gender

Just to clarify, the dates and age applied in this example are made up and does not correspond to any actual patients.

I have tried something like distinct(.., .keep_all = TRUE) and the approaches described here, but without luck. I also tried ifelse(). 

p <- structure(list(DOS = c("09.01.2018", "21.07.2015", "29.07.2013", 
"06.04.2017", "18.02.2013", "16.05.2001", "16.06.2014", "02.09.2014", 
"14.01.2013", "18.04.2017", "15.05.2017", "13.06.2019", "02.08.2016", 
"01.12.2016", "12.04.2018", "14.11.2017", "26.04.2019", "15.06.2017", 
"13.11.2017", "08.05.2018", "24.05.2018", "09.03.2017", "03.11.2014", 
"17.06.2013", "03.08.2015", "12.08.2016", "09.05.2019", "14.05.2019", 
"12.12.2017", "20.02.2018", "16.04.2019", "16.10.2018", "16.03.2017", 
"10.01.2020", "15.01.2020", "04.11.2019", "23.05.2019", "25.04.2019", 
"15.11.2018", "07.03.2019", "24.01.2019", "10.01.2019", "06.09.2018", 
"27.07.2017", "13.07.2017", "23.03.2017", "12.01.2017", "11.08.2016", 
"05.05.2016", "05.05.2016", "11.02.2016", "03.12.2015", "14.06.2018", 
"03.09.2015", "11.06.2015", "19.07.2016", "13.02.2019", "13.03.2018", 
"04.08.2015", "18.09.2019", "15.08.2017", "22.12.2015", "14.05.2019", 
"29.06.2017", "30.04.2019", "13.11.2018", "05.02.2019", "08.01.2019", 
"23.10.2018", "11.09.2018", "04.09.2018", "28.08.2018", "22.05.2018", 
"08.05.2018", "28.11.2017", "19.09.2017", "15.08.2017", "11.07.2017", 
"23.05.2017", "28.03.2017", "16.02.2016", "05.01.2016", "20.10.2015", 
"13.07.2015", "23.06.2015", "14.10.2014", "28.01.2014", "26.11.2013", 
"13.12.2018", "10.09.2019", "19.01.2016", "30.07.2013", "18.03.2019", 
"11.02.2019", "21.01.2019", "07.01.2019", "14.06.2016", "26.04.2016", 
"04.04.2016", "09.02.2016"), Age = c(80, 66, 88, 65, 54, 36, 
59, 65, 59, 54, 57, 70, 68, 64, 69, 33, 50, 69, 69, 68, 77, 57, 
42, 69, 64, 59, 45, 44, 63, 43, 57, 80, 69, 79, 72, 61, 77, 47, 
71, 65, 75, 55, 68, 61, 35, 54, 68, 53, 69, 32, 60, 72, 67, 72, 
75, 44, 48, 72, 66, 29, 79, 41, 52, 55, 89, 77, 72, 60, 74, 33, 
75, 82, 66, 61, 69, 62, 86, 77, 78, 61, 69, 39, 61, 69, 62, 70, 
57, 42, 59, 50, 74, 76, 47, 62, 75, 61, 65, 67, 73, 60)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -100L))

And

d <- structure(list(DOS = c("16.10.2018", "16.03.2017", "10.01.2020", 
"15.01.2020", "04.11.2019", "05.07.2019", "13.06.2019", "12.07.2018", 
"30.05.2019", "23.05.2019", "23.05.2019", "25.04.2019", "15.11.2018", 
"07.03.2019", "24.01.2019", "10.01.2019", "06.09.2018", "03.01.2019", 
"27.07.2017", "13.11.2018", "13.07.2017", "23.03.2017", "12.01.2017", 
"11.08.2016", "05.05.2016", "04.09.2018", "16.08.2018", "05.05.2016", 
"11.02.2016", "03.12.2015", "14.06.2018", "03.09.2015", "11.06.2015", 
"19.07.2016", "01.05.2018", "31.07.2019", "01.07.2019", "13.02.2019", 
"13.03.2018", "04.08.2015", "19.12.2017", "18.09.2019", "15.08.2017", 
"22.12.2015", "30.07.2019", "25.06.2019", "04.06.2019", "28.05.2019", 
"14.05.2019", "29.06.2017", "30.04.2019", "16.04.2019", "19.03.2019", 
"13.11.2018", "05.02.2019", "08.01.2019", "23.10.2018", "11.09.2018", 
"04.09.2018", "28.08.2018", "22.05.2018", "08.05.2018", "28.11.2017", 
"19.09.2017", "15.08.2017", "11.07.2017", "23.05.2017", "28.03.2017", 
"16.02.2016", "05.01.2016", "20.10.2015", "13.07.2015", "23.06.2015", 
"14.10.2014", "28.01.2014", "26.11.2013", "13.12.2018", "10.09.2019", 
"29.07.2019", "06.05.2019", "19.01.2016", "30.07.2013", "01.07.2019", 
"18.03.2019", "11.02.2019", "21.01.2019", "07.01.2019", "14.06.2016", 
"26.04.2016", "04.04.2016", "09.02.2016", "10.11.2015", "01.06.2015", 
"05.08.2014", "17.08.2015", "14.08.2017", "10.04.2017", "30.03.2015", 
"15.12.2014", "24.11.2014"), Age = c(80, 69, 79, 72, 61, 51, 
70, 82, 64, 77, 50, 47, 71, 65, 75, 55, 68, 80, 61, 60, 35, 54, 
68, 53, 69, 51, 72, 32, 60, 72, 67, 72, 75, 44, 75, 43, 36, 48, 
72, 66, 61, 29, 79, 41, 66, 67, 68, 79, 52, 55, 89, 62, 55, 77, 
72, 60, 74, 33, 75, 82, 66, 61, 69, 62, 86, 77, 78, 61, 69, 39, 
61, 69, 62, 70, 57, 42, 59, 50, 77, 68, 74, 76, 68, 47, 62, 75, 
61, 65, 67, 73, 60, 78, 72, 51, 73, 70, 69, 83, 20, 69), Gender = c("M", 
"F", "F", "F", "M", "F", "M", "F", "F", "F", "F", "F", "F", "F", 
"F", "F", "M", "F", "F", "M", "F", "F", "M", "F", "M", "F", "F", 
"F", "F", "F", "M", "M", "F", "F", "M", "M", "F", "F", "M", "F", 
"F", "F", "M", "M", "M", "M", "F", "M", "F", "F", "F", "F", "F", 
"M", "F", "F", "F", "M", "F", "M", "M", "F", "F", "F", "F", "F", 
"M", "F", "F", "F", "F", "F", "F", "F", "F", "F", "F", "F", "M", 
"M", "F", "M", "F", "F", "M", "F", "F", "F", "F", "F", "F", "F", 
"F", "F", "F", "F", "M", "M", "M", "F")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -100L))
cmirian
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1 Answers1

1

It seems to me that a simple left_join should work

p <- p %>%
     left_join(d, by=c('DOS','Age'))

Note that after the join, based on the data provided, some age values are empty. For example p has (09.01.2018, 80) and d has(16.10.2018, 80) and (03.01.2019, 80). So it doesn't seem that all information found in p can be found in d.

camnesia
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  • Thank you. That solved it. All data in `p` can be found in `d` in my "real" data, as mentioned, the posted data are fictional. – cmirian Feb 27 '20 at 13:23