I'm am doing some coding in R. I would want to show the rows that have duplicates for columns ID and NAME but have different values for AGE.
For example I have this table:
ID | NAME | AGE
111| Mark| 22
222| Anne| 21
333| Chery| 30
444| Megan| 16
555| Charles| 37
111| Mark| 23
222| Anne| 22
333| Chery| 30
111| Mark| 22
As of now I have this code:
readfile <- read.csv(file='/home/user/shane/names.csv')
dat <- data.frame(ID=c(readfile$ID),NAME=c(readfile$NAME),AGE=c(readfile$AGE))
nam <- duplicated(dat[,c('ID','NAME)]) | duplicated(dat[,c('ID','NAME], fromLast = TRUE)
readfile[nam,]
The output looks like this:
ID | NAME | AGE
111| Mark| 22
222| Anne| 21
333| Chery| 30
111| Mark| 23
222| Anne| 22
333| Chery| 30
111| Mark| 22
I would want the output to be:
ID | NAME | AGE
111| Mark| 22
222| Anne| 21
111| Mark| 23
222| Anne| 22
111| Mark| 22
I would want to remove the columns with the ID = 333 as they have the same value in Age. would anyone have a suggestion?
The SELECT DISTINCT statement is used to return only distinct (different) values. Inside a table, a column often contains many duplicate values; and sometimes you only want to list the different (distinct) values.
I just tweaked your code :)
library(plyr)
dat1 <- ddply(dat, .(ID, NAME, AGE), nrow)
dat2 <- merge(dat1, dat, by=c("ID", "NAME", "AGE"))
dat3 <- dat2[!(!duplicated(dat2[, 1:2], fromLast=T) & !duplicated(dat2[, 1:2])),]
dat3[dat3$ID %in% dat3[dat3$V1 == 1, 1], 1:3]
Output is:
ID NAME AGE
1 111 Mark 22
2 111 Mark 22
3 111 Mark 23
4 222 Anne 21
5 222 Anne 22
Sample data:
dat <- data.frame(ID=c(111,222,333,444,555,111,222,333,111),
NAME=c('Mark','Anne','Chery','Megan','Charles','Mark','Anne','Chery','Mark'),
AGE=c(22,21,30,16,37,23,22,30,22))
# ID NAME AGE
#1 111 Mark 22
#2 222 Anne 21
#3 333 Chery 30
#4 444 Megan 16
#5 555 Charles 37
#6 111 Mark 23
#7 222 Anne 22
#8 333 Chery 30
#9 111 Mark 22
Update: Corrected formatting for better reading
A dplyr
solution:
library(dplyr)
dat %>%
group_by(ID, NAME) %>%
filter(n() > 1, sum(duplicated(AGE)) == 0) %>%
ungroup()
# A tibble: 4 x 3
ID NAME AGE
<dbl> <fctr> <dbl>
1 111 Mark 22
2 222 Anne 21
3 111 Mark 23
4 222 Anne 22
I used the data kindly provided by @Prem.
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