Remove rows based on factor-levels

Question

I have a data.frame df in format \"long\".

df <- data.frame(site = rep(c(\"A\",\"B\",\"C\"), 1, 7),
                 time = c(11,11,11,22,22,         


        

Answer 1

Would rle work for you?

df <- df[order(df$time), ]
df <- subset(df, time != rle(df$time)$value[rle(df$time)$lengths == 1])
df <- df[order(df$site), ]
df
##   site time value
## 1    A   11    17
## 4    A   22    -3
## 2    B   11     8
## 5    B   22     5
## 3    C   11     0
## 6    C   22    13

Re-looking at your data, it seems that this solution might be too simple for your needs though....

Update

Here's an approach that should be better than the rle solution that I put above. Rather than look for a run-length of "1", will delete rows that do not match certain conditions of the results of table(df$site, df$time). To illustrate, I've also added some more fake data.

df <- data.frame(site = rep(c("A","B","C"), 1, 7),
                 time = c(11,11,11,22,22,22,33),
                 value = ceiling(rnorm(7)*10))
df2 <- data.frame(site = rep(c("A","B","C"), 1, 7),
                 time = c(14,14,15,15,16,16,16),
                 value = ceiling(rnorm(7)*10))
df <- rbind(df, df2)
df <- df[order(df$site), ]

temp <- as.numeric(names(which(colSums(with(df, table(site, time)))
                               >= length(levels(df$site)))))
df2 <- merge(df, data.frame(temp), by.x = "time", by.y = "temp")
df2 <- df2[order(df2$site), ]
df2
##   time site value
## 3   11    A    -2
## 4   16    A    -2
## 7   22    A     2
## 1   11    B   -16
## 5   16    B     3
## 8   22    B    -6
## 2   11    C     8
## 6   16    C    11
## 9   22    C   -10

Here's the result of tabulating and summing up the site/time combination:

colSums(with(df, table(site, time)))
## 11 14 15 16 22 33 
##  3  2  2  3  3  1 

Thus, if we were interested in including sites where at least two sites had the timestamp, we could change the line >= length(levels(df$site)) (in this example, 3) to >= length(levels(df$site))-1 (obviously, 2).

Not sure if this solution is useful to you at all, but I thought I would share it to show the flexibility in solutions we have with R.