CODEWITHSUNDEEP
rdata.tablesubsetr-factor
Max
Max

Reputation: 43

Remove rows based on factor-levels

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

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

df
  site time value
1    A   11    12
2    A   22   -24
3    A   33   -30
4    B   11     3
5    B   22    16
6    C   11     3
7    C   22     9

Question

How do I remove the rows where an unique element of df$time is not present for each of the levels of df$site ?

In this case I want to remove df[3,], because for df$time the timestamp 33 is only present for site A and not for site B and site C.

Desired output:

df.trimmed
  site time value
1    A   11    12
2    A   22   -24
4    B   11     3
5    B   22    16
6    C   11     3
7    C   22     9

The data.frame has easily 800k rows and 200k unique timestamps. I don't want to use loops but I don't know how to use vectorized functions like apply() or lapply() for this case.

Upvotes: 4

Views: 3475

Answers (2)

BenBarnes
BenBarnes

Reputation: 19454

Here's another possible solution using the data.table package:

unTime <- unique(df$time)

library(data.table)

DT <- data.table(df, key = "site")

(notInAll <- unique(DT[, list(ans = which(!unTime %in% time)), by = key(DT)]$ans))
# [1] 3

DT[time %in% unTime[-notInAll]]

#      site time value
# [1,]    A   11     3
# [2,]    A   22    11
# [3,]    B   11    -6
# [4,]    B   22    -2
# [5,]    C   11   -19
# [6,]    C   22   -14

EDIT from Matthew
Nice. Or a slightly more direct way :

DT = as.data.table(df)
tt = DT[,length(unique(site)),by=time]
tt
   time V1
1:   11  3
2:   22  3
3:   33  1

tt = tt[V1==max(V1)]      # See * below
tt
   time V1
1:   11  3
2:   22  3

DT[time %in% tt$time]
   site time value
1:    A   11     7
2:    A   22    -2
3:    B   11     8
4:    B   22   -10
5:    C   11     3
6:    C   22     1

In case no time is present in all sites, when final result should be empty (as Ben pointed out in comments), the step marked * above could be :

tt = tt[V1==length(unique(DT$site))]

Upvotes: 5

A5C1D2H2I1M1N2O1R2T1
A5C1D2H2I1M1N2O1R2T1

Reputation: 193687

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.

Upvotes: 2

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