Reputation: 3
Looping through two dataframes and adding columns inside of the loop
I have a problem when specifying a loop with a data frame. The general idea I have is the following: I have an area which contains a certain number of raster quadrants. These raster quadrants have been visited irregularily over several years (e.g. from 1950 -2015).
I have two data frames: 1) a data frame containing the IDs of the rasterquadrants (and one column for the year of first visit of this quadrant):
df1<- as.data.frame(cbind(c("12345","12346","12347","12348"),rep(NA,4)))
df1[,1]<- as.character(df1[,1])
df1[,2]<- as.numeric(df1[,2])
names(df1)<-c("Raster_Q","First_visit")
2) a data frame that contains the infos on the visits; this one is ordered with by 1st rasterquadrants and then 2nd years. This dataframe has the info when the rasterquadrant was visited and when.
df2<- as.data.frame(cbind(c(rep("12345",5),rep("12346",7),rep("12347",3),rep(12348,9)),
c(1950,1952,1955,1967,1951,1968,1970,
1998,2001,2014,2015,2017,1965,1986,2000,1952,1955,1957,1965,2003,2014,2015,2016,2017)))
df2[,1]<- as.character(df2[,1])
df2[,2]<- as.numeric(as.character(df2[,2]))
names(df2)<-c("Raster_Q","Year")
I want to know when and how often the full area was 'sampled'.
Scheme of what I want to do; different colors indicate different areas/regions
My rationale: I sorted the complete data in df2 according to Quadrant and Year. I then match the rasterquadrant in df1 with the name of the rasterquadrant in df2 and the first value of year from df2 is added.
For this I wrote a loop (see below)
In order not to replicate a quadrant I created a vector "visited"
visited<-c()
Every entry of df2 that matches df1 will be written into this vector, so that the second entry of e.g. rasterquadrant "12345" in df2 is ignored in the loop.
Here comes the loop:
visited<- c()
for (i in 1:nrow(df2)){
index<- which(df1$"Raster_Q"==df2$"Raster_Q"[i])
if(length(index)==0) {next()} else{
if(df1$"Raster_Q"[index] %in% visited){next()} else{
df1$"First_visit"[index]<- df2$"Year"[i]
visited[index]<- df1$"Raster_Q"[index]
}
}
}
This gives me the first full sampling period.
Raster_Q First_visit
1 12345 1950
2 12346 1968
3 12347 1965
4 12348 1952
However, I want to have all full sampling periods.
So I do:
df1$"Second_visit"<-NA
I reset the visited vector and specify the following loop:
visited <- c()
for (i in 1:nrow(df2)){
if(df2$Year[i]<=max(df1$"First_visit")){next()} else{
index<- which(df1$"Raster_Q"==df2$"Raster_Q"[i])
if(length(index)==0) {next()} else{
if(df1$"Raster_Q"[index] %in% visited){next()} else{
df1$"Second_visit"[index]<- df2$"Year"[i]
visited[index]<- df1$"Raster_Q"[index]
}
}
}
}
Which is basically the same loop as before, however, only making sure that, if df2$"Year" in a certain raster quadrant has already been included in the first visit, then it is skipped.
That gives me the second full sampling period:
Raster_Q First_visit Second_visit
1 12345 1950 NA
2 12346 1968 1970
3 12347 1965 1986
4 12348 1952 2003
Okay, so far so good. I could do that all by hand. But I have loads and loads of rasterquadrants and several areas that can and should be screened in this way. So doing all of this in a single loop for this would be really great! However, I realized that this will create a problem because the loop then gets recursive:
The added column will not be included in the subsequent iteration of the loop, because the df1 itself is not re-read for each loop, and in consequence, the new coulmn for the new sampling period will not be included in the following iterations:
visited<- c()
for (i in 1:nrow(df2)){
m<-ncol(df1)
index<- which(df1$"Raster_Q"==df2$"Raster_Q"[i])
if(length(index)==0) {next()} else{
if(df1$"Raster_Q"[index] %in% visited){next()} else{
df1[index,m]<- df2$"Year"[i]
visited[index]<- df1$"Raster_Q"[index]
#finish "first_visit"
df1[,m+1]<-NA
# add column for "second visit"
if(df2$Year[i]<=max(df1$"First_visit")){next()} else{
# make sure that the first visit year are not included
index<- which(df1$"Raster_Q"==df2$"Raster_Q"[i])
if(length(index)==0) {next()} else{
if(df1$"Raster_Q"[index] %in% visited){next()} else{
df1[index,m+1]<- df2$"Year"[i]
visited[index]<- df1$"Raster_Q"[index]
}
}
}
This won't work. Another issue is that the vector visited() is not emptied during this loop, so that basically every Raster_Q has already been visited in the second sampling period.
I am stuck.... any ideas?
Upvotes: 0
Views: 84
Answers (1)
Reputation: 2290
You can do this without a for loop by using the dplyr and tidyr packages. First, you take your df2 and use dplyr::arrange to order by raster and year. Then you can rank the years visited using the rank function inside of the dplyr::mutate function. Then using tidyr::spread you can put them all in their own columns. Here is the code:
df <- df2 %>%
arrange(Raster_Q, Year) %>%
group_by(Raster_Q) %>%
mutate(visit = rank(Year),
visit = paste0("visit_", as.character(visit))) %>%
tidyr::spread(key = visit, value = Year)
Here is the output:
> df
# A tibble: 4 x 10
# Groups: Raster_Q [4]
Raster_Q visit_1 visit_2 visit_3 visit_4 visit_5 visit_6 visit_7 visit_8 visit_9
* <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 12345 1950 1951 1952 1955 1967 NA NA NA NA
2 12346 1968 1970 1998 2001 2014 2015 2017 NA NA
3 12347 1965 1986 2000 NA NA NA NA NA NA
4 12348 1952 1955 1957 1965 2003 2014 2015 2016 2017
EDIT: So I think I understand your problem a little better now. You are looking to remove all duplicate visits to each quadrant that happened before the maximum Year of each respective "round" of visits. So to accomplish this, I wrote a short function that in essence does what the code above does, but with a slight change. Here is the function:
filter_by_round <- function(data, round) {
output <- data %>%
arrange(Raster_Q, Year) %>%
group_by(Raster_Q) %>%
mutate(visit = rank(Year, ties.method = "first")) %>%
ungroup() %>%
mutate(in_round = ifelse(Year <= max(.$Year[.$visit == round]) & visit > round,
TRUE, FALSE)) %>%
filter(!in_round) %>%
select(-c(in_round, visit))
return(output)
}
What this function does, is look through the data and if a given year is less than the max year for the specified "visit round" then it is removed. To apply this only to the first round, you would do this:
df2 %>%
filter_by_round(1) %>%
group_by(Raster_Q) %>%
mutate(visit = rank(Year, ties.method = "first")) %>%
ungroup() %>%
mutate(visit = paste0("visit_", as.character(visit))) %>%
tidyr::spread(key = visit, value = Year)
which would give you this:
# A tibble: 4 x 8
Raster_Q visit_1 visit_2 visit_3 visit_4 visit_5 visit_6 visit_7
* <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 12345 1950 NA NA NA NA NA NA
2 12346 1968 1970 1998 2001 2014 2015 2017
3 12347 1965 1986 2000 NA NA NA NA
4 12348 1952 2003 2014 2015 2016 2017 NA
However, while it does accomplish what your for loop would have, you now have other occurrences of the same problem. I have come up with a way to do this successfully but it requires you to know how many "visit rounds" you had or some trial and error. To accomplish this, you can use map and assign the change to a global variable.
# I do this so we do not lose the original dataset
df <- df2
# I chose 1:5 after some trial and error showed there are 5 unique
# "visit rounds" in your toy dataset
# However, if you overshoot your number, it should still work,
# you will just get warnings about `max` not working correctly
# however, this may casue issues, so figuring out your exact number is
# recommended
purrr::map(1:5, function(x){
# this assigns the output of each iteration to the global variable df
df <<- df %>%
filter_by_round(x)
})
# now applying the original transformation to get the spread dataset
df %>%
group_by(Raster_Q) %>%
mutate(visit = rank(Year, ties.method = "first")) %>%
ungroup() %>%
mutate(visit = paste0("visit_", as.character(visit))) %>%
tidyr::spread(key = visit, value = Year)
This will give you the following output:
# A tibble: 4 x 6
Raster_Q visit_1 visit_2 visit_3 visit_4 visit_5
* <chr> <dbl> <dbl> <dbl> <dbl> <dbl>
1 12345 1950 NA NA NA NA
2 12346 1968 1970 2014 2015 2017
3 12347 1965 1986 NA NA NA
4 12348 1952 2003 2014 2015 2016
granted, this is probably not the most elegant solution, but it works. Hopefully this solves the problem for you
Upvotes: 3
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