I'm sure this should be easier to do than the way I know how to do it.
I'd like to apply fields from a short dataframe back into a long one based on matching a common factor.
Example short dataframe, list of valid cases:
$ptid (factor) values 1,2,3,4,5...20
$valid 1/0 (to represent true/false; variable through ptid)
long dataframe has 15k rows, each level of $ptid will have several thousand rows
I want to apply $valid onto those rows when the it is 1/true from the list above
The way I know how to do it is to loop through each row of long dataframe, but this is horribly inelegant and also slow.
I have a niggling feeling there is a much better way with dply or similar and I'd really like to learn how.
Worked this out based on the comments, thank you Colonel.
combination_dataset <- Merge(short_dataframe, long_dataframe) worked (very quickly).
Thanks to those who commented.
Related
I have a small problem, which I don't think is too hard, but I couldn't find any answer here (maybe I phrased my research wrong so please excuse me if the question has already been asked!)
I am importing data from an excel sheet which is split in two columns as in the following picture:
Now, I am trying to import all the data in the second column to my R script, but by splitting it into different vectors: one vector for category A, one for category B, etc... by keeping the data points in the order they are in the file (because as it happens, they are in chronological order).
Now, the categories each have a different number of elements, however, they are ordered alphabetically (ie you'll never find an A in the B's, for example). So I guess that makes it easier, but I'm still a novice with R and I don't really know how to proceed without getting really messy with the code and I know there's probably a simple way of doing it.
Does anyone have an idea on how to treat this nicely please? :)
We can use split in base R to return a list of vectors of 'Data' based on the unique values in 'Category'
lst1 <- split(df1$Data, df1$Category)
My problem is quite simple but I not beeing able to solve him. I have a tibble dataframe and want to know how much 0's each column have. I tried to use the function sum(dataframe$column == 0) on each column but I think this is kinda inefficient since I want to apply this to a bunch of differents dataframes. Are there any other more automatic way to do it?
This is a question of a general approach in R, I'm trying to find a way into R language but the data types and loop approaches (apply, sapply, etc) are a bit unclear to me.
What is my target:
Query data from API with parameters from a config list with multiple parameters. Return the data as aggregated data.frame.
First I want to define a list of multiple vectors (colums)
site segment id
google.com Googleuser 123
bing.com Binguser 456
How to manage such a list of value groups (row by row)? data.frames are column focused, you cant write a data.frame row by row in an R script. So the only way I found to define this initial config table is a csv, which is really an approach I try to avoid, but I can't find a way to make it more elegant.
Now I want to query my data, lets say with this function:
query.data <- function(site, segment, id){
config <- define_request(site, segment, id)
result <- query_api(config)
return result
}
This will give me a data.frame as a result, this means every time I query data the same columns are used. So my result should be one big data.frame, not a list of similar data.frames.
Now sapply allows to use one parameter-list and multiple static parameters. The mapply works, but it will give me my data in some crazy output I cant handle or even understand exactly what it is.
In principle the list of data.frames is ok, the data is correct, but it feels cumbersome to me.
What core concepts of R I did not understand yet? What would be the approach?
If you have a lapply/sapply solution that is returning a list of dataframes with identical columns, you can easily get a single large dataframe with do.call(). do.call() inputs each item of a list as arguments into another function, allowing you to do things such as
big.df <- do.call(rbind, list.of.dfs)
Which would append the component dataframes into a single large dataframe.
In general do.call(rbind,something) is a good trick to keep in your back pocket when working with R, since often the most efficient way to do something will be some kind of apply function that leaves you with a list of elements when you really want a single matrix/vector/dataframe/etc.
I have a large number (over 200,000) separate files, each of which contains a single row and many columns (sometimes upwards of several hundred columns). There is one column (id) in common across all of the files. Otherwise, the column names are semi-random, with incomplete overlap between dataframes. Currently, I am using %in% to determine which columns are in common and then merging using those columns. This works perfectly well, although I am confident (maybe hopeful is a better word) that it could be done faster.
For example:
dfone<-data.frame(id="12fgh",fred="1",wilma="2",barney="1")
dftwo<-data.frame(id="36fdl",fred="5",daphne="3")
common<-names(dfone)[names(dfone) %in% names(dftwo)]
merged<-merge(dfone,dftwo,by=common,all=TRUE)
So, since I'm reading a large number of files, here is what I'm doing right now:
fls<-list.files()
first<-fls[1]
merged<-read.csv(first)
for (fl in fls) {
dffl<-read.csv(fl)
common<-names(dffl)[names(dffl) %in% names(merged)]
merged<-merge(dffl,merged,by=common,all=TRUE)
# print(paste(nrow(merged)," rows in dataframe.",sep=""))
# flush.console()
# print(paste("Just did ",fl,".",sep=""))
# flush.console()
}
Obviously, the commented out section is just a way of keeping track of it as it's running. Which it is, although profoundly slowly, and it runs ever more slowly as it is assembling the data frame.
(1) I am confident that a loop isn't the right way to do this, but I can't figure out a way to vectorize this
(2) My hope is that there is some way to do the merge that I'm missing that doesn't involve my column name comparison kludge
(3) All of which is to say that this is running way too slowly to be viable
Any thought on how to optimize this mess? Thank you very much in advance.
A much shorter and cleaner approach is to read them all into a list and then merge.
do.call(merge, lapply(list.files(), read.csv))
It will still be slow though. You could speed it up by replacing read.csv with something faster (e.g., data.table::fread) and possibly by replacing lapply with parallel::mclapply.
New to R, taking a very accelerated class with very minimal instruction. So I apologize in advance if this is a rookie question.
The assignment I have is to take a specific column that has 21 levels from a dataframe, and condense them into 4 levels, using an if, or ifelse statement. I've tried what feels like hundreds of combinations, but this is the code that seemed most promising:
> b2$LANDFORM=ifelse(b2$LANDFORM=="af","af_type",
ifelse(b2$LANDFORM=="aflb","af_type",
ifelse(b2$LANDFORM=="afub","af_type",
ifelse(b2$LANDFORD=="afwb","af_type",
ifelse(b2$LANDFORM=="afws","af_type",
ifelse(b2$LANDFORM=="bfr","bf_type",
ifelse(b2$LANDFORM=="bfrlb","bf_type",
ifelse(b2$LANDFORM=="bfrwb","bf_type",
ifelse(b2$LANDFORM=="bfrwbws","bf_type",
ifelse(b2$LANDFORM=="bfrws","bf_type",
ifelse(b2$LANDFORM=="lb","lb_type",
ifelse(bs$LANDFORM=="lbaf","lb_type",
ifelse(b2$LANDFORM=="lbub","lb_type",
ifelse(b2$LANDFORM=="lbwb","lb_type","ws_type"))))))))))))))
LANDFORM is a factor, but I tried changing it to a character too, and the code still didn't work.
"ws_type" is the catch all for the remaining variables.
the code runs without errors, but when I check it, all I get is:
> unique(b2$LANDFORM)
[1] NA "af_type"
Am I even on the right path? Any suggestions? Should I bite the bullet and make a new column with substr()? Thanks in advance.
If your new levels are just the first two letters of the old ones followed by _type you can easily achieve what you want through:
#prototype of your column
mycol<-factor(sample(c("aflb","afub","afwb","afws","bfrlb","bfrwb","bfrws","lb","lbwb","lbws","wslb","wsub"), replace=TRUE, size=100))
as.factor(paste(sep="",substr(mycol,1,2),"_type"))
After a great deal of experimenting, I consulted a co-worker, and he was able to simplify a huge amount of this. Basically, I should have made a new column composed of the first two letters of the variables in LANDFORM, and then sample from that new column and replace values in LANDFORM, in order to make the ifelse() statement much shorter. The code is:
> b2$index=as.factor(substring(b2$LANDFORM,1,2))
b2$LANDFORM=ifelse(b2$index=="af","af_type",
ifelse(b2$index=="bf","bf_type",
ifelse(b2$index=="lb","lb_type",
ifelse(b2$index=="wb","wb_type",
ifelse(b2$index=="ws","ws_type","ub_type")))))
b2$LANDFORM=as.factor(b2$LANDFORM)
Thanks to everyone who gave me some guidance!