I have two separate dataframe, one (frame1) has the general info about the location of sensors and the other one (frame2) has the time series for all the locations with the siteIDs column common between the two.
I want to add another column to frame2. I thought it would be possible to use lapply, but it is not working. I have also tried using [[ instead of $, no gain. It does not produce any warning or error. It simply does not do anything.
gaugeList<-as.list(unique(frame2$siteIDs))
frame2[['timeZone']]<-as.character(NA)
lapply(gaugeList,function(gaugeX) { frame2$timeZone[which(frame2$siteIDs==gaugeX)] <- (as.character(frame1$timeZone[which (frame1$siteIDs==gaugeX)]))})
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My goal of this code is to create a loop that aggregates each company's word frequency by a certain principle vector I created and adds it to a list. The problem is, after I run this, it only prints the 7 principles that I have rather than the word frequencies along side them. The word frequencies being the certain column of the FREQBYPRINC.AG data frame. Individually, running this code without the loop and just testing out a certain column, it works no problem. For some reason, the loop doesn't want to give me the correct data frames for the list. Any suggestions?
list.agg<-vector("list",ncol(FREQBYPRINC.AG)-2)
for (i in 1:14){
attach(FREQBYPRINC.AG)
list.agg[i]<-aggregate(FREQBYPRINC.AG[,i+1],by=list(Type=principle),FUN=sum,na.rm=TRUE)
}
I really wish I could help. After reading your statement, It seems that to you , you feel that the code should be working and it is not. Well maybe there exists a glitch.
Since you had previously specified list. agg as a list, you need to subset it with double square brackets. Try this one out:
list.agg<-vector("list",ncol(FREQBYPRINC.AG)-2)
for (i in 1:14){
list.agg[[i]]<-aggregate(FREQBYPRINC.AG[,i+1],by=list
(Type=principle),FUN=sum,na.rm=TRUE)}
I have a list of five data frames full of user responses to a survey.
In each of these data frames, the second column is the user id number. Some of the users took the survey multiple times, and I am trying to weed out the duplicate responses and just keep the first record.
The naming conventions are fairly standard, so the column in the first data frame is called akin to survey1_id and the second is survey2_id, etc. with the exception being that the column in the third data frame is called survey3a_id.
So basically what I tried to do was this:
for (i in seq(1,5)) {
newdata <- distinct(survey_list[[i]], grep(names("^survey.*_id$", survey_list[[i]]), value = TRUE))
}
But this doesn't work.
I originally thought it was just because the grep output had quotes around it, but I tried to strip them with noquote() and that didn't work. I then realized that distinct() doesn't actually evaluate the second argument, it just takes it literally, so I tried to force it to evaluate using eval(), but that didn't work. (Not sure I really expected it to.)
So now I'm kind of stuck. I don't know if the best solution is just to write five individual lines of code or, for a more generalizable solution, to sort and compare item-by-item in a loop? Was just hoping for a cleaner solution. I'm kind of new to this stuff.
I have this csv with 4000+ entries and I am trying to create a histogram of one of the variables. Because of the way the data was collected, there was a possibility that if data was uncollectable for that entry, it was coded as a period (.). I still want to create a histogram and just ignore that specific entry.
What would be the best or easiest way to go about this?
I tried making it so that the histogram would only use the data for every entry except the one with the period by doing
newlist <- data1$var[1:3722]+data1$var[3724:4282]
where 3723 is the entry with the period, but R said that + is not meaningful for factors. I'm not sure if I went about this the right way, my intention was to create a vector or list or table conjoining those two subsets above into one bigger list called newlist.
Your problem is deeper that you realize. When R read in the data and saw the lone . it interpreted that column as a factor (categorical variable).
You need to either convert the factor back to a numeric variable (this is FAQ 7.10) or reread the data forcing it to read that column as numeric, if you are using read.table or one of the functions that calls read.table then you can set the colClasses argument to specify a numeric column.
Once the column of data is a numeric variable then a negative subscript or !is.na will work (or some functions will automatically ignore the missing value).
I have a data frame with hundreds of columns whose names I want to change. I'm very new to R, so it's rather easy to think through the logic of this, but I simply can't find a relevant example online.
The closest I could sort of get was this:
projectFileAllCombinedNames <- for (i in 1:200){names(projectFileAllCombined)[i+1] <-variableNames[i]}
Basically, starting at the second column of projectFileAllCombined, I want to loop through the columns in the dataframe and assign them the data values in the second data frame. I was able to change one column name manually with this code:
colnames(projectFileAllCombined)[2]<-"newColumnName"
but I can't possibly do that for hundreds of columns. I've spent multiple hours on this and can't crack it with any number of Google searches on "change multiple columns in r" or "change column names in r". The best I can find online is examples where people change a few columns with a c() function and I get how that works, but that still seems to require typing out all the column names as parameters to the function, unless there is a way to just pass the "variableNames" file into that c() function, but I don't know of one.
Will
colnames(projectFileAllCombined)[-1] <- variableNames
not suffice?
This assumes the ordering of columns in projectFileAllCombined is the same as the ordering of the new variable names in variableNames, and that
length(variableNames) == (ncol(projectFileAllCombined) - 1)
The key point here is that the replacement function 'colnames<-'() is vectorised and can replace any number of column names in a single call if passed a vector of replacement values.
Ok, I'm stuck in a dumbness loop. I've read thru the helpful ideas at How to sort a dataframe by column(s)? , but need one more hint. I'd like a function that takes a matrix with an arbitrary number of columns, and sorts by all columns in sequence. E.g., for a matrix foo with N columns,
does the equivalent of foo[order(foo[,1],foo[,2],...foo[,N]),] . I am happy to use a with or by construction, and if necessary define the colnames of my matrix, but I can't figure out how to automate the collection of arguments to order (or to with) .
Or, I should say, I could build the entire bloody string with paste and then call it, but I'm sure there's a more straightforward way.
The most elegant (for certain values of "elegant") way would be to turn it into a data frame, and use do.call:
foo[do.call(order, as.data.frame(foo)), ]
This works because a data frame is just a list of variables with some associated attributes, and can be passed to functions expecting a list.