I am new to R and I have troubles understanding how displaying an index works.
# Find indices of NAs in Max.Gust.SpeedMPH
ind <- which(is.na(weather6$Max.Gust.SpeedMPH))
# Look at the full rows for records missing Max.Gust.SpeedMPH
weather6[ind, ]
My code here works, no problem but I don't understand why weather6[ind] won't display the same thing as weather6[ind, ] . I got very lucky and mistyped the first time.
I apologize in advance that the question might have been posted somewhere else, I searched and couldn't find a proper answer.
So [ is a function just like any other function in R, but we call it strangely. Another way to write it in this case would be:
'[.data.frame'(weather6,ind,)
or the other way:
'[.data.frame'(weather6,ind)
The first three arguments to the function are named x, i and j. If you look at the code, early on it branches with the line:
if (Narg < 3L)
Putting the extra comma tells R that you've called the function with 3 arguments, but that the j argument is "missing". Otherwise, without the comma, you have only 2 arguments, and the function code moves on the the next [ method for lists, in which it will extract the first column instead.
Related
My question refers to redundant code and a problem that I've been having with a lot of my R-Code.
Consider the following:
list_names<-c("putnam","einstein","newton","kant","hume","locke","leibniz")
combined_df_putnam$fu_time<-combined_df_putnam$age*365.25
combined_df_einstein$fu_time<-combined_einstein$age*365.25
combined_df_newton$fu_time<-combined_newton$age*365.25
...
combined_leibniz$fu_time<-combined_leibniz$age*365.25
I am trying to slim-down my code to do something like this:
list_names<-c("putnam","einstein","newton","kant","hume","locke","leibniz")
paste0("combined_df_",list_names[0:7]) <- data.frame("age"=1)
paste0("combined_df_",list_names[0:7]) <- paste0("combined_df_",list_names[0:7])$age*365.25
When I try to do that, I get "target of assignment expands to non-language object".
Basically, I want to create a list that contains descriptors, use that list to create a list of dataframes/lists and use these shortcuts again to do calculations. Right now, I am copy-pasting these assignments and this has led to various mistakes because I failed to replace the "name" from the previous line in some cases.
Any ideas for a solution to my problem would be greatly appreciated!
The central problem is that you are trying to assign a value (or data.frame) to the result of a function.
In paste0("combined_df_",list_names[0:7]) <- data.frame("age"=1), the left-hand-side returns a character vector:
> paste0("combined_df_",list_names[0:7])
[1] "combined_df_putnam" "combined_df_einstein" "combined_df_newton"
[4] "combined_df_kant" "combined_df_hume" "combined_df_locke"
[7] "combined_df_leibniz"
R will not just interpret these strings as variables that should be created and be referenced to. For that, you should look at the function assign.
Similarily, in the code paste0("combined_df_",list_names[0:7])$age*365.25, the paste0 function does not refer to variables, but simply returns a character vector -- for which the $ operator is not accepted.
There are many ways to solve your problem, but I will recommend that you create a function that performs the necessary operations of each data frame. The function should then return the data frame. You can then re-use the function for all 7 philosophers/scientists.
I have a list of identifiers as follows:
url_num <- c('85054655', '85023543', '85001177', '84988480', '84978776', '84952756', '84940316', '84916976', '84901819', '84884081', '84862066', '84848942', '84820189', '84814935', '84808144')
And from each of these I'm creating a unique variable:
for (id in url_num){
assign(paste('test_', id, sep = ""), FUNCTION GOES HERE)
}
This leaves me with my variables which are:
test_8505465, test_85023543, etc, etc
Each of them hold the correct output from the function (I've checked), however my next step is to combine them into one big vector which holds all of these created variables as a seperate element in the vector. This is easy enough via:
c(test_85054655,test_85023543,test_85001177,test_84988480,test_84978776,test_84952756,test_84940316,test_84916976,test_84901819,test_84884081,test_84862066,test_84848942,test_84820189,test_84814935,test_84808144)
However, as I update the original 'url_num' vector with new identifiers, I'd also have to come down to the above chunk and update this too!
Surely there's a more automated way I can setup the above chunk?
Maybe some sort of concat() function in the original for-loop which just adds each created variable straight into an empty vector right then and there?
So far I've just been trying to list all the variable names and somehow get the output to be in an acceptable format to get thrown straight into the c() function.
for (id in url_num){
cat(as.name(paste('test_', id, ",", sep = "")))
}
...which results in:
test_85054655,test_85023543,test_85001177,test_84988480,test_84978776,test_84952756,test_84940316,test_84916976,test_84901819,test_84884081,test_84862066,test_84848942,test_84820189,test_84814935,test_84808144,
This is close to the output I'm looking for but because it's using the cat() function it's essentially a print statement and its output can't really get put anywhere. Not to mention I feel like this method I've attempted is wrong to begin with and there must be something simpler I'm missing.
Thanks in advance for any help you guys can give me!
Troy
I am a beginner in R and apologize in advance for asking a basic question, but I couldn't find answer anywhere on Google (maybe because the question is so basic that I didn't even know how to correctly search for it.. :D)
So if I do the following in R:
v = c(50, 25)
names(v) = c("First", "Last")
v["First"]/v["Last"]
I get the output as:
First
2
Why is it that the name, "First" appears in the output and how to get rid of it?
From help("Extract"), this is because
Subsetting (except by an empty index) will drop all attributes except names, dim and dimnames.
and
The usual form of indexing is [. [[ can be used to select a single element dropping names, whereas [ keeps them, e.g., in c(abc = 123)[1].
Since we are selecting single elements, you can switch to double-bracket indexing [[ and names will be dropped.
v[["First"]] / v[["Last"]]
# [1] 2
As for which name is preserved when using single bracket indexing, it looks like it's always the first (at least with the / operator). We'd have to go digging into the C source for further explanation. If we switch the order, we still get the first name on the result.
v["Last"] / v["First"]
# Last
# 0.5
I don't know how to write for-loops in r. Here is what I want to do:
I have a df called "na" with 50 columns (ana1_1:ana50_1). I want to loop these commands over all columns. Here are the commands for the first two columns (ana1_1 and ana2_1):
t<-table(na$ana1_1)
ana1_1<-capture.output(sort(t))
cat(ana1_1,file="ana.txt",sep="\n",append=TRUE)
t<-table(na$ana2_1)
ana2_1<-capture.output(sort(t))
cat(ana2_1,file="ana.txt",sep="\n",append=TRUE)
After the loop, all tables (ana1_1:ana50_1) should be written in ana.txt Has anyone an idea, how to solve the problem? Thank you very much!
One approach would be to loop through the columns with lapply and using the same code as in the OP's post
invisible(lapply(na, function(x) {
x1 <- capture.output(sort(table(x)))
cat(x1, file='ana.txt', sep="\n", append=TRUE)
}))
Wrapping with invisible so that it won't print 'NULL' in the R console.
We can wrap with a condition to check if the file already exists so that it won't add the same lines by accidentally running the code again.
if(!file.exists('ana.txt')){
invisible( lapply(na, function(x) {
x1 <- capture.output(sort(table(x)))
cat(x1, file='ana.txt', sep="\n", append=TRUE)
}))
}
Here is a solution with a for loop. Loops tend to be slow in r so people prefer other solutions (e.g. the great answer provided by akrun). This answer is for your understanding of the loop syntax:
for(i in 1:50){
t1<-table(na[,i])
t2<-capture.output(sort(t1))
cat(t2,file="ana.txt",sep="\n",append=TRUE)
}
We are looping through i from 1 to 50 (first line). To select a column there's two (there's actually more than two, but that's for another time) ways to access it: na$ana1_1 or na[,1] both select the first column (second line). In the first case you refer by column name, in the second by column index. Here the second case is more convenient. The rest is your desired calculations.
Be aware that cat creates a new file if ana.txt is not existing yet and appends to it if it is already there.
I've got a function that uses readline to ask people to enter in data. But I'm at a loss as to the best method to insure that the data entered meet my criteria. I'm figuring "if" statements may be the best way to go to check for errors, but I'm not sure how to incorporate them. My attempt at using them is obviously flawed (see below).
As a simple example, 2 of the most likely problems I'm going to run into would be I'd like to insure that at least some value is entered in for x (and if a value is entered for x it is a number) and that V1 and V2 contain the same number of values.
fun<-function(){
T<-readline("What is x" )
if(T=="" | typeof(x)!=numeric)
{print("Input non-aceptable")
T<-readline("What is x ")}
else
V<-readline("Enter 4 values" )
V2<-readline("Enter 4 more values ")
if(length(V1)!=length(V2))
{print("V1 & V2 do not contain equal # of values")
V<-readline("Enter 4 values ")
V<-readline("Enter 4 more values ")}
else
T<-as.numeric(T)
V<-as.numeric(V)
V2<-as.numeric(V2)
return(list(x,V1,V2)
}
As you can see, my hope is to try and spot potential errors before they cause an actual error to happen, and then to give the person an opportunity to re-enter the data. If "if" statements are the way to go, can I get some help on using the correctly?
Thanks!
In R the boolean types TRUE and FALSE can also be represented by T and F. So first off try changing the variables that you have named T to something sensible... like x maybe???
Secondly, in your typeof(x) argument, you called the variable T, so that won't work. In addition there were no quotes around numeric. Try if(!(is.numeric(x)))
Thirdly, your variables are inconsistently named, V and V, and then V1 and V2. Aside from hard to read, it also just won't work.
Lastly, your return statement needs a second closing parenthesis, the function code block needs a closing curly brace.