I am trying to convert a set of variables in my data file from numbers to letters, such that every 2 is b every 3 is c and so on.
I have tried the following code and receive the error listed below:
B3_cfile$Row <- as.character(LETTERS(B3_cfile$Row))
Error: could not find function LETTERS
My code may also be incorrect as I am very new to R. However my understanding is that LETTERS is a function built into R that does not require an additional package.
As Phil writes in the comment, LETTERS is not a function, but a vector (an array), to access the content, you need to use brackets: LETTERS[B3_cfile$Row]
if you are new to R, you can check the types easily as:
str(LETTERS)
Related
I want to learn how to access data from a nested list in R. I am relatively new to the R programming language, so I am unsure how to proceed.
The data is a 'large list(947 elements, 654.9mb) and takes the form:
The numbers within the datalist refer to station numbers and when I click on one (in Rstudio) it looks like this:
I want to kow how I can access the data within 'doy' for example. I have tried:
data[[1]]
which returns all the data for the first element of the list (site, location, doy,ltm etc). So clearly the number used within the square brackets is interpreted as an index for the list, as opposed to an identifier for the elements/station in the list.
Then I tried:
data$1
but it returned the error:
Error: unexpected numeric constant in "data$1"
Then I tried:
data[data$1==doy]
But was returned this:
Error: unexpected numeric constant in "data[data$1"
So at this point, I realise that it is not construing the number of the station as a category/factor within the list. It's just reading it as a number. So I thought I'd put some quotes around it to see if that changed what happened:
data[data$"1"=="doy"]
This returned
named list()
But when I looked at it in the environment, it was a list of 0.
I looked at some of the similar question here on Stack (like: accessing nested lists in R) and tried:
data[data$"1"=="doy",][[1]]
But just got:
Error in data[data$"1" == "doy", ] : incorrect number of dimensions
How can I access this data? It reminds me of a structure in Matlab, but it doesn't seem to be indexed in a similar fashion in R.
Let's look at some ways to do what you want:
data[[1]]
This returns the first element of the list, which is itself a list. You can use the $ subsetting shorthand, but the name of the first element is nonstandard. R prefers names that start with letters and include only alphanumeric characters, periods and underscores. You can escape this behavior with backticks:
data$`1`
If you want to access one of the elements of list 1 in your list of lists, you need to further subset. To get to doy, which is the third element of 1. You can do that four ways.
data[[1]][[3]]
data$`1`[[3]]
data[[1]]$doy
data$`1`$doy
One way (in addition to what Ben Norris has shown):
our_list[[c("1", "doy")]]
Reproducible example data (please provide next time)
our_list <- list(`1` = list(site = "x", doy = 3))
Here is a challenge for you: I was trying to make a tic tac toe based on R. First, the players have to configure putting in the name of the players, and the game should check if the name exists in a file called "Players.txt" (if not, the game will create one), if the name exists, the game will ask for a new one. The last part of the game is that the game should record all the punctuation of the players (each gambling chip used will subtract 5 points of 100 that the player has at the beginning of the game). The problem is when a player wins, the game shows the following error: "Error in table[location_name1, 3]: Incorrect number of dimension in R".
A vector can either be atomic or a list. Atomic vectors can only contain elements of one and the same data type. That means, you are "accidentally" creating a list with
vector=c(win,name1,name2,table)
with the result that each column of the data frame should become an entry.
You can solve it with
vector <- list(win, name1, name2, table)
vector is still a list but now it has the format I believe you want.
Having done that you still get errors. The reason is that these assignments fail.
location_name1=which(grepl(name1,table$gamers))
location_name2=which(grepl(name2,table$gamers))
They return an empty vector because earlier in the code you set win=vector[1]... table=vector[4]. Since vector is now a list, you have to subset it accordingly. That means you have to chance the statements to table=vector[[4]].
Now you are going to get another problem. The reason is that you treat the columns table$scores as text. When you read the data you need to make sure that this columns is not interpreted as text. You also have to eliminate all statements that coerce the column into text. Otherwise table[location_name1,3]=table[location_name1,3]+pointsx will obviously fail because you cannot add a number to a string.
For example, you coerce the column into a character column with this statement:
name1 <- data.frame(gamers=name1,games="1",scores="100")
games and scores are strings not numbers. Another example is the assigment after reading the table from the file. You can make sure that scoresare numeric by doing this.
scores <- as.numeric(table[,3])
Please get familiar with Rstudio debugging capabilities (https://support.rstudio.com/hc/en-us/articles/205612627-Debugging-with-RStudio). This way you can go through your code line by line and check consequences of each assignment to the data frame.
I'm a total newbie with R, and I'm trying to create a histogram (with value and frequency as the axises) from a csv file (just one row of values). Any idea how I can do this?
I'm also an R newbie, and I ran into the same thing. I made two separate mistakes, actually, so I'll describe them both here.
Mistake 1: Passing a frequency table to hist(). Originally I was trying to pass a frequency table to hist() instead of passing in the raw data. One way to fix this is to use the rep() ("replicate") function to explode your frequency table back into a raw dataset, as described here:
Creating a histogram using aggregated data
Simple R (histogram) from counted csv file
Instead of that, though, I just decided to read in my original dataset instead of the frequency table.
Mistake 2: Wrong data type. My raw data CSV file contains two columns: hostname and bookings (idea is to count the number of bookings each host generated during some given time period). I read it into a table.
> tbl <- read.csv('bookingsdata.csv')
Then when I tried to generate a histogram off the second column, I did this:
> hist(tbl[2])
This gave me the "'x' must be numeric" error you mention in a comment. (It was trying to read the "bookings" column header in as a data value.)
This fixed it:
> hist(tbl$bookings)
You should really start to read some basic R manual...
CRAN offers a lot of them (look into the Manuals and Contributed sections)
In any case:
setwd("path/to/csv/file")
myvalues <- read.csv("filename.csv")
hist(myvalues, 100) # Example: 100 breaks, but you can specify them at will
See the manual pages for those functions for more help (accessible through ?read.table, ?read.csv and ?hist).
To plot the histogram, the values must be of numeric class i.e the data must be of numeric value. Here the value of x seems to be of some other class.
Run the following command and see:
sapply(myvalues[1,],class)
First time poster here, so I'll try and make myself as clear as possible on the help I need. I'm fairly new to R, and this is my first real independent programming experience.
I have stock tick data for about 2.5 years, each day has its own file. The files are .txt and consist of approximately 20-30 million rows, and averaging I guess 360mb each. I am working one file at a time for now. I don't need all the data these files contain, and I was hoping that I could use the programming to minimize my files a bit.
Now my problem is that I am having some difficulties with writing the proper code so R understands what I need it to do.
Let me first show you some of the data so you can get an idea of the formatting.
M977
R 64266NRE1VEW107 FI0009653869 2EURXHEL 630 1
R 64516SSA0B 80SHB SE0002798108 8SEKXSTO 40 1
R 645730BBREEW750 FR0010734145 8EURXHEL 640 1
R 64655OXS1C 900SWE SE0002800136 8SEKXSTO 40 1
R 64663OXS1P 450SWE SE0002800219 8SEKXSTO 40 1
R 64801SSIEGV LU0362355355 11EURXCSE 160 1
M978
Another snip of data:
M732
D 3547742
A 3551497B 200000 67110 02800
D 3550806
D 3547743
A 3551498S 250000 69228 09900
So as you can see each line begins with a letter. Each letter denotes what the line means. For instance R means order book directory message, M means milliseconds after last second, H means stock trading action message. There are 14 different letters used in total.
I have used the readLines function to import the data into R. This however seems to take a very long time for R to process when I want to work with the data.
Now I would like to write some sort of If function that says if the first letter is R then from offset 1 to 4 the code means Market Segment Identifier etc., and have R add columns to these so I can work with the data in a more structured fashion.
What is the best way of importing such data, and also creating some form of structure - i.e. use unique ID information in the line of data to analyze 1 stock at a time for instance.
You can try something like this :
options(stringsAsFactors = FALSE)
f_A <- function(line,tab_A){
values <- unlist(strsplit(line," "))[2:5]
rbind(tab_A,list(name_1=as.character(values[1]),name_2=as.numeric(values[2]),name_3=as.numeric(values[3]),name_4=as.numeric(values[4])))
}
tab_A <- data.frame(name_1=character(),name_2=numeric(),name_3=numeric(),name_4=numeric(),stringsAsFactors=F)
for(i in readLines(con="/home/data.txt")){
switch(strsplit(x=i,split="")[[1]][1],M=cat("1\n"),R=cat("2\n"),D=cat("3\n"),A=(tab_A <- f_A(i,tab_A)))
}
And replace cat() by different functions that add values to each type of data.frame. Use the pattern of the function f_A() to construct others functions and same things for the table structure.
You can combine your readLines() command with regular expressions. To get more information about regular expressions, look at the R help site for grep()
> ?grep
So you can go through all the lines, check for each line what it means, and then handle or store the content of the line however you like. (Regular Expressions are also useful to split the data within one line...)
I am a relatively novice r user and am attempting to use the partimat() function within the klaR package to plot decision boundaries for a linear discriminant analysis but I keep encountering the same error. I have tried inputing the arguments multiple different ways according to the manual, but keep getting the following error:
Error in partimat.default(x, grouping, ...) :
at least two classes required
Here is an example of the input I've given:
partimat(sources1[,c(3:19)],grouping=sources1[,2],method="lda",prec=100)
where my data table is loaded in under the name "sources1" with columns 3 through 19 containing the explanatory variables and column 2 containing the classes. I have also tried doing it by entering the formula like so:
partimat(sources1$group~sources1$tio2+sources1$v+sources1$cr+sources1$co+sources1$ni+sources1$rb+sources1$sr+sources1$y+sources1$zr+sources1$nb+sources1$la+sources1$gd+sources1$yb+sources1$hf+sources1$ta+sources1$th+sources1$u,data=sources1)
with these being the column heading.
I have successfully run an LDA on this same data set without issue so I'm not quite sure what is wrong.
From the source code of the partimat.default function getAnywhere(partimat.default) it states
if (nlevels(grouping) < 2)
stop("at least two classes required")
Therefore maybe you haven't defined your grouping column as a factor variable. If you try summary(sources1[,2]) what do you get? If it's not a factor, try
sources1[,2] <- as.factor(sources1[,2])
Or in method 2 try removing the "sources1$"on each of your variable names in the formula as you specify the data frame in which to look for these variable names in the data argument. I think you are effectively specifying the dataframe twice and it might be looking, for instance, for
"sources1$sources1$groups"
Rather than
"sources1$groups"
Without further error messages or a reproducible example (i.e. include some data in your post) it's hard to say really.
HTH