I'm using R to compute the best fit of a sequence of initializations, and I named them Initialization1, Initialization2, etc.. I compared the best fit with the largest result_probs value. And I want to use the one, say Initialization1, with the best property I want again.
best_fit <- paste("Initialization", which.max(results_probObs), sep = "")
best_estimated <- somefunction(best_fit, string1)
However, best_fit here is a character and can't be used as the existing Initialization1 (which is a list). I've tried as.name() too. It gave me a symbol and couldn't be used as a list as well.
Thank you very much for helping.
Related
I'm quite new to R and I've been learning with the available resources on the internet.
I came across this issue where I have a vector (a) with vars "1", "2", and "3". I want to use the count function to generate a new df with the categories for each of those variables and its frequencies.
The function I want to use in a loop is this
b <- count(mydata, var1)
However, when I use this loop below;
for (i in (a)) {
'j' <- count(mydata[, i])
print (j)
}
The loop happens but the frequencies which gets saved on j is only of the categorical variable "var 3".
Can someone assist me on this code please?
TIA!
In R there are generally better ways than to use loops to process data. In your particular case, the “straightforward” way fails, because the idea of the “tidyverse” is to have the data in tidy format (I highly recommend you read this article; it’s somewhat long but its explanation is really fundamental for any kind of data processing, even beyond the tidyverse). But (from the perspective of your code) your data is spread across multiple columns (wide format) rather than being in a single column (long form).
The other issue is that count (like many other tidyverse functions) expect an unevaluated column name. It does not accept the column name via a variable. akrun’s answer shows how you can work around this (using tidy evaluation and the bang-bang operator) but that’s a workaround that’s not necessary here.
The usual solution, instead of using a loop, would first require you to bring your data into long form, using pivot_longer.
After that, you can perform a single count on your data:
result <- mydata %>%
pivot_longer(all_of(a), names_to = 'Var', values_to = 'Value') %>%
count(Var, Value)
Some comments regarding your current approach:
Be wary of cryptic variable names: what are i, j and a? Use concise but descriptive variable names. There are some conventions where i and j are used but, if so, they almost exclusively refer to index variables in a loop over vector indices. Using them differently is therefore quite misleading.
There’s generally no need to put parentheses around a variable name in R (except when that name is the sole argument to a function call). That is, instead of for (i in (a)) it’s conventional to write for (i in a).
Don’t put quotes around your variable names! R happens to accept the code 'j' <- … but since quotes normally signify string literals, its use here is incredibly misleading, and additionally doesn’t serve a purpose.
I try to display the first 20 rows of a data frame by using stargazer. But some of the variable names are so long (such as Prevelance of unnourishment (% of population)) that the table just cannot fit in. I understand that renaming the variables with shorter names will work but that's not the way I'm looking for. I also thought about changing the latex codes that has been produced but turned out those cannot be changed. I guess the best way is to do something with the R command. Mine is:
stargazer(as.matrix(data[1:20,]), type='latex')
How should I change it to make the table fit in?
Thanks a lot!
use abbreviate to shorten names. You can control the length of names by adjusting minlength argument. For more info, please read ?abbreviate
By following this, sometimes non-unique names may appear, so to take care of it, you would use make.unique on the abbreviated names.
colnames(data) <- abbreviate( colnames(data), minlength = 3, strict = TRUE )
stargazer(as.matrix(data[1:20,]), type='latex')
I'll start off by admitting that I'm terrible at the apply functions, and function writing in general, in R. I am working on a course project to clean and model some text data, and I would like to include a step that cleans up contractions.
The qdapDictionaries package includes a contractions data frame with two columns, the first column is the contraction and the second is the expanded version. For example:
contraction expanded
5 aren't are not
I want to use the values in here to run a gsub function on my text, which I still have in a large character element. Something like gsub(contr,expd,text).
Here's an example vector that I am using to test things out:
vct <- c("I've got a problem","it shouldn't be that hard","I'm having trouble 'cause I'm dumb")
I'm stumped on how to loop through the data frame (without actually writing a loop, because it seems like the least efficient way to do it) so I can run all the gsubs that I need.
There's probably a simple answer, but here's what I tried: first, I created a function that would return the expanded version if passed a contraction:
expand <- function(contr) {
expd <- contractions[which(contractions[1]==contr),2]
}
I can use sapply with this and it does work, more or less; looping over the first column in contractions, sapply(contractions[,1],expand) returns a named vector of characters with the expanded phrases.
I can't figure out how to combine this vector with gsub though. I tried writing a second function gsub_expand and changing the expand function to return both the contraction and the expansion:
gsub_expand <- function(list, text) {
text <- gsub(list[[1]],list[[2]],text)
return(text)
}
When I ran gsub_expand(sapply(contractions[,1],expand),vct) it only corrected a portion of my vector.
[1] "I've got a problem" "it shouldn't be that hard" "I'm having trouble because I'm dumb"
The first entry in the contractions data frame is 'cause and because, so the interior sapply doesn't seem to actually be looping. I'm stuck in the logic of what I want to pass to what, and what I'm supposed to loop over.
Thanks for any help.
Two options:
stringr::str_replace_all
The stringr package does mostly the same things you can do with base regex functions, but sometimes in a dramatically simpler way. This is one of those times. You can pass str_replace_all a named list or character vector, and it will use the names as patterns and the values as replacements, so all you need is
library(stringr)
contractions <- c("I've" = 'I have', "shouldn't" = 'should not', "I'm" = 'I am')
str_replace_all(vct, contractions)
and you get
[1] "I have got a problem" "it should not be that hard"
[3] "I am having trouble 'cause I am dumb"
No muss, no fuss, just works.
lapply/mapply/Map and gsub
You can, of course, use lapply or a for loop to repeat gsub. You can formulate this call in a few ways, depending on how your data is stored, and how you want to get it out. Let's first make a copy of vct, because we're going to overwrite it:
vct2 <- vct
Now we can use any of these three:
lapply(1:length(contractions),
function(x){vct2 <<- gsub(names(contractions[x]), contractions[x], vct2)})
# `mapply` is a multivariate version of `sapply`
mapply(function(x, y){vct2 <<- gsub(x, y, vct2)}, names(contractions), contractions)
# `Map` is a multivariate version of `lapply`
Map(function(x, y){vct2 <<- gsub(x, y, vct2)}, names(contractions), contractions)
each of which will return slightly different useless data, but will also save the changes to vct2, which now looks the same as the results of str_replace_all above.
These are a little complicated, mostly because you need to save the internal version of vct as you go with each change made. The vct <<- writes to the initialized vct2 outside the function's environment, allowing us to capture the successive changes. Be a little careful with <<-; it's powerful. See ?assignOps for more info.
I am using R for text mining and have data that have been concatenated from different text columns. There are cases where words have been split by a space like"functi oning". I want to detect all such cases and remove space in between by doing dictionary check. I know splitWords function in aspell, I want a function exactly opposite of what this does.
Here is an approach, based on some code I found, but you need to provide some example text and even just pseudo code to help others respond.
First create an object that has a huge set of words spelled correctly. Then you compare your vector of words to that set with adist and an argument set to a single difference -- ideally, the internal spaces you would like to remove. I doubt that this will solve everything, but it may help.
sorted_words <- comments(sort(table(strsplit(tolower(paste(readLines("http://www.norvig.com/big.txt"), collapse = " ")), "[^a-z]+")), decreasing = TRUE))
correct <- function(*your vector*) { c(sorted_words[adist(*your vector*, sorted_words) <= min(adist(word, sorted_words), 2)], word)[1] }
Then use the correct function.
I am confused. I input a .csv file in R and want to fit a linear multivariate regression model.
However, R declares all my obvious numeric variables to be factors and my categorial variables to be integers. Therefore, I cannot fit the model.
Does anyone know how to resolve this?
I know this is probably so basic. But I really need to know this. Elsewhere, I found only posts concerning how to declare factors. But this does not apply here.
Any suggestions very much appreciated!
The easiest way, imo, to handle this is to just tell R what type of data your columns contain when you read them into the workspace. For example, if you have a csv file where the first column should be characters, columns 2-21 should be numeric, and column 22 should be a factor, here's how I would read that csv file into the workspace:
Data <- read.csv("MyData.csv", colClasses=c("character", rep("numeric", 20), "factor"))
Sometimes (with certain versions of R, as Andrew points out) float entries in a CSV are long enough that it thinks they are strings and not floats. In this case, you can do the following
data <- read.csv("filename.csv")
data$some.column <- as.numeric(as.character(data$some.column))
Or you could pass stringsAsFactors=F to the read.csv call, and just apply as.numeric in the next line. That might be a bad idea though if you have a lot of data.
It's a little harder to say what's going on with the categorical variables. You might want to try just treating those as strings and see how that works. Sometimes R will treat factor vectors as being of numeric type, so this is a good first sanity check. If that doesn't work, you can also see if the regression functions in question will let you declare how the variables should be treated.
It is hard to tell without a sample of your data file and the commands that you have been using to try and work with the data, but here are some general problems that can lead to what you describe (though there could be other possibilities as well).
The read.csv and read.table (which is called by read.csv) function will try and guess the types of data when they are not told what each column should be (the colClasses argument). If everything looks like a number then it will convert to a number, but if it sees anything in the first lines that does not look like part of a number then it will read it in as character and convert to a factor. Some of the common reasons why what you think should be a number but R sees something non-numeric include: a finger slip results in a letter somewhere in the column; similar looking substitutions, O for 0 or l for 1; a comma where one is not expected, many European files use , where R expects . (but there are options to tell R what you want here) or if you use read.table without setting sep when it really is a comma separated file.
If you have a categorical variable represented by integers, then R will convert it to integers unless you tell it to make a factor. If you use as.numeric on a factor then it will return the integers used to represent the factor internally. How to convert a factor with labels that are numbers to a numeric is a question (and answer) in the FAQ.
If this does not point you in the right direction then give us a sample of your data and what commands you are using.