R encoding ASCII backtick - r

I have the following backtick on my list's names. Prior lists did not have this backtick.
$`1KG_1_14106394`
[1] "PRDM2"
$`1KG_20_16729654`
[1] "OTOR"
I found out that this is a 'ASCII grave accent' and read the R page on encoding types. However what to do about it ? I am not clear if this will effect some functions (such as matching on list names) or is it OK leave it as is ?
Encoding help page: https://stat.ethz.ch/R-manual/R-devel/library/base/html/Encoding.html
Thanks!

My understanding (and I could be wrong) is that the backticks are just a means of escaping a list name which otherwise could not be used if left unescaped. One example of using backticks to refer to a list name is the case of a name containing spaces:
lst <- list(1, 2, 3)
names(lst) <- c("one", "after one", "two")
If you wanted to refer to the list element containing the number two, you could do this using:
lst[["after one"]]
But if you want to use the dollar sign notation you will need to use backticks:
lst$`after one`
Update:
I just poked around on SO and found this post which discusses a similar question as yours. Backticks in variable names are necessary whenever a variable name would be forbidden otherwise. Spaces is one example, but so is using a reserved keyword as a variable name.
if <- 3 # forbidden because if is a keyword
`if` <- 3 # allowed, because we use backticks
In your case:
Your list has an element whose name begins with a number. The rules for variable names in R is pretty lax, but they cannot begin with a number, hence:
1KG_1_14106394 <- 3 # fails, variable name starts with a number
KG_1_14106394 <- 3 # allowed, starts with a letter
`1KG_1_14106394` <- 3 # also allowed, since escaped in backticks

Related

Subsetting a string vector based on a partial match of unknown characters

I have a vector of 8-character file names of the format
"/relative/path/to/folder/a(bc|de|fg)...[xy]1.sav"
where the brackets hold one of two-three known characters, and the '...' are three unknown characters. I want to match all character vectors that has the same unknown sequence XXX and sort into a list of character vectors.
I am not sure how to proceed on this. I am thinking about a way to extract the letters in the fourth to sixth position (...), and put into a vector then use `grep to get all the files with the matching string.
E.g.
# Pseudo-code. Not functioning code, but sort of the thing I want to do
> char.extr <- str_extract(file.vector, !"a(bc|de|fg)...[xy]1.sav")
> char.extr
"JKL", "MNO" ,"PQR" ...
# Use grep and lapply to put matched strings into list
> path.list <- lapply(char.extr, grep, file.vector)
> path.list
1. "/relative/path/to/folder/abcJKLx1.sav"
"/relative/path/to/folder/adeJKLy1.sav"
2. "/relative/path/to/folder/afgMNOx1.sav"
"/relative/path/to/folder/abcMNOy1.sav"
Since we know the name structure, I'd imaging extracting the 3 letter substring and then using split to get individual lists is what you're looking for.
split(path.list, substr(basename(path.list), 4, 6))

read.csv ;check.names=F; R;Look at the picture,why it works a treat?

please see the the column name "if" in the second column,the deifference is :when check.name=F,"." beside "if" disappear
Sorry for the code,because I try to type some codes to generate this data.frame like in the picture,but i failed due to the "if".We know that "if" is a reserved word in R(like else,for, while ,function).And here, i deliberately use the "if" as the column name (the 2nd column),and see whether R will generate some novel things.
So using another way, I type the "if" in the excel and save as the format of csv in order to use read.csv.
Question is:
Why "if." changes to "if"?(After i use check.names=FALSE)
enter image description here
?read.csv describes check.names= in a similar fashion:
check.names: logical. If 'TRUE' then the names of the variables in the
data frame are checked to ensure that they are syntactically
valid variable names. If necessary they are adjusted (by
'make.names') so that they are, and also to ensure that there
are no duplicates.
The default action is to allow you to do something like dat$<column-name>, but unfortunately dat$if will fail with Error: unexpected 'if' in "dat$if", ergo check.names=TRUE changing it to something that the parser will not trip over. Note, though, that dat[["if"]] will work even when dat$if will not.
If you are wondering if check.names=FALSE is ever a bad thing, then imagine this:
dat <- read.csv(text = "a,a\n2,3")
dat
# a a.1
# 1 2 3
dat <- read.csv(text = "a,a\n2,3", check.names = FALSE)
dat
# a a
# 1 2 3
In the second case, how does one access the second column by-name? dat$a returns 2 only. However, if you don't want to use $ or [[, and instead can rely on positional indexing for columns, then dat[,colnames(dat) == "a"] does return both of them.

Legal column names in R and consequences of syntactically invalid column names

df <- read.csv(
text = '"2019-Jan","2019-Feb",
"3","1"',
check.names = FALSE
)
OK, so I use check.names = FALSE and now my column names are not syntactically valid. What are the practical consequences?
df
#> 2019-Jan 2019-Feb
#> 1 3 1 NA
And why is this NA appearing in my data frame? I didn't put that in my code. Or did I?
Here's the check.names man page for reference:
check.names logical. If TRUE then the names of the variables in the
data frame are checked to ensure that they are syntactically valid
variable names. If necessary they are adjusted (by make.names) so that
they are, and also to ensure that there are no duplicates.
The only consequence is that you need to escape or quote the names to work with them. You either string-quote and use standard evaluation with the [[ column subsetting operator:
df[['2019-Jan']]
… or you escape the identifier name with backticks (R confusingly also calls this quoting), and use the $ subsetting:
df$`2019-Jan`
Both work, and can be used freely (as long as they don’t lead to exceedingly unreadable code).
To make matters more confusing, R allows using '…' and "…" instead of `…` in certain contexts:
df$'2019-Jan'
Here, '2019-Jan' is not a character string as far as R is concerned! It’s an escaped identifier name.1
This last one is a really bad idea because it confuses names2 with character strings, which are fundamentally different. The R documentation advises against this. Personally I’d go further: writing 'foo' instead of `foo` to refer to a name should become a syntax error in future versions of R.
1 Kind of. The R parser treats it as a character string. In particular, both ' and " can be used, and are treated identically. But during the subsequent evaluation of the expression, it is treated as a name.
2 “Names”, or “symbols”, in R refer to identifiers in code that denote a variable or function parameter. As such, a name is either (a) a function name, (b) a non-function variable name, (c) a parameter name in a function declaration, or (d) an argument name in a function call.
The NA issue is unrelated to the names. read.csv is expecting an input with no comma after the last column. You have a comma after the last column, so read.csv reads the blank space after "2019-Feb", as the column name of the third column. There is no data for this column, so an NA value is assigned.
Remove the extra comma and it reads properly. Of course, it may be easier to just remove the last column after using read.csv.
df <- read.csv(
text = '"2019-Jan","2019-Feb"
"3","1"',
check.names = FALSE
)
df
# 2019-Jan 2019-Feb
# 1 3 1
Consider df$foo where foo is a column name. Syntactically invalid names will not work.
As for the NA it’s a consequence of there being three columns in your first line and only two in your second.

Assign names to list elements without titled quotes

I am interested to assign names to list elements. To do so I execute the following code:
file_names <- gsub("\\..*", "", doc_csv_names)
print(file_names)
"201409" "201412" "201504" "201507" "201510" "201511" "201604" "201707"
names(docs_data) <- file_names
In this case the name of the list element appears with ``.
docs_data$`201409`
However, in this case the name of the list element appears in the following way:
names(docs_data) <- paste("name", 1:8, sep = "")
docs_data$name1
How can I convert the gsub() result to receive the latter naming pattern without quotes?
gsub() and paste () seem to produce the same class () object. What is the difference?
Both gsub and paste return character objects. They are different because they are completely different functions, which you seem to know based on their usage (gsub replaces instances of your pattern with a desired output in a string of characters, while paste just... pastes).
As for why you get the quotations, that has nothing to do with gsub and everything to do with the fact that you are naming variables/columns with numbers. Indeed, try
names(docs_data) <- paste(1:8)
and you'll realize you have the same problem when invoking the naming pattern. It basically has to do with the fact that R doesn't want to be confused about whether a number is really a number or a variable because that would be chaos (how can 1 refer to a variable and also the number 1?), so what it does in such cases is change a number 1 into the character "1", which can be given names. For example, note that
> 1 <- 3
Error in 1 <- 3 : invalid (do_set) left-hand side to assignment
> "1" <- 3 #no problem!
So R is basically correcting that for you! This is not a problem when you name something using characters. Finally, an easy fix: just add a character in front of the numbers of your naming pattern, and you'll be able to invoke them without the quotations. For example:
file_names <- paste("file_",gsub("\\..*", "", doc_csv_names),sep="")
Should do the trick (or just change the "file_" into whatever you want as long as it's not empty, cause then you just have numbers left and the same problem)!

Data Frame containing hyphens using R

I have created a list (Based on items in a column) in order to subset my dataset into smaller datasets relating to a particular variable. This list contains strings with hyphens in them -.
dim.list <- c('Age_CareContactDate-Gender', 'Age_CareContactDate-Group',
'Age_ServiceReferralReceivedDate-Gender',
'Age_ServiceReferralReceivedDate-Gender-0-18',
'Age_ServiceReferralReceivedDate-Group',
'Age_ServiceReferralReceivedDate-Group-ReferralReason')
I have then written some code to loop through each item in this list subsetting my main data.
for (i in dim.list) {assign(paste("df1.",i,sep=""),df[df$Dimension==i,])}
This works fine, however when I come to aggregate this in order to get some summary statistics I can't reference the dataset as R stops reading after the hyphen (I assume that the hyphen is some special character)
If I use a different list without hyphens e.g.
dim.list.abr <- c('ACCD_Gen','ACCD_Grp',
'ASRRD_Gen',
'ASRRD_Gen_0_18',
'ASRRD_Grp',
'ASRRD_Grp_RefRsn')
When my for loop above executes I get 6 data.frames with no observations.
Why is this happening?
Comment to answer:
Hyphens aren't allowed in standard variable names. Think of a simple example: a-b. Is it a variable name with a hyphen or is it a minus b? The R interpreter assumes a minus b, because it doesn't require spaces for binary operations. You can force non-standard names to work using backticks, e.g.,
# terribly confusing names:
`a-b` <- 5
`x+y` <- 10
`mean(x^2)` <- "this is awful"
but you're better off following the rules and using standard names without special characters like + - * / % $ # # ! & | ^ ( [ ' " in them. At ?quotes there is a section on Names and Identifiers:
Identifiers consist of a sequence of letters, digits, the period (.) and the underscore. They must not start with a digit nor underscore, nor with a period followed by a digit. Reserved words are not valid identifiers.
So that's why you're getting an error, but what you're doing isn't good practice. I completely agree with Axeman's comments. Use split to divide up your data frame into a list. And keep it in a list rather than use assign, it will be much easier to loop over or use lapply with that way. You might want to read my answer at How to make a list of data frames for a lot of discussion and examples.
Regarding your comment "dim.list is not the complete set of unique entries in the Dimensions column", that just means you need to subset before you split:
nice_list = df[df$Dimension %in% dim.list, ]
nice_list = split(nice_list, nice_list$Dimension)

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