I have a discrete variable with scores from 1-3. I would like to change it so 1=2, 2=1, 3=3.
I have tried
recode(Data$GEB43, "c(1=2; 2=1; 3=3")
But that doesn't work.
I know this is an overly stupid question that can be solved in excel within seconds but trying to learn how to do basics like this in R.
We should always provide a minimal reproducible example:
df <- data.frame(x=c(1,1,2,2,3,3))
You didn't specifiy the package for recode so I assumed dplyr. ?dplyr::recode tells us how the arguments should be passed to the function. In the original question "c(1=2; 2=1; 3=3" is a string (i.e. not an R expression but a character string "c(1=2; 2=1; 3=3"). To make it an R expression we have to get rid of the double quotes and replace the ; with ,. Additionally, we need a closing bracket i.e. c(1=2, 2=1, 3=3). But still, as ?dplyr::recode tells us, this is not the way to pass this information to recode:
Solution using dplyr::recode:
dplyr::recode(df$x, "1"=2, "2"=1, "3"=3)
Returns:
[1] 2 2 1 1 3 3
Assuming, you mean dplyr::recode, the syntax is
recode(.x, ..., .default = NULL, .missing = NULL)
From the documentation it says
.x - A vector to modify
... - Replacements. For character and factor .x, these should be named and replacement is based only on their name. For numeric .x, these can be named or not. If not named, the replacement is done based on position i.e. .x represents positions to look for in replacements
So when you have numeric value you can replace based on position directly
recode(1:3, 2, 1, 3)
#[1] 2 1 3
Related
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.
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$foo where foo is a column name. Syntactically invalid names will not work.
As for the NA it’s a consequence of there being three columns in your first line and only two in your second.
I have a dataframe, wineSA, with two columns. One of these columns is populated with a character string, like so:
> summary(wineSA$description)
Length Class Mode
129971 character character
An example of a typical entry would be:
review <- "Aromas include tropical fruit, broom, brimstone and dried herb. The palate isn't overly expressive, offering unripened apple, citrus and dried sage alongside brisk acidity."
I also have a function, that when applied to a string returns a sentiment score, like so:
> Getting_Sentimental(review)
[1] 0.4317412
I want to apply this function to every element in the wineSA$description column and add the sentiment score, as a separate column, to the data frame wineSA.
I have tried the following method, which uses apply(), but I get this message:
> wineSA$reviewSentiment <- apply(wineSA$description, FUN = Getting_Sentimental)
Error in apply(wineSA$description, FUN = Getting_Sentimental) :
dim(X) must have a positive length
I'm not sure apply() is appropriate here, but when I use either sapply() or lappy() it populates the new column with the same value for the sentiment.
Is there a special way of handling functions on string characters? Is there anything I'm missing?
Thanks
If you load the pracma package into the r console and type
gammainc(2,2)
you get
lowinc uppinc reginc
0.5939942 0.4060058 0.5939942
This looks like some kind of a named tuple or something.
But, I can't work out how to extract the number below the lowinc, namely 0.5939942. The code (gammainc(2,2))[1] doesn't work, we just get
lowinc
0.5939942
which isn't a number.
How is this done?
As can be checked with str(gammainc(2,2)[1]) and class(gammainc(2,2)[1]), the output mentioned in the OP is in fact a number. It is just a named number. The names used as attributes of the vector are supposed to make the output easier to understand.
The function unname() can be used to obtain the numerical vector without names:
unname(gammainc(2,2))
#[1] 0.5939942 0.4060058 0.5939942
To select the first entry, one can use:
unname(gammainc(2,2))[1]
#[1] 0.5939942
In this specific case, a clearer version of the same might be:
unname(gammainc(2,2)["lowinc"])
Double brackets will strip the dimension names
gammainc(2,2)[[1]]
gammainc(2,2)[["lowinc"]]
I don't claim it to be intuitive, or obvious, but it is mentioned in the manual:
For vectors and matrices the [[ forms are rarely used, although they
have some slight semantic differences from the [ form (e.g. it drops
any names or dimnames attribute, and that partial matching is used for
character indices).
The partial matching can be employed like this
gammainc(2, 2)[["low", exact=FALSE]]
In R vectors may have names() attribute. This is an example:
vector <- c(1, 2, 3)
names(vector) <- c("first", "second", "third")
If you display vector, you should probably get desired output:
vector
> vector
first second third
1 2 3
To ensure what type of output you get after the function you can use:
class(your_function())
I hope this helps.
I have been working with R for about 2 months and have had a little bit of trouble getting a hold of how the $ and %% terms.
I understand I can use the $ term to pull a certain value from a function (e.g. t.test(x)$p.value), but I'm not sure if this is a universal definition. I also know it is possible to use this to specify to pull certain data.
I'm also curious about the use of the %% term, in particular, if I am placing a value in between it (e.g. %x%) I am aware of using it as a modulator or remainder e.g. 7 %% 5 returns 2. Perhaps I am being ignorant and this is not real?
Any help or links to literature would be greatly appreciated.
Note: I have been searching for this for a couple hours so excuse me if I couldn't find it!
You are not really pulling a value from a function but rather from the list object that the function returns. $ is actually an infix that takes two arguments, the values preceding and following it. It is a convenience function designed that uses non-standard evaluation of its second argument. It's called non-standard because the unquoted characters following $ are first quoted before being used to extract a named element from the first argument.
t.test # is the function
t.test(x) # is a named list with one of the names being "p.value"
The value can be pulled in one of three ways:
t.test(x)$p.value
t.test(x)[['p.value']] # numeric vector
t.test(x)['p.value'] # a list with one item
my.name.for.p.val <- 'p.value'
t.test(x)[[ my.name.for.p.val ]]
When you surround a set of characters with flanking "%"-signs you can create your own vectorized infix function. If you wanted a pmax for which the defautl was na.rm=TRUE do this:
'%mypmax%' <- function(x,y) pmax(x,y, na.rm=TRUE)
And then use it without quotes:
> c(1:10, NA) %mypmax% c(NA,10:1)
[1] 1 10 9 8 7 6 7 8 9 10 1
First, the $ operator is for selecting an element of a list. See help('$').
The %% operator is the modulo operator. See help('%%').
The '$' operator is used to select particular element from a list or any other data component which contains sub data components.
For example: data is a list which contains a matrix named MATRIX and other things too.
But to get the matrix we write,
Print(data$MATRIX)
The %% operator is a modulus operator ; which provides the remainder.
For example: print(7%%3)
Will print 1 as an output