# [R] extract parts of a list before symbol

@vi@e@gross m@iii@g oii gm@ii@com @vi@e@gross m@iii@g oii gm@ii@com
Fri May 26 18:09:13 CEST 2023

```Evan,

Yes, once you know a bit about the details, all kinds of functions are
available to solve problems without going the hard way.

But the names() function is taught fairly widely and did you also pick up on
the fact that it can be used on both sides so it also sets the names?

> # Create a list with mixed case names
> named <- list(age=69, Month="October", YEAR=1492)

> # Save the names to be reset later
> old_names <- names(named)

> # Show current names for the items in named
> print(names(named))
[1] "age"   "Month" "YEAR"

> # Replace the current names with an upper case version starting with
"var_"
> names(named) <- paste0("var_", toupper(names(named)))

> # Show current namesfor the items in named
> print(names(named))
[1] "var_AGE"   "var_MONTH" "var_YEAR"

> # Reset the names to the original names
> names(named) <- old_names

> # Show current namesfor the items in named
> print(names(named))
[1] "age"   "Month" "YEAR"

Look out for other such functions that work on both the LHS and RHS.

> evens <- seq(0,20,2)
> evens
[1]  0  2  4  6  8 10 12 14 16 18 20
> length(evens)
[1] 11
> length(evens) <- 5
> evens
[1] 0 2 4 6 8

If you look just a bit under the hood, R stores all kinds of things about a
variable that you can change with the right function. You can change the
dimensions of a vector for example with dim() as well as get the current
dimensions.

-----Original Message-----
From: R-help <r-help-bounces using r-project.org> On Behalf Of Evan Cooch
Sent: Friday, May 26, 2023 10:38 AM
To: r-help using r-project.org
Subject: Re: [R] extract parts of a list before symbol

Many thanks to all. Wasn't even aware of the names function. That does
the trick for present purposes.

On 5/26/2023 12:02 AM, avi.e.gross using gmail.com wrote:
> All true Jeff, but why do things the easy way! LOL!
>
> My point was that various data structures, besides the list we started
with,
> store the names as an attribute. Yes, names(listname) works fine to
extract
> whatever parts they want. My original idea of using a data.frame was
because
> it creates names when they are absent.
>
> And you are correct that if the original list was not as shown with only
all
> items of length 1, converting to a data.frame fails.
>
> >From what you say, it is a harder think to write a function that returns
a
> "name" for column N given a list. As you note, you get a null when there
are
> no names.  You get empty strings when one or more (but not all) have no
> names. But it can be done.
>
> The OP initially was looking at a way to get a text version of a variable
> they could use using perhaps regular expressions to parse.  Of course that
> is not as easy as just looking at the names attribute in one of several
> ways. But it may help in a sense to deal with the cases mentioned above.
> The problem is that str() does not return anything except to stdout so it
> must be captured to do silly things.
>
>> test <- list(a=3,b=5,c=11)
>> str(test)
> List of 3
>   \$ a: num 3
>   \$ b: num 5
>   \$ c: num 11
>
>> str(test[1])
> List of 1
>   \$ a: num 3
>
>> str(test[2])
> List of 1
>   \$ b: num 5
>
>> str(list(a=1, 2, c=3))
> List of 3
>   \$ a: num 1
>   \$  : num 2
>   \$ c: num 3
>
>> str(list(1, 2, 3))
> List of 3
>   \$ : num 1
>   \$ : num 2
>   \$ : num 3
>
>> text <- str(list(a=1, 2, c=3)[1])
> List of 1
>   \$ a: num 1
>
>> text <- capture.output(str(list(a=1, 2, c=3)))
>> text
> [1] "List of 3"   " \$ a: num 1" " \$  : num 2" " \$ c: num 3"
> So you could use some imaginative code that extracts what you want. I
> repeat, this is not a suggested way nor the best, just something that
seems
> to work:
>
>> sub("(^[\\\$ ]*)(\\w+|)(:.*\$)", "\\2", text[2:length(text)])
> [1] "a" ""  "c"
>
> Obviously the first line of output needs to be removed as it does not fit
> the pattern.
>
> Perhaps in this case a way less complex way is to use summary() rather
than
> str as it does return the output as text.
>
>> summary(list(a=1, 2, c=3)) -> text
>> text
>    Length Class  Mode
> a 1      -none- numeric
>    1      -none- numeric
> c 1      -none- numeric
>
> This puts the variable name, if any, at the start but parsing that is not
> trivial as it is not plain text.
>
> Bottom line, try not to do things the hard way. Just carefully use names()
> ...
>
> -----Original Message-----
> From: R-help <r-help-bounces using r-project.org> On Behalf Of Jeff Newmiller
> Sent: Thursday, May 25, 2023 10:32 PM
> To: r-help using r-project.org
> Subject: Re: [R] extract parts of a list before symbol
>
> What a remarkable set of detours, Avi, all deriving apparently from a few
> gaps in your understanding of R.
>
> As Rolf said, "names(test)" is the answer.
>
> a) Lists are vectors. They are not atomic vectors, but they are vectors,
so
> as.vector(test) is a no-op.
>
> test <- list( a = 1, b = 2, c=3 )
> attributes(test)
> attributes(as.vector(test))
>
> (Were you thinking of the unlist function? If so, there is no reason to
> convert the value of the list to an atomic vector in order to look at the
> value of an attribute of that list.)
>
> b) Data frames are lists, with the additional constraint that all elements
> have the same length, and that a names attribute and a row.names attribute
> are both required. Converting a list to a data frame to get the names is
> expensive in CPU cycles and breaks as soon as the list elements have a
> variety of lengths.
>
> c) All data in R is stored as vectors. Worrying about whether a data value
> is a vector is pointless.
>
> d) All objects can have attributes, including the name attribute. However,
> not all objects must have a name attribute... including lists. Omitting a
> name for any of the elements of a list in the constructor will lead to
> having a zero-length character values in the name attribute where the
names
> were omitted. Omitting all names in the list constructor will cause no
names
> attribute to be created for that list.
>
> test2 <- list( 1, 2, 3 )
> attributes(test2)
>
> e) The names() function returns the value of the names attribute. If that
> attribute is missing, it returns NULL. For dataframes, the colnames
function
> is equivalent to the names function (I rarely use the colnames function).
> For lists, colnames returns NULL... there are no "columns" in a list,
> because there is no constraint on the (lengths of the) contents of a list.
>
> names(test2)
>
> f) The names attribute, if it exists, is just a character vector. It is
> never necessary to convert the output of names() to a character vector. If
> the names attribute doesn't exist, then it is up to the user to write code
> that creates it.
>
> names(test2) <- c( "A", "B", "C" )
> attributes(test2)
> names(test2)
> # or use the argument names in the list function
>
> names(test2) <- 1:3 # integer
> names(test2) # character
> attributes(test2)\$names <- 1:3 # integer
> attributes(test2) # character
> test2[[ "2" ]] == 2  # TRUE
> test2\$`2`  == 2 # TRUE
>
>
>
> On May 25, 2023 6:17:37 PM PDT, avi.e.gross using gmail.com wrote:
>> Evan,
>>
>> List names are less easy than data.frame column names so try this:
>>
>>> test <- list(a=3,b=5,c=11)
>>> colnames(test)
>> NULL
>>> colnames(as.data.frame(test))
>> [1] "a" "b" "c"
>>
>> But note an entry with no name has one made up for it.
>>
>>
>>> test2 <- list(a=3,b=5, 666, c=11)
>>> colnames(data.frame(test2))
>> [1] "a"    "b"    "X666" "c"
>>
>> But that may be overkill as simply converting to a vector if ALL parts
are
>> of the same type will work too:
>>
>>> names(as.vector(test))
>> [1] "a" "b" "c"
>>
>> To get one at a time:
>>
>>> names(as.vector(test))[1]
>> [1] "a"
>>
>> You can do it even simple by looking at the attributes of your list:
>>
>>> attributes(test)
>> \$names
>> [1] "a" "b" "c"
>>
>>> attributes(test)\$names
>> [1] "a" "b" "c"
>>> attributes(test)\$names[3]
>> [1] "c"
>>
>>
>> -----Original Message-----
>> From: R-help <r-help-bounces using r-project.org> On Behalf Of Evan Cooch
>> Sent: Thursday, May 25, 2023 1:30 PM
>> To: r-help using r-project.org
>> Subject: [R] extract parts of a list before symbol
>>
>> Suppose I have the following list:
>>
>> test <- list(a=3,b=5,c=11)
>>
>> I'm trying to figure out how to extract the characters to the left of
>> the equal sign (i.e., I want to extract a list of the variable names, a,
>> b and c.
>>
>> I've tried the permutations I know of involving sub - things like
>> sub("\\=.*", "", test), but no matter what I try, sub keeps returning
>> (3, 5, 11). In other words, even though I'm trying to extract the
>> 'stuff' before the = sign, I seem to be successful only at grabbing the
>> stuff after the equal sign.
>>
>> Pointers to the obvious fix? Thanks...
>>
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