[R] Row exclude
Rui Barradas
ru|pb@rr@d@@ @end|ng |rom @@po@pt
Sat Jan 29 19:33:22 CET 2022
Hello,
Thanks for the comments, a few others inline.
Às 18:04 de 29/01/2022, Avi Gross escreveu:
> There are many creative ways to solve problems and some may get you in
> trouble if you present them in class while even in some work situations,
> they may be hard for most to understand, let alone maintain and make
> changes.
>
> This group is amorphous enough that we have people who want "help" who
> are new to the language, but also people who know plenty and encounter a
> new kind of problem, and of course people who want to make use of what
> they see as free labor.
>
> Rui presented a very interesting idea and I like some aspects. But if
> presented to most people, they might have to start looking up things.
>
> But I admit I liked some of the ideas he uses and am adding them to my
> bag of tricks. Some were overkill for this particular requirement but
> that also makes them more general and useful.
>
> First, was the use of locale-independent regular expressions like
> [[:alpha:]] that match any combination of [:lower:] and [:upper:] and
> thus are not restricted to ASCII characters. Since I do lots of my
> activities in languages other than English and well might include names
> with characters not normally found in English, or not even using an
> overlapping alphabet, I can easily encounter items in the Name column
> that might not match [A-Za-z] but will match with [:alpha:].
>
> I don't know if using [:digit:] has benefits over [0-9] and I do note
> there was no requirement to match more complex numbers than integers so
> no need to allow periods or scientific notation and so on.
Yes, I used locale-independent regular expressions. It's a habit I
aquired a while ago. It took some time to stop using character ranges
but once gone I'm more comfortable with the use of classes like
[:alpha:] and [:digit:].
[After all my native language, (Portuguese) has
cedillas(ES)/cedilhas(PT) and accented letters].
>
> Then there is the use of mapply. The more general version of the problem
> presented would include a data.frame with any number of columns, where a
> subset of the columns might need to be checked for conditions that vary
> across the columns but may include some broad categories of conditions
> that might be re-used. If all the conditions are regular expression
> matches you can build, then you can extend the list Rui used to have
> more items and also include expressions that always match so that some
> columns are effectively ignored:
>
>
> regex <- list("[[:digit:]]", "[[:alpha:]]", "[[:alpha:]]", "[.*])
>
>
> So this generalizes to N columns as long as you supply exactly N
> patterns in the list, albeit mapply does recycle arguments if needed as
> in the simplest case where you want all columns checked the same way.
>
> Rui then uses an anonymous function to pass to mapply() and that is a
> newish feature added recently to R, I think. It was perhaps meant
> specifically to be used with the new pipe symbol, but can be used
> anywhere but perhaps not in older versions of R.
>
>
> \(x, r) grepl(r, x)
>
No, the new anonymous function wasn't specifically meant to be used with
the new pipe operator, it was meant to be a short-hand notation for
anonymous functions and used interchangeably with the old notation.
mapply(\(x, r), etc)
mapply(function(x, r) etc)
>
> I note Rui also uses grepl() which returns a logical vector. I will show
> my first attempt at the end where I used grep() to return index numbers
> of matches instead. For this context, though, he made use of the fact
> that mapply in this case returns a matrix of type logical:
>
> i <- mapply(\(x, r) grepl(r, x), dat1, regex)
>
>> i
>
> Name Age Weight
> [1,] FALSE FALSE TRUE
> [2,] FALSE FALSE FALSE
> [3,] FALSE FALSE FALSE
> [4,] FALSE TRUE FALSE
> [5,] FALSE FALSE FALSE
> [6,] TRUE FALSE FALSE
>
> And since R treats TRUE as 1 and FALSE as 0, then summing the rows gives
> you a small integer between 0 and the number of columns, inclusive, and
> only rows with no TRUE in them are wanted for this purpose:
And rowSums is a fast function.
>
>
> dat1[rowSums(i) == 0L, ]
>
> All I all, nicely done, but not trivial to read without comments, LOL!
>
> And, yes, it could be made even more obscure as a one-liner.
>
> My first attempt was a bit more focused on the specific needs described.
> I am not sure how the HTML destroyer in this mailing list might wreck
> it, but I made it a two-statement version that is formatted on multiple
> lines. An explanation first.
>
> I looked at using grep() on one column at a time to look for what should
> NOT be there and ask it to invert the answer so it effectively tells me
> which rows to keep. So it tests column 1 ($Name) to see if it has digits
> in it and returns FALSE if it finds them which later means toss this
> row. It returns TRUE if that entry, so far, makes the row valid. But
> note since I am not using grepl() it does not return TRUE/FALSE at all.
> Rather it returns index numbers of the ones that now inverted are TRUE.
> What goes in is a vector of individual items from a column of the data.
> What goes out is the indices of which ones I want to keep that can be
> used to index the entire data.frame. Based on the ample data, it returns
> 1:5 as row 6 has a digit in "Jack3".
>
>
> grep("[0-9]", dat1$Name, invert = TRUE)
>
>
> Similarly, two other grep() statements test if the second and third
> columns contain any characters in "[a-zA-Z]" and return a similar index
> vector if they are OK.
>
> What I would then have are three numeric vectors, not a matrix. Each
> contains a subset of all the indices:
>
>
>> grep("[0-9]", dat1$Name, invert = TRUE)
> [1] 1 2 3 4 5
>> grep("[a-zA-Z]", dat1$Age, invert = TRUE)
> [1] 1 2 3 5 6
>> grep("[a-zA-Z]", dat1$Weight, invert = TRUE)
> [1] 2 3 4 5 6
>
> This set of data was designed to toss out one of each column so they all
> are of the same length but need not be. Like Rui, my condition for
> deciding which rows to keep is that all three of the index vectors have
> a particular entry. He summed them as logicals, but my choice has small
> integers so the way I combine them to exclude any not in all three is to
> use a sort of set intersect method. The one built-in to R only handles
> two at a time so I nested two calls to intersect but in a more general
> case, I would use some package (or build my own function) that handles
> intersecting any number of such items.
>
> Here is the full code, minus the initialization.
>
>
> rows.keep <-
> intersect(intersect(grep("[0-9]", dat1$Name, invert = TRUE),
> grep("[a-zA-Z]", dat1$Age, invert = TRUE)),
> grep("[a-zA-Z]", dat1$Weight, invert = TRUE))
> result <- dat1[rows.keep,]
>
>
Using the same idea, another two options, both with Reduce.
The 1st uses Avi's grep and regex's, the latter could be the character
classes "[[:alpha:]]" and "[[:digit:]]" but this code is inspired in
his. The results are put on a list and Reduce intersects the list
members. Then subsetting is as usual.
The 2nd uses the fact that Mapis a wrapper for mapply that defaults to
not simplifying its output. grep/invert will find the non-matches and
Reduce intersects the result list, as above.
From ?Map:
Map is a simple wrapper to mapply which does not attempt to simplify the
result, similar to Common Lisp's mapcar (with arguments being recycled,
however). Future versions may allow some control of the result type.
# 1st
grep_list <- list(
grep("[0-9]", dat1$Name, invert = TRUE),
grep("[a-zA-Z]", dat1$Age, invert = TRUE),
grep("[a-zA-Z]", dat1$Weight, invert = TRUE)
)
keep1 <- Reduce(intersect, grep_list)
dat1[keep1,]
# 2nd
keep2 <- Map(\(x, r) grep(r, x, invert = TRUE), dat1, regex)
keep2 <- Reduce(intersect, keep2)
identical(keep1, keep2)
#[1] TRUE
Hope this helps,
Rui Barradas
>
>
>
>
>
>
>
>
>
> -----Original Message-----
> From: Rui Barradas <ruipbarradas using sapo.pt>
> To: David Carlson <dcarlson using tamu.edu>; Bert Gunter <bgunter.4567 using gmail.com>
> Cc: r-help using R-project.org (r-help using r-project.org) <r-help using r-project.org>
> Sent: Sat, Jan 29, 2022 3:46 am
> Subject: Re: [R] Row exclude
>
> Hello,
>
> Getting creative, here is another way with mapply.
>
>
> regex <- list("[[:digit:]]", "[[:alpha:]]", "[[:alpha:]]")
>
> i <- mapply(\(x, r) grepl(r, x), dat1, regex)
> dat1[rowSums(i) == 0L, ]
>
> # Name Age Weight
> #2 Bob 25 142
> #3 Carol 24 120
> #5 Katy 35 160
>
>
> Hope this helps,
>
> Rui Barradas
>
>
> Às 06:30 de 29/01/2022, David Carlson via R-help escreveu:
> > Given that you know which columns should be numeric and which should be
> > character, finding characters in numeric columns or numbers in character
> > columns is not difficult. Your data frame consists of three character
> > columns so you can use regular expressions as Bert mentioned. First you
> > should strip the whitespace out of your data:
> >
> > dat1 <-read.table(text="Name, Age, Weight
> > Alex, 20, 13X
> > Bob, 25, 142
> > Carol, 24, 120
> > John, 3BC, 175
> > Katy, 35, 160
> > Jack3, 34, 140",sep=",", header=TRUE, stringsAsFactors=FALSE,
> > strip.white=TRUE)
> >
> > Now check to see if all of the fields are character as expected.
> >
> > sapply(dat1, typeof)
> > # Name Age Weight
> > # "character" "character" "character"
> >
> > Now identify character variables containing numbers and numeric variables
> > containing characters:
> >
> > BadName <- which(grepl("[[:digit:]]", dat1$Name))
> > BadAge <- which(grepl("[[:alpha:]]", dat1$Age))
> > BadWeight <- which(grepl("[[:alpha:]]", dat1$Weight))
> >
> > Next remove those rows:
> >
> > (dat2 <- dat1[-unique(c(BadName, BadAge, BadWeight)), ])
> > # Name Age Weight
> > # 2 Bob 25 142
> > # 3 Carol 24 120
> > # 5 Katy 35 160
> >
> > You still need to convert Age and Weight to numeric, e.g. dat2$Age <-
> > as.numeric(dat2$Age).
> >
> > David Carlson
> >
> >
> > On Fri, Jan 28, 2022 at 11:59 PM Bert Gunter <bgunter.4567 using gmail.com
> <mailto:bgunter.4567 using gmail.com>> wrote:
> >
> >> As character 'polluted' entries will cause a column to be read in (via
> >> read.table and relatives) as factor or character data, this sounds
> like a
> >> job for regular expressions. If you are not familiar with this subject,
> >> time to learn. And, yes, ZjQcmQRYFpfptBannerStart
> >> This Message Is From an External Sender
> >> This message came from outside your organization.
> >> ZjQcmQRYFpfptBannerEnd
> >>
> >> As character 'polluted' entries will cause a column to be read in (via
> >> read.table and relatives) as factor or character data, this sounds
> like a
> >> job for regular expressions. If you are not familiar with this subject,
> >> time to learn. And, yes, some heavy lifting will be required.
> >> See ?regexp for a start maybe? Or the stringr package?
> >>
> >> Cheers,
> >> Bert
> >>
> >>
> >>
> >>
> >> On Fri, Jan 28, 2022, 7:08 PM Val <valkremk using gmail.com
> <mailto:valkremk using gmail.com>> wrote:
> >>
> >>> Hi All,
> >>>
> >>> I want to remove rows that contain a character string in an integer
> >>> column or a digit in a character column.
> >>>
> >>> Sample data
> >>>
> >>> dat1 <-read.table(text="Name, Age, Weight
> >>> Alex, 20, 13X
> >>> Bob, 25, 142
> >>> Carol, 24, 120
> >>> John, 3BC, 175
> >>> Katy, 35, 160
> >>> Jack3, 34, 140",sep=",",header=TRUE,stringsAsFactors=F)
> >>>
> >>> If the Age/Weight column contains any character(s) then remove
> >>> if the Name column contains an digit then remove that row
> >>> Desired output
> >>>
> >>> Name Age weight
> >>> 1 Bob 25 142
> >>> 2 Carol 24 120
> >>> 3 Katy 35 160
> >>>
> >>> Thank you,
> >>>
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