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.
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)
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:
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,]
-----Original Message-----
From: Rui Barradas <ruipbarra...@sapo.pt>
To: David Carlson <dcarl...@tamu.edu>; Bert Gunter <bgunter.4...@gmail.com>
Cc: r-help@R-project.org (r-help@r-project.org) <r-help@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.4...@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 <valkr...@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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______________________________________________
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