[R] [FORGED] Re: Help needed for one question (Urgent)

[Wang Jiefei]

Agree, especially there is an "Urgent" on the title. He must be too
"urgent" to think about your answer. I will wonder if your effort will be
in vain.

Best,
Jiefei



On Tue, Nov 5, 2019 at 4:52 PM Rolf Turner <r.turner using auckland.ac.nz> wrote:

>
> Richard:  I know that you mean well, but *please* don't do people's
> homework for them!!!  (They are *cheating* by asking R-help to do their
> homework.)
>
> cheers,
>
> Rolf Turner
>
> On 6/11/19 4:27 AM, Richard O'Keefe wrote:
> > This looks vaguely like something from exercism.
> > Let's approach it logically.
> >   xa xb xc ya yb zc
> > We see two patterns here:
> > A:  x x x y y z
> > B: a b c a b c
> > If only we had these two character vectors, we could use
> >   paste(A, B, sep = "")
> > to get the desired result.  So now we have reduced the
> > problem to two simpler subproblems.  We have been given
> > a clue that rep() might be useful.
> > A: rep(c("x", "y", "z"), c(1, 2, 3))
> > B: rep(c("a", "b", "c"), 3)
> > But you were told not to use c().  So now we have three
> > simpler subsubproblems:
> > C: "x" "y" "z"
> > D: 3 2 1
> > E: "a" "b" "c"
> > You were given another hint.  seq().  That builds a vector of numbers.
> > Reading ?seq will give you
> > D: seq(from = 3, to = 1, by = -1)
> > or using ":" syntax,
> > D: 3:1
> >
> > What about C and E?  This needs two more pieces of knowledge:
> > - the variable letters,whose value is c("a","b",...,"y","z")
> > - how vector indexing works in R.
> > E: letters[1:3]
> > C: letters[24:26]
> > So now we can put all the pieces together:
> > paste(rep(letters[24:26], 3:1), rep(letters[1:3], 2), sep = "")
> >
> > You were given
> >   - seq
> >   - rep
> > as hints.  You were expected to look up string handling in R
> > and find things like paste(), substr(), and nchar().
> >
> > What about the variable 'letters'?
> > Well, you were expected to know or find out about substr.
> > You were certainly expected to know about "vectorising".
> > So you would naturally try substr("abc", 1:3, 1:3).
> > And that would not work.
> > So you would be expected to read the documentation:
> > ?substr
> > And then you would find that substr() *doesn't* do what
> > you expect, but substring() *does*.  So
> > C: substring("xyz", 1:3, 1:3)
> > E: substring("abc", 1:3, 1:3)
> >
> > This is not really an exercise in R programming.
> > In real R programming you *don't* avoid arbitrary aspects of the
> > language and library, but use whatever is appropriate.
> > So what *is* this exercise about?
> >
> > (1) It is an exercise in working backwards.  (See the classic book
> > "How to Solve It" by Polya.)  You know what you must construct,
> > you have been given some directions about what to use.  It's
> > about saying "well, I could *finish* this task by doing this action,
> > so what would I have to set up for that?"  In this case, the key
> > step for me was seeing xa xb xc ya yb yc as (x,x,x,y,y,z)++(a,b,c,a,b,c).
> > The mention of rep had me *looking* for repetitions like that.
> >
> > (2) It is an exercise in using the R documentation to figure out how to
> > use rep and seq and what is available for splitting and pasting strings.
> >
> > There is of course no unique answer to this.
> > substring("xaxbxcyaybzc", seq(from=1,to=11,by=2), seq(from=2,to=12,by=2))
> > is another solution.  You didn't say you *had* to use rep.
> >
> > It's not the answer that matters for an exercise like this.
> > It's how you get there.
> >
> >
> >
> >
> >
> > On Tue, 5 Nov 2019 at 23:40, Chandeep Kaur <chandeep.virdi using gmail.com>
> wrote:
> >>
> >> Dear Team,
> >>
> >> Could you please help me with the below question? How can I get the
> desired
> >> output?
> >>
> >> Produce the following sequence using only rep(), seq() and potentially
> >> other functions/operators. You must not use c() nor explicit loops
> >>
> >> “xa” “xb” “xc” “ya” “yb” “zc”
> >>
> >> Thanks & Regards,
> >>
> >> Chandeep Kaur
>
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