26 Feb
2008
26 Feb
'08
7:01 p.m.
Hi 2001,
Solutions to problem set 1 are posted to the course website. We'll hand
back your hardcopies in section on Thursday, and for this problem set I'll
email a brief note and grade to those of you who submitted the entire
problem set electronically.
A couple of notes:
Please use the solutions on the website as a guide for future problem sets.
In particular, note how there are two files, one with R code (run-able; it's
a file we can open and run as-is in R) and one with solutions and
explanations. Submitting a separate R file is important so we can easily
run your code and see where things went awry (or well, of course).
If you use R to calculate something (such as a probability), include the
actual value that R estimates in your write-up; don't just include the code
without its output.
Many of you had clever approaches to the birthday problem- good work. In
the solutions we show two ways of estimating the room sizes in problem 2.
Even if you generated all correct answers, take a look at the solution
code. All of you can benefit from making your code more efficient.
Common mistakes on this problem set include 1d, 2a, 2b and 3d. In 1d, a
common error resulted from forgetting which gender corresponded to 1's and
which to 0's. In both parts of problem 2, a common difficulty was figuring
out how to find triplicate birthdays without counting two pairs of
birthdays. (We discuss this problem in the R file to the solutions).
Another mistake on 2 involved testing a few room sizes without resetting the
counter for 'samedays' so the estimate for the number of required people was
too low. In 3d, some struggled with drawing simulated datasets- you can
review the solutions and also the R code from section 2 for a slick way to
sample from a dataset.
And finally, don't worry if your solutions were not exactly those posted on
the solutions- these are all simulated estimates, so some variance is
expected and perfectly fine.
One problem set down...
Jenn
27 Feb
27 Feb
2:36 a.m.
New subject: [gov2001-l] Qu2
Can anyone help?
I'm trying to define the function for qu2
This works fine for individual values of x:
g <- function(x){
if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y <-
sin(x)*cos(x))
return(y)
}
But when I input a vector (in order to draw the graph) like this:
ruler <- seq(-2*pi, 2*pi, by=.01)
g(ruler)
I get this error message:
Warning message:
In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) * cos(x))
else (y <- sin(x) * :
the condition has length > 1 and only the first element will be used
It seems I've got too many conditions for 'if' when the input is a
vector. When I draw the graph of g(ruler)~ruler, I get sinxcosx on
the domain [-2pi, 2pi]. Any ideas on what I could do differently?
Thanks
Jeremy
Dr Jeremy Hodgen
Senior Lecturer in Mathematics Education
King's College London
Department of Education and Professional Studies
Franklin-Wilkins Building
Waterloo Bridge Wing
150 Stamford Street
London SE1 9NH
Tel: 020 7848 3102
Fax: 020 7848 3182
E-mail: jeremy.hodgen at kcl.ac.uk
3:05 a.m.
New subject: [gov2001-l] Qu2
Hi Jeremy,
One way to evaluate a function at multiple points is to write a for loop
that will evaluate your function for every point in ruler and store it in an
empty vector. Then you can plot the vector that your loop filled against
the ruler.
Cheers,
Jenn
On Wed, Feb 27, 2008 at 2:36 AM, Jeremy Hodgen <jeremy.hodgen at kcl.ac.uk>
wrote:
Can anyone help?
I'm trying to define the function for qu2
This works fine for individual values of x:
g <- function(x){
if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y <-
sin(x)*
cos(x))
return(y)
}
But when I input a vector (in order to draw the graph) like this:
ruler <- seq(-2*pi, 2*pi, by=.01)
g(ruler)
I get this error message:
Warning message:
In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) * cos(x)) else (y
<- sin(x) * :
the condition has length > 1 and only the first element will be used
It seems I've got too many conditions for 'if' when the input is a
vector. When I draw the graph of g(ruler)~ruler, I get sinxcosx on the
domain [-2pi, 2pi]. Any ideas on what I could do differently?
Thanks
Jeremy
Dr Jeremy Hodgen
Senior Lecturer in Mathematics Education
King's College London
Department of Education and Professional Studies
Franklin-Wilkins Building
Waterloo Bridge Wing
150 Stamford Street
London SE1 9NH
Tel: 020 7848 3102
Fax: 020 7848 3182
E-mail: jeremy.hodgen at kcl.ac.uk
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
7:12 a.m.
New subject: [gov2001-l] Qu2
The same thing happened to me when I wrote my function as:
g <- function(x){
ifelse(x>=(-pi/2) & x<=(pi/2), out <- 10*sin(x)*cos(x), out <-
sin(x)*cos(x))
out
}
As you said, it worked well for individual values - in fact, it worked fine for the
optimization
as well, but not for the graph.
Then I changed my program to
g <- function(x){
out <- ifelse(x>=(-pi/2) & x<=(pi/2), 10*sin(x)*cos(x), sin(x)*cos(x))
out
}
For no reason I know of, it works. It seems that in the previous case, R somehow always
treats
x>=(-pi/2) & x<=(pi/2) as false if any x in a vector falls out of this range.
Why moving "out" out
of the ifelse function would fix the problem is beyond me.
Qian
On Wed, 27 Feb 2008 07:36:57 +0000
Jeremy Hodgen <jeremy.hodgen at kcl.ac.uk> wrote:
Can anyone help?
I'm trying to define the function for qu2
This works fine for individual values of x:
g <- function(x){
if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y <-
sin(x)*cos(x))
return(y)
}
But when I input a vector (in order to draw the graph) like this:
ruler <- seq(-2*pi, 2*pi, by=.01)
g(ruler)
I get this error message:
Warning message:
In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) * cos(x))
else (y <- sin(x) * :
the condition has length > 1 and only the first element will be used
It seems I've got too many conditions for 'if' when the input is a
vector. When I draw the
graph of g(ruler)~ruler, I get sinxcosx on the domain [-2pi, 2pi]. Any ideas on what I
could do
differently?
Thanks
Jeremy
Dr Jeremy Hodgen
Senior Lecturer in Mathematics Education
King's College London
Department of Education and Professional Studies
Franklin-Wilkins Building
Waterloo Bridge Wing
150 Stamford Street
London SE1 9NH
Tel: 020 7848 3102
Fax: 020 7848 3182
E-mail: jeremy.hodgen at kcl.ac.uk 
8:13 a.m.
New subject: [gov2001-l] Qu2
Qian,
range. Why moving "out" out
of the ifelse function would fix the problem is beyond me.
This is a subtle point. Consider the two statements:
# 1
ifelse(x>=1, out <- x^2, out <- -x)
# 2
out <- ifelse(x>=1,x^2,-x)
Now 1 will lead to unanticipated behavior if the condition x>1 is true for
some elements of x but false for others. As long as any elements I true,
"out" will be defined by only the "yes" condition x^2. The reasons is
that R
out put R in a dilemma, it wants to evaluates both x^2 or -x but the output
is called "out" in both cases so out should be subsequently overwritten. In
this case R decides to evaluate all elements of x with only one condition,
the yes condition.
Expression 2 is not affected by this because the out object is defined as
the output of the whole ifelse() command, so R does what you want it to do,
evaluate elements by elements and puts it all together in the out vector
object at the end.
Notice that 1 will still work as you think if the condition x>1 is either
true or false for ALL elements of x, because then only one of the out
objects will be defined.
Always use 2, when in doubt.
Hope this helps.
Jens
For no reason I know of, it works. It seems that in
the previous case,
R somehow always treats
x>=(-pi/2) & x<=(pi/2) as false if any x in a vector falls out of this
> -----Original Message-----
> From: gov2001-l-bounces at lists.fas.harvard.edu [mailto:gov2001-l-
> bounces at lists.fas.harvard.edu] On Behalf Of Qian Guo
> Sent: Wednesday, February 27, 2008 7:12 AM
> To: gov2001-l at lists.fas.harvard.edu
> Subject: Re: [gov2001-l] Qu2
>
> The same thing happened to me when I wrote my function as:
> g <- function(x){
> ifelse(x>=(-pi/2) & x<=(pi/2), out <- 10*sin(x)*cos(x), out <-
> sin(x)*cos(x))
> out
> }
>
> As you said, it worked well for individual values - in fact, it worked
> fine for the optimization
> as well, but not for the graph.
>
> Then I changed my program to
> g <- function(x){
> out <- ifelse(x>=(-pi/2) & x<=(pi/2), 10*sin(x)*cos(x),
> sin(x)*cos(x))
> out
> }
>
For no reason I know of, it works. It seems that in
the previous case,
R somehow always treats
x>=(-pi/2) & x<=(pi/2) as false if any x in a vector falls out of this
range. Why moving "out" out
of the ifelse function would fix the problem is beyond me.
>
> Qian
>
>
>
> On Wed, 27 Feb 2008 07:36:57 +0000
> Jeremy Hodgen <jeremy.hodgen at kcl.ac.uk> wrote:
> > Can anyone help?
> >
> > I'm trying to define the function for qu2
> >
> > This works fine for individual values of x:
> >
> > g <- function(x){
> > if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y
> <- sin(x)*cos(x))
> > return(y)
> > }
> > But when I input a vector (in order to draw the graph) like this:
> >
> > ruler <- seq(-2*pi, 2*pi, by=.01)
> > g(ruler)
> >
> > I get this error message:
> >> Warning message:
> >> In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) *
cos(x))
> >> else (y <- sin(x) * :
> >> the condition has length > 1 and only the first element will be
> used
> >>
> >
> >
> > It seems I've got too many conditions for 'if' when the input is
a
> vector. When I draw the
> >graph of g(ruler)~ruler, I get sinxcosx on the domain [-2pi, 2pi].
> Any ideas on what I could do
> >differently?
> >
> > Thanks
> >
> > Jeremy
> >
> >
> >
> > Dr Jeremy Hodgen
> > Senior Lecturer in Mathematics Education
> > King's College London
> > Department of Education and Professional Studies
> >Franklin-Wilkins Building
> > Waterloo Bridge Wing
> > 150 Stamford Street
> > London SE1 9NH
> >
> > Tel: 020 7848 3102
> >Fax: 020 7848 3182
> > E-mail: jeremy.hodgen at kcl.ac.uk
> _______________________________________________
> gov2001-l mailing list
> gov2001-l at lists.fas.harvard.edu
> http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
8:31 a.m.
New subject: [gov2001-l] Qu2
Jens,
This is very helpful. Thanks a lot.
Qian
On Wed, 27 Feb 2008 08:13:18 -0500
"Jens Hainmueller" <jhainmueller at gmail.com> wrote:
Qian,
range. Why moving "out" out
of the ifelse function would fix the problem is beyond me.
This is a subtle point. Consider the two statements:
# 1
ifelse(x>=1, out <- x^2, out <- -x)
# 2
out <- ifelse(x>=1,x^2,-x)
Now 1 will lead to unanticipated behavior if the condition x>1 is true for
some elements of x but false for others. As long as any elements I true,
"out" will be defined by only the "yes" condition x^2. The reasons is
that R
out put R in a dilemma, it wants to evaluates both x^2 or -x but the output
is called "out" in both cases so out should be subsequently overwritten. In
this case R decides to evaluate all elements of x with only one condition,
the yes condition.
Expression 2 is not affected by this because the out object is defined as
the output of the whole ifelse() command, so R does what you want it to do,
evaluate elements by elements and puts it all together in the out vector
object at the end.
Notice that 1 will still work as you think if the condition x>1 is either
true or false for ALL elements of x, because then only one of the out
objects will be defined.
Always use 2, when in doubt.
Hope this helps.
Jens
For no reason I know of, it works. It seems that
in the previous case,
R somehow always treats
x>=(-pi/2) & x<=(pi/2) as false if any x in a vector falls out of this
> -----Original Message-----
> From: gov2001-l-bounces at lists.fas.harvard.edu [mailto:gov2001-l-
> bounces at lists.fas.harvard.edu] On Behalf Of Qian Guo
> Sent: Wednesday, February 27, 2008 7:12 AM
> To: gov2001-l at lists.fas.harvard.edu
> Subject: Re: [gov2001-l] Qu2
>
> The same thing happened to me when I wrote my function as:
> g <- function(x){
> ifelse(x>=(-pi/2) & x<=(pi/2), out <- 10*sin(x)*cos(x), out <-
> sin(x)*cos(x))
> out
> }
>
> As you said, it worked well for individual values - in fact, it worked
> fine for the optimization
> as well, but not for the graph.
>
> Then I changed my program to
> g <- function(x){
> out <- ifelse(x>=(-pi/2) & x<=(pi/2), 10*sin(x)*cos(x),
> sin(x)*cos(x))
> out
> }
>
For no reason I know of, it works. It seems that
in the previous case,
R somehow always treats
x>=(-pi/2) & x<=(pi/2) as false if any x in a vector falls out of this
range. Why moving "out" out
of the ifelse function would fix the problem is beyond me.
>
> Qian
>
>
>
> On Wed, 27 Feb 2008 07:36:57 +0000
> Jeremy Hodgen <jeremy.hodgen at kcl.ac.uk> wrote:
> > Can anyone help?
> >
> > I'm trying to define the function for qu2
> >
> > This works fine for individual values of x:
> >
> > g <- function(x){
> > if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y
> <- sin(x)*cos(x))
> > return(y)
> > }
> > But when I input a vector (in order to draw the graph) like this:
> >
> > ruler <- seq(-2*pi, 2*pi, by=.01)
> > g(ruler)
> >
> > I get this error message:
> >> Warning message:
> >> In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) *
cos(x))
> >> else (y <- sin(x) * :
> >> the condition has length > 1 and only the first element will be
> used
> >>
> >
> >
> > It seems I've got too many conditions for 'if' when the input is
a
> vector. When I draw the
> >graph of g(ruler)~ruler, I get sinxcosx on the domain [-2pi, 2pi].
> Any ideas on what I could do
> >differently?
> >
> > Thanks
> >
> > Jeremy
> >
> >
> >
> > Dr Jeremy Hodgen
> > Senior Lecturer in Mathematics Education
> > King's College London
> > Department of Education and Professional Studies
> >Franklin-Wilkins Building
> > Waterloo Bridge Wing
> > 150 Stamford Street
> > London SE1 9NH
> >
> > Tel: 020 7848 3102
> >Fax: 020 7848 3182
> > E-mail: jeremy.hodgen at kcl.ac.uk
> _______________________________________________
> gov2001-l mailing list
> gov2001-l at lists.fas.harvard.edu
> http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
http://lists.fas.harvard.edu/mailman/listinfo/gov2001-l
8:21 a.m.
New subject: [gov2001-l] Qu2
Jeremy,
The issue here is similar to my previous email, but slightly different. The
error:
the condition has length > 1 and only the first
element will be used
always occurs when you use an if(candition) statement, but the condition
evaluates to a vector of length k with k>1 (like c(TRUE,FALSE) or so). In
this case it is unclear which element R should use (ie. the TRUE or the
FALSE) so by default it goes with the first statement and also issues a
warning.
In this case, the ifelse() command (that triggers a different action
depending on element by element checking of the condition) is your best bet
to make the function work with vectors. See me previous email and Quan?s on
this.
Jens
From: gov2001-l-bounces at lists.fas.harvard.edu
[mailto:gov2001-l-bounces at lists.fas.harvard.edu] On Behalf Of Jeremy Hodgen
Sent: Wednesday, February 27, 2008 2:37 AM
To: gov2001-l at lists.fas.harvard.edu
Subject: Re: [gov2001-l] Qu2
Can anyone help?
I'm trying to define the function for qu2
This works fine for individual values of x:
g <- function(x){
if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y <-
sin(x)*cos(x))?
return(y)
}
But when I input a vector (in order to draw the graph) like this:
ruler <- seq(-2*pi, 2*pi, by=.01)?
g(ruler)
I get this error message:
Warning message:
In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) * cos(x)) else (y
<-
sin(x) * ?:
??the condition has length > 1 and only the first element will be used
It seems I've got too many conditions for 'if' when the input is a vector.
When I draw the graph of g(ruler)~ruler, I get sinxcosx on the domain [-2pi,
2pi]. Any ideas on what I could do differently?
Thanks
Jeremy
Dr Jeremy Hodgen
Senior Lecturer in Mathematics Education
King's College London
Department of Education and Professional Studies
Franklin-Wilkins Building
Waterloo Bridge Wing
150 Stamford Street
London SE1 9NH
Tel: 020 7848 3102
Fax: 020 7848 3182
E-mail: jeremy.hodgen at kcl.ac.uk
8:36 a.m.
New subject: [gov2001-l] Qu2
No wonder I didn't get the warning message (but the same wrong plot).
Qian
On Wed, 27 Feb 2008 08:21:10 -0500
"Jens Hainmueller" <jhainmueller at gmail.com> wrote:
Jeremy,
The issue here is similar to my previous email, but slightly different. The
error:
the condition has length > 1 and only the
first element will be used
always occurs when you use an if(candition) statement, but the condition
evaluates to a vector of length k with k>1 (like c(TRUE,FALSE) or so). In
this case it is unclear which element R should use (ie. the TRUE or the
FALSE) so by default it goes with the first statement and also issues a
warning.
In this case, the ifelse() command (that triggers a different action
depending on element by element checking of the condition) is your best bet
to make the function work with vectors. See me previous email and Quan?s on
this.
Jens
From: gov2001-l-bounces at lists.fas.harvard.edu
[mailto:gov2001-l-bounces at lists.fas.harvard.edu] On Behalf Of Jeremy Hodgen
Sent: Wednesday, February 27, 2008 2:37 AM
To: gov2001-l at lists.fas.harvard.edu
Subject: Re: [gov2001-l] Qu2
Can anyone help?
I'm trying to define the function for qu2
This works fine for individual values of x:
g <- function(x){
if ((x < pi/2) & (x > -1*pi/2)) (y <- 10*sin(x)*cos(x)) else (y <-
sin(x)*cos(x))
return(y)
}
But when I input a vector (in order to draw the graph) like this:
ruler <- seq(-2*pi, 2*pi, by=.01)
g(ruler)
I get this error message:
Warning message:
In if ((x < pi/2) & (x > -1 * pi/2)) (y <- 10 * sin(x) * cos(x)) else (y
<-
sin(x) * :
the condition has length > 1 and only the first element will be used
It seems I've got too many conditions for 'if' when the input is a
vector.
When I draw the graph of g(ruler)~ruler, I get sinxcosx on the domain [-2pi,
2pi]. Any ideas on what I could do differently?
Thanks
Jeremy
Dr Jeremy Hodgen
Senior Lecturer in Mathematics Education
King's College London
Department of Education and Professional Studies
Franklin-Wilkins Building
Waterloo Bridge Wing
150 Stamford Street
London SE1 9NH
Tel: 020 7848 3102
Fax: 020 7848 3182
E-mail: jeremy.hodgen at kcl.ac.uk
_______________________________________________
gov2001-l mailing list
gov2001-l at lists.fas.harvard.edu
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7 comments
4 participants
participants (4)
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guoqi＠gse.harvard.edu -
jeremy.hodgen＠kcl.ac.uk -
jhainmueller＠gmail.com -
jmlarson＠fas.harvard.edu