My rough interpretation (which is by no means the official one, as I think
this is still up for debate), is as follows:,
1) convert(T,x) is the most specific: it will convert x to type T, throwing
an error if outside the range (though it can round, for instance when
converting to a Rational
On Saturday, 21 March 2015 09:38:23 UTC+1, Simon Byrne wrote:
* in some cases, x can be a string: at the moment this seems to only work
for BigFloat and BigInt, but perhaps this should apply to other types.
See https://github.com/JuliaLang/julia/issues/10594 in case anyone has an
opinion.
Thanks for pointing me in the right direction. I fixed it. It turns out
that when I set the unitrange (using 1:n), the 1 was being interpreted as
Int64, which caused the UnitRange to be Int64, which messed things up. I
now do a convert(T,1) in the construction and everything's good:
julia
Sadly it looks like this doesn't entirely solve the problem. The issue is
that while it does indeed insert the function into the Expr tree, the
namespace for the variables is for the module that the macro is defined in,
and not the user's file (the standard issue with eval'ing).
For example,
Have you read the macro hygiene section in the manual chapter about
metaprogramming? The esc function is what you want to be able to pass through
symbols to be evaluated in the invoker's name space.
This seems to happen on OS X occasionally, for some reason. What you can do
is try the open a terminal repl command which will open a plain copy of
the version of Julia that Juno is running from. You should be able to
`Pkg.update()` from there, but if you see another error then let us know.
On 17
We may want to consider doing the minimal number of backports now for 0.3.7
and then doing a bunch more backports right afterwards. Or at least that's
what we should do for anything even a little bit risky.
On Fri, Mar 20, 2015 at 11:03 PM, Tony Kelman t...@kelman.net wrote:
There have been
Thanks for the explanation!
One related question, in case anyone's still reading here: I'm planning to
run my code on a cluster, where the size of the machine I'm being allocated
varies with the overall usage level. That is, my code might run on a 12
core machine one day, a 24 core machine the
I was able to get ODBC to work, but I had to give up Juno. I
installed Julia-0.3.6 and ran the following script:
using DataFrames
using ODBC
using Gadfly
conn = ODBC.connect(bigsur)
df = query(Select * from IBM)
names(df)
closes = df[6]
p = plot(x = collect(1:100), y = closes[1:100])
This
I am trying to solve Laplace's equation using ApproxFun with the following
boundary conditions
{
\phi(x, 0) = 0 \\
\phi(x, 1) = \frac{1}{(1 + x)^2 + 1} \\
\phi(0, y) = \frac{y}{1 + y^2} \\
\phi(1, y) = \frac{y}{4 + y^2}
\end{aligned}
}
I'm not clear how to express these. I've tried
{
d =
The first plot does a bunch of compilation. Second time you evaluate it
will be much quicker.
On Sat, Mar 21, 2015 at 7:28 PM, Charles Brauer charlesbra...@gmail.com
wrote:
I was able to get ODBC to work, but I had to give up Juno. I
installed Julia-0.3.6 and ran the following script:
using
Hello @all,
I'm basically interested in the rationale behind using @ for macros like
@assert and @time and so on.
Personally, I think that using lots of macros clutters the otherwise very
likable Julia code.
Also what is the purpose behind using @? I mean macro expansion is a pre
compilation
An error occurred upon my first attempt to use Julia for solving a
meaningful convex optimization problem:
using Convex
x = Variable(3)
t = rand(3)
s = t+rand(3)
constraints = [x=0, x=s]
p = minimize(sum(1./(x-t)), constraints)
`./` has no method matching ./(::Int64, ::AdditionAtom)
Searched
Glad to help. There is also a function Base.one(T) which is defined to give
the multiplicative identity of T which for all subtypes of Real should be
the same as convert(T, 1). You aren't really using it as the multiplicative
identity though so I am not sure which is better.
On Sat, Mar 21, 2015,
There is a pretty good set of answers on StackOverflow now. First-posts
need to be manually approved, which is why this message appears to be
time-delayed. (Please don't cross-post in the future because it is the same
set of people answering everywhere).
On Sat, Mar 21, 2015 at 12:53 PM, Gregor
It's expanding f1-f4 as bivariate functions on [-1,1]^2, but instead you
want to expand them as univariate functions on [-1,1]:
d = Interval()^2
f1 = Fun(x-0)
f2 = Fun(x-1 / ((1 + x)^2 + 1))
f3 = Fun(y-y / (1 + y^2))
f4 = Fun(y-y / (4 + y^2))
u = [dirichlet(d),lap(d)]\[f1,f2,f3,f4]
On
I have, but it's not entirely clear to me how I would use it in this
particular case. The code in question is being called from a function
(which while ultimately is being called from a macro) doesn't directly
return an expression in this case. Specifically, I have something that
looks like:
Yeah, that makes sense to me. I'll do a couple more of the simple ones that
have recently been flagged, then run through Elliot's checklist.
On Saturday, March 21, 2015 at 6:19:35 AM UTC-7, Stefan Karpinski wrote:
We may want to consider doing the minimal number of backports now for
0.3.7
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