I have two code snippets that I assumed to both result in alerts. However the first results none while the second performs the alerts.
(map #(.alert js/window %) ["hey1" "hey2"])
This slight modification prints (nil nil) as expected, as well as fixing the alert issue. The question being WHY?
(print (map #(.alert js/window %) ["hey1" "hey2"]))
Another weird observation is that the first snippet works from a browser-repl, but not when typed into code.
Is the map function side effect free, but print is not? Maybe some core code optimization I do not know about?
Work-arounds and answers are both appreciated. If you need more info please let me know in a comment.
[org.clojure/clojurescript "1.8.51"]
BOOT_CLOJURE_VERSION=1.7.0
BOOT_VERSION=2.5.5
java version "1.8.0_101"
Description: Ubuntu 14.04.4 LTS
You don't want to use map for a side-effecty operation like alert. The issue you are seeing is a result of map being lazy, so it won't actually do the work until you consume the elements of the resulting sequence. If you really want to do side-effect sort of things, doseq might be a better option, especially if you don't need a sequence of results:
(doseq [msg ["hey1" "hey2"]]
(.alert js/window msg))
Or you can use doall to force the evaluation of the whole sequence:
(doall (map #(.alert js/window %) ["hey1" "hey2"]))
Related
I'm porting FFT code from Java to Lua, and I'm starting to worry a bit about the fact that in Lua the array part of a table starts indexing at 1 while in Java array indexing starts at 0.
For the input array this causes no problem because the Java code is set up to handle the possibility that the data under consideration is not located at the start of the array. However, all of the working arrays internal to the code are assumed to starting indexing at 0. I know that the code will work as written -- Lua tables are awesome like that -- but I have no sense at all about the performance hit I might incur by having the "0" element of the array going into the hash table part of the underlying C structure (or indeed, if that is what will happen).
My question: is this something worth worrying about? Should I be planning to profile and hand-optimize the code? (The code will eventually be used to transform many relatively small (> 100 time points) signals of varying lengths not known in advance.)
I have made small, probably not that reliable, test:
local arr = {}
for i=0,10000000 do
arr[i] = i*2
end
for k, v in pairs(arr) do
arr[k] = v*v
end
And similar version with 1 as the first index. On my system:
$ time lua example0.lua
real 2.003s
$ time lua example1.lua
real 2.014s
I was also interested how table.insert will perform
for i=1,10000000 do
table.insert(arr, 2*i)
...
and, suprisingly
$ time lua example2.lua
real 6.012s
Results:
Of course, it depends on what system you're running it, probably also whic lua version, but it seems that it makes little to no difference between zero-start and one-start. Bigger difference is caused by the way you insert things to array.
I think the correct answer in this case is changing the algorithm so that everything is indexed with 1. And consider that part of the conversion.
Your FFT will be less surprising to another Lua user (like me), given that all "array-like" tables are indexed by one.
It might not be as stressful as you might think, given the way numeric loops are structured in Lua (where the "start" and the "end" are "inclusive"). You would be exchanging this:
for i=0,#array-1 do
... (do stuff with i)
end
By this:
for i=1,#array do
... (do stuff with i)
end
The non-numeric loops would remain unchanged (except that you will be able to use ipairs too, if you so desire).
I have been struggling with parallel and async constructs in F# for the last couple days and not sure where to go at this point. I have been programming with F# for about 4 months - certainly no expert - and I currently have a series of calculations that are implemented in F# (asp.net 4.5) and are working correctly when executed sequentially. I am running the calculations on a multi-core server and since there are millions of inputs to perform the same calculation on, I am hoping to take advantage of parallelism to speed it up.
The calculations are extremely data parallel - basically the exact calculation on different input data. I have tried a number of different avenues and I continually run into the same issue - it seems as if the parallel looping never gets to the end of the input data set. I have tried TPL, ConcurrentQueues, Parallel.Array.map/iter and all the same result: the program starts out fine and then somewhere in the middle (indeterminate) it just hangs and never completes. For simplicity I actually removed the calculation from the program and I am just calling a print method, and Here is where the code is currently at:
let runParallel =
let ids = query {for c in db.CustTable do select c.id} |> Seq.take(5)
let customerInputArray= getAllObservations ids
Array.Parallel.iter(fun c -> testParallel c) customerInputArray
let key = System.Console.ReadKey()
0
A few points...
I limited the results above to only 5 just for debugging. The actual program does not apply the Take(5).
The testParallel method is just a printfn "test".
The customerInputArray is a complex data type. It is a tuple of lists that contain records. So I am pretty sure my problem must be there...but I added exception handling and no exception is getting raised, so have no idea how to go about finding the problem.
Any help is appreciated. Thanks in advance.
EDIT: Thanks for the advice...I think it is definitely deadlock. When I remove all of the printfn, sprintfn, and string concat operations, it completes. (of course, I need those things in there.)
Is printfn, sprintfn, and string ops not thread-safe?
Another EDIT: Iteration always stops on the last item..So if my input array has 15 items, the processing stops on item 14, or seems to never get to item 15. Then everything just hangs. Does not matter what the size of the input array is..Any ideas what can be causing this? I even switched over to Parallel.ForEach (instead of Array.Parallel) and same behavior.
Update on the situation and how I resolved this issue.
I was unable to upload code from my example due to my company's firewall policy, so in the end my question did not have enough details. I failed to mention that I was using a type provider which was important information in this situation. But here is what I figured out.
I am using the F# type provider for SQL Server and was passing around its Service Types which I suspect are not thread-safe. When I replaced the ServiceTypes with plain old F# Records, the code worked fine - no more deadlocks and everything completed without error.
Detecting cycles in a single linked list is a well known problem. I know that this question has been asked a zillion times all over the internet. The reason why I am asking it again is I thought of a solution which I did not encounter at other places. (I admit I haven't searched that deeply either).
My solution is:
Given a linked list and pointer to some node, break the link between node and node->next();
Then start at node->next() and traverse till either you hit an end (which means there was no loop) or till you reach at node which means there was a loop.
Is there anything wrong/good about above solution ?
Note: Do join the link back once you are done.
That will work to detect complete cycles (i.e., cycles with a period of the whole list), e.g.:
A -> B -> C -> D -> A
But what if we have a cycle somewhere else in the list?
e.g.,
A -> B -> C -> D -> E -> C
I can't see that your algorithm will detect the cycle in this case.
Keep in mind that to detect the first case, we need not even break the link. We could just traverse the list and keep comparing the next link for each node with the head element to see if we'd started back at the start yet (or hit the end).
I guess the most trivial approach (not necessarily the best, but one that everybody should know how to implement in Java in a few lines of code) is to build a Hash Set of the nodes, start adding them until you find one that you already saw before. Takes extra memory though.
If you can mark nodes, start marking them until you find one you marked before (the hash map is essentially an external marker).
And check the usual graph theory books...
You are not allowed to break a link, even if you join it back at the end. What if other programs read the list at the same time ?
The algorithm must not damage the list while working on it.
I am trying to run this code but it keeps crashing:
log10(x):=log(x)/log(10);
char(x):=floor(log10(x))+1;
mantissa(x):=x/10**char(x);
chop(x,d):=(10**char(x))*(floor(mantissa(x)*(10**d))/(10**d));
rnd(x,d):=chop(x+5*10**(char(x)-d-1),d);
d:5;
a:10;
Ibwd:[[30,rnd(integrate((x**60)/(1+10*x^2),x,0,1),d)]];
for n from 30 thru 1 step -1 do Ibwd:append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd);
Maxima crashes when it evaluates the last line. Any ideas why it may happen?
Thank you so much.
The problem is that the difference becomes negative and your rounding function dies horribly with a negative argument. To find this out, I changed your loop to:
for n from 30 thru 1 step -1 do
block([],
print (1/(2*n-1)-a*last(first(Ibwd))),
print (a*last(first(Ibwd))),
Ibwd: append([[n-1,rnd(1/(2*n-1)-a*last(first(Ibwd)),d)]],Ibwd),
print (Ibwd));
The last difference printed before everything fails miserably is -316539/6125000. So now try
rnd(-1,3)
and see the same problem. This all stems from the fact that you're taking the log of a negative number, which Maxima interprets as a complex number by analytic continuation. Maxima doesn't evaluate this until it absolutely has to and, somewhere in the evaluation code, something's dying horribly.
I don't know the "fix" for your specific example, since I'm not exactly sure what you're trying to do, but hopefully this gives you enough info to find it yourself.
If you want to deconstruct a floating point number, let's first make sure that it is a bigfloat.
say z: 34.1
You can access the parts of a bigfloat by using lisp, and you can also access the mantissa length in bits by ?fpprec.
Thus ?second(z)*2^(?third(z)-?fpprec) gives you :
4799148352916685/140737488355328
and bfloat(%) gives you :
3.41b1.
If you want the mantissa of z as an integer, look at ?second(z)
Now I am not sure what it is that you are trying to accomplish in base 10, but Maxima
does not do internal arithmetic in base 10.
If you want more bits or fewer, you can set fpprec,
which is linked to ?fpprec. fpprec is the "approximate base 10" precision.
Thus fpprec is initially 16
?fpprec is correspondingly 56.
You can easily change them both, e.g. fpprec:100
corresponds to ?fpprec of 335.
If you are diddling around with float representations, you might benefit from knowing
that you can look at any of the lisp by typing, for example,
?print(z)
which prints the internal form using the Lisp print function.
You can also trace any function, your own or system function, by trace.
For example you could consider doing this:
trace(append,rnd,integrate);
If you want to use machine floats, I suggest you use, for the last line,
for n from 30 thru 1 step -1 do :
Ibwd:append([[n-1,rnd(1/(2.0*n- 1.0)-a*last(first(Ibwd)),d)]],Ibwd);
Note the decimal points. But even that is not quite enough, because integration
inserts exact structures like atan(10). Trying to round these things, or compute log
of them is probably not what you want to do. I suspect that Maxima is unhappy because log is given some messy expression that turns out to be negative, even though it initially thought otherwise. It hands the number to the lisp log program which is perfectly happy to return an appropriate common-lisp complex number object. Unfortunately, most of Maxima was written BEFORE LISP HAD COMPLEX NUMBERS.
Thus the result (log -0.5)= #C(-0.6931472 3.1415927) is entirely unexpected to the rest of Maxima. Maxima has its own form for complex numbers, e.g. 3+4*%i.
In particular, the Maxima display program predates the common lisp complex number format and does not know what to do with it.
The error (stack overflow !!!) is from the display program trying to display a common lisp complex number.
How to fix all this? Well, you could try changing your program so it computes what you really want, in which case it probably won't trigger this error. Maxima's display program should be fixed, too. Also, I suspect there is something unfortunate in simplification of logs of numbers that are negative but not obviously so.
This is probably waaay too much information for the original poster, but maybe the paragraph above will help out and also possibly improve Maxima in one or more places.
It appears that your program triggers an error in Maxima's simplification (algebraic identities) code. We are investigating and I hope we have a bug fix soon.
In the meantime, here is an idea. Looks like the bug is triggered by rnd(x, d) when x < 0. I guess rnd is supposed to round x to d digits. To handle x < 0, try this:
rnd(x, d) := if x < 0 then -rnd1(-x, d) else rnd1(x, d);
rnd1(x, d) := (... put the present definition of rnd here ...);
When I do that, the loop runs to completion and Ibwd is a list of values, but I don't know what values to expect.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is R's apply family more than syntactic sugar
Just what the title says. Stupid question, perhaps, but my understanding has been that when using an "apply" function, the iteration is performed in compiled code rather than in the R parser. This would seem to imply that lapply, for instance, is only faster than a "for" loop if there are a great many iterations and each operation is relatively simple. For instance, if a single call to a function wrapped up in lapply takes 10 seconds, and there are only, say, 12 iterations of it, I would imagine that there's virtually no difference at all between using "for" and "lapply".
Now that I think of it, if the function inside the "lapply" has to be parsed anyway, why should there be ANY performance benefit from using "lapply" instead of "for" unless you're doing something that there are compiled functions for (like summing or multiplying, etc)?
Thanks in advance!
Josh
There are several reasons why one might prefer an apply family function over a for loop, or vice-versa.
Firstly, for() and apply(), sapply() will generally be just as quick as each other if executed correctly. lapply() does more of it's operating in compiled code within the R internals than the others, so can be faster than those functions. It appears the speed advantage is greatest when the act of "looping" over the data is a significant part of the compute time; in many general day-to-day uses you are unlikely to gain much from the inherently quicker lapply(). In the end, these all will be calling R functions so they need to be interpreted and then run.
for() loops can often be easier to implement, especially if you come from a programming background where loops are prevalent. Working in a loop may be more natural than forcing the iterative computation into one of the apply family functions. However, to use for() loops properly, you need to do some extra work to set-up storage and manage plugging the output of the loop back together again. The apply functions do this for you automagically. E.g.:
IN <- runif(10)
OUT <- logical(length = length(IN))
for(i in IN) {
OUT[i] <- IN > 0.5
}
that is a silly example as > is a vectorised operator but I wanted something to make a point, namely that you have to manage the output. The main thing is that with for() loops, you always allocate sufficient storage to hold the outputs before you start the loop. If you don't know how much storage you will need, then allocate a reasonable chunk of storage, and then in the loop check if you have exhausted that storage, and bolt on another big chunk of storage.
The main reason, in my mind, for using one of the apply family of functions is for more elegant, readable code. Rather than managing the output storage and setting up the loop (as shown above) we can let R handle that and succinctly ask R to run a function on subsets of our data. Speed usually does not enter into the decision, for me at least. I use the function that suits the situation best and will result in simple, easy to understand code, because I'm far more likely to waste more time than I save by always choosing the fastest function if I can't remember what the code is doing a day or a week or more later!
The apply family lend themselves to scalar or vector operations. A for() loop will often lend itself to doing multiple iterated operations using the same index i. For example, I have written code that uses for() loops to do k-fold or bootstrap cross-validation on objects. I probably would never entertain doing that with one of the apply family as each CV iteration needs multiple operations, access to lots of objects in the current frame, and fills in several output objects that hold the output of the iterations.
As to the last point, about why lapply() can possibly be faster that for() or apply(), you need to realise that the "loop" can be performed in interpreted R code or in compiled code. Yes, both will still be calling R functions that need to be interpreted, but if you are doing the looping and calling directly from compiled C code (e.g. lapply()) then that is where the performance gain can come from over apply() say which boils down to a for() loop in actual R code. See the source for apply() to see that it is a wrapper around a for() loop, and then look at the code for lapply(), which is:
> lapply
function (X, FUN, ...)
{
FUN <- match.fun(FUN)
if (!is.vector(X) || is.object(X))
X <- as.list(X)
.Internal(lapply(X, FUN))
}
<environment: namespace:base>
and you should see why there can be a difference in speed between lapply() and for() and the other apply family functions. The .Internal() is one of R's ways of calling compiled C code used by R itself. Apart from a manipulation, and a sanity check on FUN, the entire computation is done in C, calling the R function FUN. Compare that with the source for apply().
From Burns' R Inferno (pdf), p25:
Use an explicit for loop when each
iteration is a non-trivial task. But a
simple loop can be more clearly and
compactly expressed using an apply
function. There is at least one
exception to this rule ... if the result will
be a list and some of the components
can be NULL, then a for loop is
trouble (big trouble) and lapply gives
the expected answer.