I am doing an iterative computation using flink dataset API.
But the result of each iteration is a part of my complete solution.
(If more details required: I am computing lattice nodes level-wise starting from top towards bottom in each iteration, see Formal Concept Analysis)
If I use flink dataset API with bulk iteration without saving my result, the code will look like below:
val start = env.fromElements((0, BitSet.empty))
val end = start.iterateWithTermination(size) { inp =>
val result = ObjData.mapPartition(new MyMapPartition).withBroadcastSet(inp, "concepts").groupBy(0).reduceGroup(new MyReduceGroup)
(result,result)
}
end.count()
But, if I try to write partial results within iteration (_.writeAsText()) or any action, I will get error:
org.apache.flink.api.common.InvalidProgramException: A data set that is part of an iteration was used as a sink or action. Did you forget to close the iteration?
The alternative without bulk iteration seems to be below:
var start = env.fromElements((0, BitSet.empty))
var count = 1L
var all = count
while (count > 0){
start = ObjData.mapPartition(new MyMapPartition).withBroadcastSet(start, "concepts").groupBy(0).reduceGroup(new MyReduceGroup)
count = start.count()
all = all + count
}
println("total nodes: " + all)
But this approach is exceptionally slow on smallest input data, iteration version takes <30 seconds and loop version takes >3 minutes.
I guess flink is not able to create optimal plan to execute the loop.
Any workaround I should try? Is some modification to flink is possible to be able to save partial results on hadoop etc.?
Unfortunately, it is not currently possible to output intermediate results from a bulk iteration. You can only output the final result at the end of the iteration.
Also, as you correctly noticed, Flink cannot efficiently unroll a while-loop or for-loop, so that won't work either.
If your intermediate results are not that big, one thing you can try is appending your intermediate results in the partial solution and then output everything in the end of the iteration. A similar approach is implemented in the TransitiveClosureNaive example, where paths discovered in an iteration are accumulated in the next partial solution.
Related
The problem happened when I try to keep my cached result in a List and try to calculate new DataFrame by all the data from the last list in each iteration. However, Even I use an empty DataFrame and get an empty result each time, the function will suddenly get very slow after about 8~12 round.
Here is my code
testLoop(Nil)
def testLoop(lastDfList:List[DataFrame]){
// do some dummy transformation like union and cache the result
val resultDf = lastDfList.foldLeft(Seq[Data]().toDF){(df, lastDf) => df.union(lastDf)}.cache
// always get 0, of course
println(resultDf.count)
// benchmark action
benchmark(resultDf.count)
testLoop(resultDf::lastDfList)
}
the benchmark result
1~6 round : < 200ms
7 round : 367ms
8 round : 918ms
9 round : 2476ms
10 round : 7833ms
11 round : 24231ms
I don't think GC or Block eviction is the problem in my case since I already use an empty DataFrame, but I don't know what is the cause? Do I misunderstand the meaning of cache or something?
Thanks!
After reading ImDarrenG's solution, I changed my code to be the following:
spark.sparkContext.setCheckpointDir("/tmp")
testLoop(Nil)
def testLoop(lastDfList:List[DataFrame]){
// do some dummy transformation like union and cache the result
val resultDf = lastDfList.foldLeft(Seq[Data]().toDF){(df, lastDf) => df.union(lastDf)}.cache
resultDf.checkpoint()
// always get 0, of course
println(resultDf.count)
// benchmark action
benchmark(resultDf.count)
testLoop(resultDf::lastDfList)
}
But it still become very slow after a few iterations.
Here you create a list of DataFrames by adding resultDf to the beginning of lastDfList and pass that to the next iteration of testLoop:
testLoop(resultDf::lastDfList)
So lastDfList gets longer each pass.
This line creates a new DataFrame by unioning each member of lastDfList:
val resultDf = lastDfList.foldLeft(Seq[Data]().toDF){(df, lastDf) => df.union(lastDf))}.cache
Each member of lastDfList is a union of it's predecessors, therefore, Spark is maintaining a lineage that becomes exponentially larger with each pass of testLoop.
I expect that the increase in time is caused by the housekeeping of the DAG. Caching the dataframes removes the need to repeat transformations, but the lineage must still be maintained by spark.
Cached data or no, it looks like you are building a really complex DAG by unioning each DataFrame with all of it's predecessors with each pass of testLoop.
You could use checkpoint to trim the lineage, and introduce some check to prevent infinite recursion.
According to API and code, checkpoint will return a new Dataset instead of changing original Dataset.
So I'm trying to iterate over the list of partitions of something, say 1:n for some n between 13 and 21. The code that I ideally want to run looks something like this:
valid_num = #parallel (+) for p in partitions(1:n)
int(is_valid(p))
end
println(valid_num)
This would use the #parallel for to map-reduce my problem. For example, compare this to the example in the Julia documentation:
nheads = #parallel (+) for i=1:200000000
Int(rand(Bool))
end
However, if I try my adaptation of the loop, I get the following error:
ERROR: `getindex` has no method matching getindex(::SetPartitions{UnitRange{Int64}}, ::Int64)
in anonymous at no file:1433
in anonymous at multi.jl:1279
in run_work_thunk at multi.jl:621
in run_work_thunk at multi.jl:630
in anonymous at task.jl:6
which I think is because I am trying to iterate over something that is not of the form 1:n (EDIT: I think it's because you cannot call p[3] if p=partitions(1:n)).
I've tried using pmap to solve this, but because the number of partitions can get really big, really quickly (there are more than 2.5 million partitions of 1:13, and when I get to 1:21 things will be huge), constructing such a large array becomes an issue. I left it running over night and it still didn't finish.
Does anyone have any advice for how I can efficiently do this in Julia? I have access to a ~30 core computer and my task seems easily parallelizable, so I would be really grateful if anyone knows a good way to do this in Julia.
Thank you so much!
The below code gives 511, the number of partitions of size 2 of a set of 10.
using Iterators
s = [1,2,3,4,5,6,7,8,9,10]
is_valid(p) = length(p)==2
valid_num = #parallel (+) for i = 1:30
sum(map(is_valid, takenth(chain(1:29,drop(partitions(s), i-1)), 30)))
end
This solution combines the takenth, drop, and chain iterators to get the same effect as the take_every iterator below under PREVIOUS ANSWER. Note that in this solution, every process must compute every partition. However, because each process uses a different argument to drop, no two processes will ever call is_valid on the same partition.
Unless you want to do a lot of math to figure out how to actually skip partitions, there is no way to avoid computing partitions sequentially on at least one process. I think Simon's answer does this on one process and distributes the partitions. Mine asks each worker process to compute the partitions itself, which means the computation is being duplicated. However, it is being duplicated in parallel, which (if you actually have 30 processors) will not cost you time.
Here is a resource on how iterators over partitions are actually computed: http://www.informatik.uni-ulm.de/ni/Lehre/WS03/DMM/Software/partitions.pdf.
PREVIOUS ANSWER (More complicated than necessary)
I noticed Simon's answer while writing mine. Our solutions seem similar to me, except mine uses iterators to avoid storing partitions in memory. I'm not sure which would actually be faster for what size sets, but I figure it's good to have both options. Assuming it takes you significantly longer to compute is_valid than to compute the partitions themselves, you can do something like this:
s = [1,2,3,4]
is_valid(p) = length(p)==2
valid_num = #parallel (+) for i = 1:30
foldl((x,y)->(x + int(is_valid(y))), 0, take_every(partitions(s), i-1, 30))
end
which gives me 7, the number of partitions of size 2 for a set of 4. The take_every function returns an iterator that returns every 30th partition starting with the ith. Here is the code for that:
import Base: start, done, next
immutable TakeEvery{Itr}
itr::Itr
start::Any
value::Any
flag::Bool
skip::Int64
end
function take_every(itr, offset, skip)
value, state = Nothing, start(itr)
for i = 1:(offset+1)
if done(itr, state)
return TakeEvery(itr, state, value, false, skip)
end
value, state = next(itr, state)
end
if done(itr, state)
TakeEvery(itr, state, value, true, skip)
else
TakeEvery(itr, state, value, false, skip)
end
end
function start{Itr}(itr::TakeEvery{Itr})
itr.value, itr.start, itr.flag
end
function next{Itr}(itr::TakeEvery{Itr}, state)
value, state_, flag = state
for i=1:itr.skip
if done(itr.itr, state_)
return state[1], (value, state_, false)
end
value, state_ = next(itr.itr, state_)
end
if done(itr.itr, state_)
state[1], (value, state_, !flag)
else
state[1], (value, state_, false)
end
end
function done{Itr}(itr::TakeEvery{Itr}, state)
done(itr.itr, state[2]) && !state[3]
end
One approach would be to divide the problem up into pieces that are not too big to realize and then process the items within each piece in parallel, e.g. as follows:
function my_take(iter,state,n)
i = n
arr = Array[]
while !done(iter,state) && (i>0)
a,state = next(iter,state)
push!(arr,a)
i = i-1
end
return arr, state
end
function get_part(npart,npar)
valid_num = 0
p = partitions(1:npart)
s = start(p)
while !done(p,s)
arr,s = my_take(p,s,npar)
valid_num += #parallel (+) for a in arr
length(a)
end
end
return valid_num
end
valid_num = #time get_part(10,30)
I was going to use the take() method to realize up to npar items from the iterator but take() appears to be deprecated so I've included my own implementation which I've called my_take(). The getPart() function therefore uses my_take() to obtain up to npar partitions at a time and carry out a calculation on them. In this case, the calculation just adds up their lengths, because I don't have the code for the OP's is_valid() function. get_part() then returns the result.
Because the length() calculation isn't very time-consuming, this code is actually slower when run on parallel processors than it is on a single processor:
$ julia -p 1 parpart.jl
elapsed time: 10.708567515 seconds (373025568 bytes allocated, 6.79% gc time)
$ julia -p 2 parpart.jl
elapsed time: 15.70633439 seconds (548394872 bytes allocated, 9.14% gc time)
Alternatively, pmap() could be used on each piece of the problem instead of the parallel for loop.
With respect to the memory issue, realizing 30 items from partitions(1:10) took nearly 1 gigabyte of memory on my PC when I ran Julia with 4 worker processes so I expect realizing even a small subset of partitions(1:21) will require a great deal of memory. It may be desirable to estimate how much memory would be needed to see if it would be at all possible before trying such a computation.
With respect to the computation time, note that:
julia> length(partitions(1:10))
115975
julia> length(partitions(1:21))
474869816156751
... so even efficient parallel processing on 30 cores might not be enough to make the larger problem solvable in a reasonable time.
I have a large vector of vectors of strings:
There are around 50,000 vectors of strings,
each of which contains 2-15 strings of length 1-20 characters.
MyScoringOperation is a function which operates on a vector of strings (the datum) and returns an array of 10100 scores (as Float64s). It takes about 0.01 seconds to run MyScoringOperation (depending on the length of the datum)
function MyScoringOperation(state:State, datum::Vector{String})
...
score::Vector{Float64} #Size of score = 10000
I have what amounts to a nested loop.
The outer loop typically would runs for 500 iterations
data::Vector{Vector{String}} = loaddata()
for ii in 1:500
score_total = zeros(10100)
for datum in data
score_total+=MyScoringOperation(datum)
end
end
On one computer, on a small test case of 3000 (rather than 50,000) this takes 100-300 seconds per outer loop.
I have 3 powerful servers with Julia 3.9 installed (and can get 3 more easily, and then can get hundreds more at the next scale).
I have basic experience with #parallel, however it seems like it is spending a lot of time copying the constant (It more or less hang on the smaller testing case)
That looks like:
data::Vector{Vector{String}} = loaddata()
state = init_state()
for ii in 1:500
score_total = #parallel(+) for datum in data
MyScoringOperation(state, datum)
end
state = update(state, score_total)
end
My understanding of the way this implementation works with #parallel is that it:
For Each ii:
partitions data into a chuck for each worker
sends that chuck to each worker
works all process there chunks
main procedure sums the results as they arrive.
I would like to remove step 2,
so that instead of sending a chunk of data to each worker,
I just send a range of indexes to each worker, and they look it up from their own copy of data. or even better, only giving each only their own chunk, and having them reuse it each time (saving on a lot of RAM).
Profiling backs up my belief about the functioning of #parellel.
For a similarly scoped problem (with even smaller data),
the non-parallel version runs in 0.09seconds,
and the parallel runs in
And the profiler shows almost all the time is spent 185 seconds.
Profiler shows almost 100% of this is spend interacting with network IO.
This should get you started:
function get_chunks(data::Vector, nchunks::Int)
base_len, remainder = divrem(length(data),nchunks)
chunk_len = fill(base_len,nchunks)
chunk_len[1:remainder]+=1 #remained will always be less than nchunks
function _it()
for ii in 1:nchunks
chunk_start = sum(chunk_len[1:ii-1])+1
chunk_end = chunk_start + chunk_len[ii] -1
chunk = data[chunk_start: chunk_end]
produce(chunk)
end
end
Task(_it)
end
function r_chunk_data(data::Vector)
all_chuncks = get_chunks(data, nworkers()) |> collect;
remote_chunks = [put!(RemoteRef(pid)::RemoteRef, all_chuncks[ii]) for (ii,pid) in enumerate(workers())]
#Have to add the type annotation sas otherwise it thinks that, RemoteRef(pid) might return a RemoteValue
end
function fetch_reduce(red_acc::Function, rem_results::Vector{RemoteRef})
total = nothing
#TODO: consider strongly wrapping total in a lock, when in 0.4, so that it is garenteed safe
#sync for rr in rem_results
function gather(rr)
res=fetch(rr)
if total===nothing
total=res
else
total=red_acc(total,res)
end
end
#async gather(rr)
end
total
end
function prechunked_mapreduce(r_chunks::Vector{RemoteRef}, map_fun::Function, red_acc::Function)
rem_results = map(r_chunks) do rchunk
function do_mapred()
#assert r_chunk.where==myid()
#pipe r_chunk |> fetch |> map(map_fun,_) |> reduce(red_acc, _)
end
remotecall(r_chunk.where,do_mapred)
end
#pipe rem_results|> convert(Vector{RemoteRef},_) |> fetch_reduce(red_acc, _)
end
rchunk_data breaks the data into chunks, (defined by get_chunks method) and sends those chunks each to a different worker, where they are stored in RemoteRefs.
The RemoteRefs are references to memory on your other proccesses(and potentially computers), that
prechunked_map_reduce does a variation on a kind of map reduce to have each worker first run map_fun on each of it's chucks elements, then reduce over all the elements in its chuck using red_acc (a reduction accumulator function). Finally each worker returns there result which is then combined by reducing them all together using red_acc this time using the fetch_reduce so that we can add the first ones completed first.
fetch_reduce is a nonblocking fetch and reduce operation. I believe it has no raceconditions, though this maybe because of a implementation detail in #async and #sync. When julia 0.4 comes out, it is easy enough to put a lock in to make it obviously have no race conditions.
This code isn't really battle hardened. I don;t believe the
You also might want to look at making the chuck size tunable, so that you can seen more data to faster workers (if some have better network or faster cpus)
You need to reexpress your code as a map-reduce problem, which doesn't look too hard.
Testing that with:
data = [float([eye(100),eye(100)])[:] for _ in 1:3000] #480Mb
chunk_data(:data, data)
#time prechunked_mapreduce(:data, mean, (+))
Took ~0.03 seconds, when distributed across 8 workers (none of them on the same machine as the launcher)
vs running just locally:
#time reduce(+,map(mean,data))
took ~0.06 seconds.
This is similar to a question I asked before, but is slightly different:
So I have a very large structure array in matlab. Suppose, for argument's sake, to simplify the situation, suppose I have something like:
structure(1).name, structure(2).name, structure(3).name structure(1).returns, structure(2).returns, structure(3).returns (in my real program I have 647 structures)
Suppose further that structure(i).returns is a vector (very large vector, approximately 2,000,000 entries) and that a condition comes along where I want to delete the jth entry from structure(i).returns for all i. How do you do this? or rather, how do you do this reasonably fast? I have tried some things, but they are all insanely slow (I will show them in a second) so I was wondering if the community knew of faster ways to do this.
I have parsed my data two different ways; the first way had everything saved as cell arrays, but because things hadn't been working well for me I parsed the data again and placed everything as vectors.
What I'm actually doing is trying to delete NaN data, as well as all data in the same corresponding row of my data file, and then doing the very same thing after applying the Hampel filter. The relevant part of my code in this attempt is:
for i=numStock+1:-1:1
for j=length(stock(i).return):-1:1
if(isnan(stock(i).return(j)))
for k=numStock+1:-1:1
stock(k).return(j) = [];
end
end
end
stock(i).return = sort(stock(i).return);
stock(i).returnLength = length(stock(i).return);
stock(i).medianReturn = median(stock(i).return);
stock(i).madReturn = mad(stock(i).return,1);
end;
for i=numStock:-1:1
for j = length(stock(i+1).volume):-1:1
if(isnan(stock(i+1).volume(j)))
for k=numStock:-1:1
stock(k+1).volume(j) = [];
end
end
end
stock(i+1).volume = sort(stock(i+1).volume);
stock(i+1).volumeLength = length(stock(i+1).volume);
stock(i+1).medianVolume = median(stock(i+1).volume);
stock(i+1).madVolume = mad(stock(i+1).volume,1);
end;
for i=numStock+1:-1:1
for j=stock(i).returnLength:-1:1
if (abs(stock(i).return(j) - stock(i).medianReturn) > 3*stock(i).madReturn)
for k=numStock+1:-1:1
stock(k).return(j) = [];
end
end;
end;
end;
for i=numStock:-1:1
for j=stock(i+1).volumeLength:-1:1
if (abs(stock(i+1).volume(j) - stock(i+1).medianVolume) > 3*stock(i+1).madVolume)
for k=numStock:-1:1
stock(k+1).volume(j) = [];
end
end;
end;
end;
However, this returns an error:
"Matrix index is out of range for deletion.
Error in Failure (line 110)
stock(k).return(j) = [];"
So instead I tried by parsing everything in as vectors. Then I decided to try and delete the appropriate entries in the vectors prior to building the structure array. This isn't returning an error, but it is very slow:
%% Delete bad data, Hampel Filter
% Delete bad entries
id=strcmp(returns,'');
returns(id)=[];
volume(id)=[];
date(id)=[];
ticker(id)=[];
name(id)=[];
permno(id)=[];
sp500(id) = [];
id=strcmp(returns,'C');
returns(id)=[];
volume(id)=[];
date(id)=[];
ticker(id)=[];
name(id)=[];
permno(id)=[];
sp500(id) = [];
% Convert returns from string to double
returns=cellfun(#str2double,returns);
sp500=cellfun(#str2double,sp500);
% Delete all data for which a return is not a number
nanid=isnan(returns);
returns(nanid)=[];
volume(nanid)=[];
date(nanid)=[];
ticker(nanid)=[];
name(nanid)=[];
permno(nanid)=[];
% Delete all data for which a volume is not a number
nanid=isnan(volume);
returns(nanid)=[];
volume(nanid)=[];
date(nanid)=[];
ticker(nanid)=[];
name(nanid)=[];
permno(nanid)=[];
% Apply the Hampel filter, and delete all data corresponding to
% observations deleted by the filter.
medianReturn = median(returns);
madReturn = mad(returns,1);
for i=length(returns):-1:1
if (abs(returns(i) - medianReturn) > 3*madReturn)
returns(i) = [];
volume(i)=[];
date(i)=[];
ticker(i)=[];
name(i)=[];
permno(i)=[];
end;
end
medianVolume = median(volume);
madVolume = mad(volume,1);
for i=length(volume):-1:1
if (abs(volume(i) - medianVolume) > 3*madVolume)
returns(i) = [];
volume(i)=[];
date(i)=[];
ticker(i)=[];
name(i)=[];
permno(i)=[];
end;
end
As I said, this is very slow, probably because I'm using a for loop on a very large data set; however, I'm not sure how else one would do this. Sorry for the gigantic post, but does anyone have a suggestion as to how I might go about doing what I'm asking in a reasonable way?
EDIT: I should add that getting the vector method to work is probably preferable, since my aim is to put all of the return vectors into a matrix and get all of the volume vectors into a matrix and perform PCA on them, and I'm not sure how I would do that using cell arrays (or even if princomp would work on cell arrays).
EDIT2: I have altered the code to match your suggestion (although I did decide to give up speed and keep with the for-loops to keep with the structure array, since reparsing this data will be way worse time-wise). The new code snipet is:
stock_return = zeros(numStock+1,length(stock(1).return));
for i=1:numStock+1
for j=1:length(stock(i).return)
stock_return(i,j) = stock(i).return(j);
end
end
stock_return = stock_return(~any(isnan(stock_return)), : );
This returns an Index exceeds matrix dimensions error, and I'm not sure why. Any suggestions?
I could not find a convenient way to handle structures, therefore I would restructure the code so that instead of structures it uses just arrays.
For example instead of stock(i).return(j) I would do stock_returns(i,j).
I show you on a part of your code how to get rid of for-loops.
Say we deal with this code:
for j=length(stock(i).return):-1:1
if(isnan(stock(i).return(j)))
for k=numStock+1:-1:1
stock(k).return(j) = [];
end
end
end
Now, the deletion of columns with any NaN data goes like this:
stock_return = stock_return(:, ~any(isnan(stock_return)) );
As for the absolute difference from medianVolume, you can write a similar code:
% stock_return_length is a scalar
% stock_median_return is a column vector (eg. [1;2;3])
% stock_mad_return is also a column vector.
median_return = repmat(stock_median_return, stock_return_length, 1);
is_bad = abs(stock_return - median_return) > 3.* stock_mad_return;
stock_return = stock_return(:, ~any(is_bad));
Using a scalar for stock_return_length means of course that the return lengths are the same, but you implicitly assume it in your original code anyway.
The important point in my answer is using any. Logical indexing is not sufficient in itself, since in your original code you delete all the values if any of them is bad.
Reference to any: http://www.mathworks.co.uk/help/matlab/ref/any.html.
If you want to preserve the original structure, so you stick to stock(i).return, you can speed-up your code using essentially the same scheme but you can only get rid of one less for-loop, meaning that your program will be substantially slower.
Update: I've provided a brief analysis of the three answers at the bottom of the question text and explained my choices.
My Question: What is the most efficient method of building a fixed interval dataset from a random interval dataset using stale data?
Some background: The above is a common problem in statistics. Frequently, one has a sequence of observations occurring at random times. Call it Input. But one wants a sequence of observations occurring say, every 5 minutes. Call it Output. One of the most common methods to build this dataset is using stale data, i.e. set each observation in Output equal to the most recently occurring observation in Input.
So, here is some code to build example datasets:
TInput = 100;
TOutput = 50;
InputTimeStamp = 730486 + cumsum(0.001 * rand(TInput, 1));
Input = [InputTimeStamp, randn(TInput, 1)];
OutputTimeStamp = 730486.002 + (0:0.001:TOutput * 0.001 - 0.001)';
Output = [OutputTimeStamp, NaN(TOutput, 1)];
Both datasets start at close to midnight at the turn of the millennium. However, the timestamps in Input occur at random intervals while the timestamps in Output occur at fixed intervals. For simplicity, I have ensured that the first observation in Input always occurs before the first observation in Output. Feel free to make this assumption in any answers.
Currently, I solve the problem like this:
sMax = size(Output, 1);
tMax = size(Input, 1);
s = 1;
t = 2;
%#Loop over input data
while t <= tMax
if Input(t, 1) > Output(s, 1)
%#If current obs in Input occurs after current obs in output then set current obs in output equal to previous obs in input
Output(s, 2:end) = Input(t-1, 2:end);
s = s + 1;
%#Check if we've filled out all observations in output
if s > sMax
break
end
%#This step is necessary in case we need to use the same input observation twice in a row
t = t - 1;
end
t = t + 1;
if t > tMax
%#If all remaining observations in output occur after last observation in input, then use last obs in input for all remaining obs in output
Output(s:end, 2:end) = Input(end, 2:end);
break
end
end
Surely there is a more efficient, or at least, more elegant way to solve this problem? As I mentioned, this is a common problem in statistics. Perhaps Matlab has some in-built function I'm not aware of? Any help would be much appreciated as I use this routine a LOT for some large datasets.
THE ANSWERS: Hi all, I've analyzed the three answers, and as they stand, Angainor's is the best.
ChthonicDaemon's answer, while clearly the easiest to implement, is really slow. This is true even when the conversion to a timeseries object is done outside of the speed test. I'm guessing the resample function has a lot of overhead at the moment. I am running 2011b, so it is possible Mathworks have improved it in the intervening time. Also, this method needs an additional line for the case where Output ends more than one observation after Input.
Rody's answer runs only slightly slower than Angainor's (unsurprising given they both employ the histc approach), however, it seems to have some problems. First, the method of assigning the last observation in Output is not robust to the last observation in Input occurring after the last observation in Output. This is an easy fix. But there is a second problem which I think stems from having InputTimeStamp as the first input to histc instead of the OutputTimeStamp adopted by Angainor. The problem emerges if you change OutputTimeStamp = 730486.002 + (0:0.001:TOutput * 0.001 - 0.001)'; to OutputTimeStamp = 730486.002 + (0:0.0001:TOutput * 0.0001 - 0.0001)'; when setting up the example inputs.
Angainor's appears robust to everything I threw at it, plus it was the fastest.
I did a lot of speed tests for different input specifications - the following numbers are fairly representative:
My naive loop: Elapsed time is 8.579535 seconds.
Angainor: Elapsed time is 0.661756 seconds.
Rody: Elapsed time is 0.913304 seconds.
ChthonicDaemon: Elapsed time is 22.916844 seconds.
I'm +1-ing Angainor's solution and marking the question solved.
This "stale data" approach is known as a zero order hold in signal and timeseries fields. Searching for this quickly brings up many solutions. If you have Matlab 2012b, this is all built in to the timeseries class by using the resample function, so you would simply do
TInput = 100;
TOutput = 50;
InputTimeStamp = 730486 + cumsum(0.001 * rand(TInput, 1));
InputData = randn(TInput, 1);
InputTimeSeries = timeseries(InputData, InputTimeStamp);
OutputTimeStamp = 730486.002 + (0:0.001:TOutput * 0.001 - 0.001);
OutputTimeSeries = resample(InputTimeSeries, OutputTimeStamp, 'zoh'); % zoh stands for zero order hold
Here is my take on the problem. histc is the way to go:
% find Output timestamps in Input bins
N = histc(Output(:,1), Input(:,1));
% find counts in the non-empty bins
counts = N(find(N));
% find Input signal value associated with every bin
val = Input(find(N),2);
% now, replicate every entry entry in val
% as many times as specified in counts
index = zeros(1,sum(counts));
index(cumsum([1 counts(1:end-1)'])) = 1;
index = cumsum(index);
val_rep = val(index)
% finish the signal with last entry from Input, as needed
val_rep(end+1:size(Output,1)) = Input(end,2);
% done
Output(:,2) = val_rep;
I checked against your procedure for a few different input models (I changed the number of Output timestamps) and the results are the same. However, I am still not sure I understood your problem, so if something is wrong here let me know.