I was trying to find an optimal d for my model by checking whether sts.adfuller(df_new.delta_prices[1:]) is statistically significant.
Here is my code
df_new['delta_prices'] = df_new.brent_price.diff(1)
model_delta_ar_1_i_1_ma_1 = ARIMA(df_new.delta_prices[1:],order=(1,0,1))
results_delta_ar_1_i_1_ma_1 = model_delta_ar_1_i_1_ma_1.fit()
results_delta_ar_1_i_1_ma_1.summary()
model_ar_1_i_1_ma_1 = ARIMA(df_new.brent_price,order=(1,1,1))
results_ar_1_i_1_ma_1 = model_ar_1_i_1_ma_1.fit()
results_ar_1_i_1_ma_1.summary()
So basically as I see it these two parts of the code should produce the same output, since it's the same just written in different ways.
However, the results are different
First code results ARIMA(1,0,1)
Second code results ARIMA(1,1,1)
I can't figure out why they are different?
Related
I am trying to do prompting using T0. I want the probability scores of the query outputs. For example, for the given prompt: Is this review positive or negative? Review: this is the best cast iron skillet you will ever buy, I wanted the following values: P(positive|prompt)
Is there a way I can condition the language model on "positive" such that I get the probability score for positive for multiple such reviews?
I am trying to do something like this:
`
inputs = tokenizer.encode(test_prompt, return_tensors="pt")
decoder_input_ids = tokenizer.encode("positive",return_tensors="pt")
outputs = model(inputs,decoder_input_ids=decoder_input_ids)
pred_logits = outputs.logits
probs=nn.functional.softmax(pred_logits,dim=-1)
`
The output dimension of probs is [1,2,32128]. I am confused and wanted to know if this is the right way to get the scores. Also, is this the right way to condition on "positive" ? by passing the positive encoding in decoder_input_ids?
I´m having serious performance issues with a job that is running everyday and I think i cannot improve the algorithm; so I´m gonnga explain you what is the problem to solve and the algorithm we have, and maybe you have some other ideas to solve the problem better.
So the problem we have to solve is:
There is a set of Rules, ~ 120.000 Rules.
Every rule has a set of combinations of Codes. Codes are basically strings. So we have ~8 combinations per rule. Example of a combination: TTAAT;ZZUHH;GGZZU;WWOOF;SSJJW;FFFOLL
There is a set of Objects, ~800 objects.
Every object has a set of ~200 codes.
We have to check for every Rule, if there is at least one Combination of Codes that is fully contained in the Objects. It means =>
loop in Rules
Loop in Combinations of the rule
Loop in Objects
every code of the combination found in the Object? => create relationship rule/object and continue with the next object
end of loop
end of loop
end of loop
For example, if we have the Rule with this combination of two codes: HHGGT; ZZUUF
And let´s say we have an object with this codes: HHGGT; DHZZU; OIJUH; ZHGTF; HHGGT; JUHZT; ZZUUF; TGRFE; UHZGT; FCDXS
Then we create a relationship between the Object and the Rule because every code of the combination of the rule is contained in the codes of the object => this is what the algorithm has to do.
As you can see this is quite expensive, because we need 120.000 x 8 x 800 = 750 millions of times in the worst-case scenario.
This is a simplified scenario of the real problem; actually what we do in the loops is a little bit more complicated, that´s why we have to reduce this somehow.
I tried to think in a solution but I don´t have any ideas!
Do you see something wrong here?
Best regards and thank you for the time :)
Something like this might work better if I'm understanding correctly (this is in python):
RULES = [
['abc', 'def',],
['aaa', 'sfd',],
['xyy', 'eff',]]
OBJECTS = [
('rrr', 'abc', 'www', 'def'),
('pqs', 'llq', 'aaa', 'sdr'),
('xyy', 'hjk', 'fed', 'eff'),
('pnn', 'rrr', 'mmm', 'qsq')
]
MapOfCodesToObjects = {}
for obj in OBJECTS:
for code in obj:
if (code in MapOfCodesToObjects):
MapOfCodesToObjects[code].add(obj)
else:
MapOfCodesToObjects[code] = set({obj})
RELATIONS = []
for rule in RULES:
if (len(rule) == 0):
continue
if (rule[0] in MapOfCodesToObjects):
ValidObjects = MapOfCodesToObjects[rule[0]]
else:
continue
for i in range(1, len(rule)):
if (rule[i] in MapOfCodesToObjects):
codeObjects = MapOfCodesToObjects[rule[i]]
else:
ValidObjects = set()
break
ValidObjects = ValidObjects.intersection(codeObjects)
if (len(ValidObjects) == 0):
break
for vo in ValidObjects:
RELATIONS.append((rule, vo))
for R in RELATIONS:
print(R)
First you build a map of codes to objects. If there are nObj objects and nCodePerObj codes on average per object, this takes O(nObj*nCodePerObj * log(nObj*nCodePerObj).
Next you iterate through the rules and look up each code in each rule in the map you built. There is a relation if a certain object occurs for every code in the rule, i.e. if it is in the set intersection of the objects for every code in the rule. Since hash lookups have O(1) time complexity on average, and set intersection has time complexity O(min of the lengths of the 2 sets), this will take O(nRule * nCodePerRule * nObjectsPerCode), (note that is nObjectsPerCode, not nCodePerObj, the performance gets worse when one code is included in many objects).
it run corectly but it should have around 500 matches but it only has around 50 and I dont know why!
This is a probelm for my comsci class that I am having isues with
we had to make a function that checks a list for duplication I got that part but then we had to apply it to the birthday paradox( more info here http://en.wikipedia.org/wiki/Birthday_problem) thats where I am runing into problem because my teacher said that the total number of times should be around 500 or 50% but for me its only going around 50-70 times or 5%
duplicateNumber=0
import random
def has_duplicates(listToCheck):
for i in listToCheck:
x=listToCheck.index(i)
del listToCheck[x]
if i in listToCheck:
return True
else:
return False
listA=[1,2,3,4]
listB=[1,2,3,1]
#print has_duplicates(listA)
#print has_duplicates(listB)
for i in range(0,1000):
birthdayList=[]
for i in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x= has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
else:
pass
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000)*100),3),"%"
This code gave me a result in line with what you were expecting:
import random
duplicateNumber=0
def has_duplicates(listToCheck):
number_set = set(listToCheck)
if len(number_set) is not len(listToCheck):
return True
else:
return False
for i in range(0,1000):
birthdayList=[]
for j in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x = has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000.0)*100),3),"%"
The first change I made was tidying up the indices you were using in those nested for loops. You'll see I changed the second one to j, as they were previously bot i.
The big one, though, was to the has_duplicates function. The basic principle here is that creating a set out of the incoming list gets the unique values in the list. By comparing the number of items in the number_set to the number in listToCheck we can judge whether there are any duplicates or not.
Here is what you are looking for. As this is not standard practice (to just throw code at a new user), I apologize if this offends any other users. However, I believe showing the OP a correct way to write a program should be could all do us a favor if said user keeps the lack of documentation further on in his career.
Thus, please take a careful look at the code, and fill in the blanks. Look up the python doumentation (as dry as it is), and try to understand the things that you don't get right away. Even if you understand something just by the name, it would still be wise to see what is actually happening when some built-in method is being used.
Last, but not least, take a look at this code, and take a look at your code. Note the differences, and keep trying to write your code from scratch (without looking at mine), and if it messes up, see where you went wrong, and start over. This sort of practice is key if you wish to succeed later on in programming!
def same_birthdays():
import random
'''
This is a program that does ________. It is really important
that we tell readers of this code what it does, so that the
reader doesn't have to piece all of the puzzles together,
while the key is right there, in the mind of the programmer.
'''
count = 0
#Count is going to store the number of times that we have the same birthdays
timesToRun = 1000 #timesToRun should probably be in a parameter
#timesToRun is clearly defined in its name as well. Further elaboration
#on its purpose is not necessary.
for i in range(0,timesToRun):
birthdayList = []
for j in range(0,23):
random_birthday = random.randint(1,365)
birthdayList.append(random_birthday)
birthdayList = sorted(birthdayList) #sorting for easier matching
#If we really want to, we could provide a check in the above nester
#for loop to check right away if there is a duplicate.
#But again, we are here
for j in range(0, len(birthdayList)-1):
if (birthdayList[j] == birthdayList[j+1]):
count+=1
break #leaving this nested for-loop
return count
If you wish to find the percent, then get rid of the above return statement and add:
return (count/timesToRun)
Here's a solution that doesn't use set(). It also takes a different approach with the array so that each index represents a day of the year. I also removed the hasDuplicate() function.
import random
sim_total=0
birthdayList=[]
#initialize an array of 0's representing each calendar day
for i in range(365):
birthdayList.append(0)
for i in range(0,1000):
first_dup=True
for n in range(365):
birthdayList[n]=0
for b in range(0, 23):
r = random.randint(0,364)
birthdayList[r]+=1
if (birthdayList[r] > 1) and (first_dup==True):
sim_total+=1
first_dup=False
avg = float(sim_total) / 1000 * 100
print "after 1000 simulations with 23 students there were", sim_total,"simulations with atleast one duplicate. The approximate problibility is", round(avg,3),"%"
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.