When using JAGS, how does one receive output from a model in the format:
Inference for Bugs model at "model.txt", fit using jags,
3 chains, each with 10000 iterations (first 5000 discarded)
n.sims = 15000 iterations saved
mu.vect sd.vect 2.5% 25% 50% 75% 97.5% Rhat n.eff
mu 9.950 0.288 9.390 9.755 9.951 10.146 10.505 1.001 11000
sd.obs 3.545 0.228 3.170 3.401 3.534 3.675 3.978 1.001 13000
deviance 820.611 3.460 818.595 819.132 819.961 821.366 825.871 1.001 15000
I assumed, as with BUGS, it would appear when the model completes however I only get something in the format:
Compiling model graph
Resolving undeclared variables
Allocating nodes
Graph information:
Observed stochastic nodes: 1785
Unobserved stochastic nodes: 1843
Total graph size: 61542
Initializing model
|++++++++++++++++++++++++++++++++++++++++++++++++++| 100%
Apologies for the basic question. If anyone can provide useful JAGS introductory material that would also be useful.
Kind regards.
If you only get the 'plus' signs, it means you only initialized the model. When jags really runs, it typically produces '***' signs after. So you are missing a line here (would have been nice to see your code). For instance if you use r2jags, you would write:
out <- jags(data = data, parameters.to.save = params, n.chains = 3, n.iter = 90000,n.burnin = 5000,
model.file = modFile)
out.upd <- update(abundance.out.mod, n.iter=10000)
Related
I’m trying to get feature vectors from the encoder model using pre-trained ALBERT v2 weights. i have a nvidia 1650ti gpu (4 GB) , and sufficient RAM(8GB) but for some reason I’m getting Runtime error saying -
RuntimeError: [enforce fail at …\c10\core\CPUAllocator.cpp:75] data.
DefaultCPUAllocator: not enough memory: you tried to allocate
491520000 bytes. Buy new RAM!
I’m really new to pytorch and deep learning in general. Can anyone please tell me what is wrong?
My entire code -
encoded_test_data = tokenized_test_values[‘input_ids’]
encoded_test_masks = tokenized_test_values[‘attention_mask’]
encoded_train_data = torch.from_numpy(encoded_train_data).to(device)
encoded_masks = torch.from_numpy(encoded_masks).to(device)
encoded_test_data = torch.from_numpy(encoded_test_data).to(device)
encoded_test_masks = torch.from_numpy(encoded_test_masks).to(device)
config = EncoderDecoderConfig.from_encoder_decoder_configs(BertConfig(),BertConfig())
EnD_model = EncoderDecoderModel.from_pretrained(‘albert-base-v2’,config=config)
feature_extractor = EnD_model.get_encoder()
feature_vector = feature_extractor.forward(input_ids=encoded_train_data,attention_mask = encoded_masks)
feature_test_vector = feature_extractor.forward(input_ids = encoded_test_data, attention_mask = encoded_test_masks)
Also 491520000 bytes is about 490 MB which should not be a problem.
I tried reducing the number of training examples and also the length of the maximum padded input . The OOM error still exists even though the required space now is 153 MB , which should easily be managable.
I also have maxed out the RAM limit of the heap of pycharm software to 2048 MB. I really dont know what to do now…
Is there any way to detect the highest peaks using a python library without setting any parameter?. I'm developing a user interface and I want the algorithm to be able to detect highest peaks automatically...
I want it to be able to detect these peaks in picture below:
graph here
Data looks like this:
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The task you are describing could be treated like anomaly/outlier detection.
One possible solution is to use a Z-score transformation and treat every value with a z score above a certain threshold as an outlier. Because there is no clear definition of an outlier it won't be able to detect such peaks without setting any parameters (threshold).
One possible solution could be:
import numpy as np
def detect_outliers(data):
outliers = []
d_mean = np.mean(data)
d_std = np.std(data)
threshold = 3 # this defines what you would consider a peak (outlier)
for point in data:
z_score = (point - d_mean)/d_std
if np.abs(z_score) > threshold:
outliers.append(point)
return outliers
# create normal data
data = np.random.normal(size=100)
# create outliers
outliers = np.random.normal(100, size=3)
# combine normal data and outliers
full_data = data.tolist() + outliers.tolist()
# print outliers
print(detect_outliers(full_data))
If you only want to detect peaks, remove the np.abs function call from the code.
This code snippet is based on a Medium Post, which also provides another way of detecting outliers.
I'm trying to train my own NLP model with CreateML with Xcode playground, and going through the tutorial by Apple: https://developer.apple.com/documentation/createml/creating_a_text_classifier_model
but the program terminated by EXC_BAD_ACCESS (code=1, address=0x0)
I found some solution from the Internet, they stated that the pointer is pointing to NULL when trying to access the variable
import Foundation
import CreateML
let source = "icecream"
let data = try MLDataTable(contentsOf: URL(fileURLWithPath: "/path/to/\(source).csv"))
let (trainingData, testingData) = data.randomSplit(by: 0.8, seed: 0)
// program stopped here
let sentimentClassifier = try MLTextClassifier(trainingData: trainingData, textColumn: "text", labelColumn: "sentiment")
// error
error: Execution was interrupted, reason: EXC_BAD_ACCESS (code=1, address=0x0).
// output
Finished parsing file /path/to/icecream.csv
Parsing completed. Parsed 100 lines in 0.03412 secs.
Finished parsing file /path/to/icecream.csv
Parsing completed. Parsed 188 lines in 0.008235 secs.
Automatically generating validation set from 10% of the data.
Tokenizing data and extracting features
Starting MaxEnt training with 146 samples
Iteration 1 training accuracy 0.650685
Iteration 2 training accuracy 0.869863
Iteration 3 training accuracy 0.945205
Iteration 4 training accuracy 0.986301
Iteration 5 training accuracy 0.993151
Finished MaxEnt training in 0.04 seconds
I am trying to learn linearK estimates on a small linnet object from the CRC spatstat book (chapter 17) and when I use the linearK function, spatstat throws an error. I have documented the process in the comments in the r code below. The error is as below.
Error in seq.default(from = 0, to = right, length.out = npos + 1L) : 'to' cannot be NA, NaN or infinite
I do not understand how to resolve this. I am following this process:
# I have data of points for each data of the week
# d1 is district 1 of the city.
# I did the step below otherwise it was giving me tbl class
d1_data=lapply(split(d1, d1$openDatefactor),as.data.frame)
# I previously create a linnet and divided it into districts of the city
d1_linnet = districts_linnet[["d1"]]
# I create point pattern for each day
d1_ppp = lapply(d1_data, function(x) as.ppp(x, W=Window(d1_linnet)))
plot(d1_ppp[[1]], which.marks="type")
# I am then converting the point pattern to a point pattern on linear network
d1_lpp <- as.lpp(d1_ppp[[1]], L=d1_linnet, W=Window(d1_linnet))
d1_lpp
Point pattern on linear network
3 points
15 columns of marks: ‘status’, ‘number_of_’, ‘zip’, ‘ward’,
‘police_dis’, ‘community_’, ‘type’, ‘days’, ‘NAME’,
‘DISTRICT’, ‘openDatefactor’, ‘OpenDate’, ‘coseDatefactor’,
‘closeDate’ and ‘instance’
Linear network with 4286 vertices and 6183 lines
Enclosing window: polygonal boundary
enclosing rectangle: [441140.9, 448217.7] x [4640080, 4652557] units
# the errors start from plotting this lpp object
plot(d1_lpp)
"show.all" is not a graphical parameter
Show Traceback
Error in plot.window(...) : need finite 'xlim' values
coords(d1_lpp)
x y seg tp
441649.2 4649853 5426 0.5774863
445716.9 4648692 5250 0.5435492
444724.6 4646320 677 0.9189631
3 rows
And then consequently, I also get error on linearK(d1_lpp)
Error in seq.default(from = 0, to = right, length.out = npos + 1L) : 'to' cannot be NA, NaN or infinite
I feel lpp object has the problem, but I find it hard to interpret the errors and how to resolve them. Could someone please guide me?
Thanks
I can confirm there is a bug in plot.lpp when trying to plot the marked point pattern on the linear network. That will hopefully be fixed soon. You can plot the unmarked point pattern using
plot(unmark(d1_lpp))
I cannot reproduce the problem with linearK. Which version of spatstat are you running? In the development version on my laptop spatstat_1.51-0.073 everything works. There has been changes to this code recently, so it is likely that this will be solved by updating to development version (see https://github.com/spatstat/spatstat).
Can I get your help on some Maths and possibly Excel?
I have benchmarked my app increasing the number of iterations and number of obligors recording the time taken in seconds with the following result:
200 400 600 800 1000 1200 1400 1600 1800 2000
20000 15.627681 30.0968663 44.7592684 60.9037558 75.8267358 90.3718977 105.8749983 121.0030672 135.9191249 150.3331682
40000 31.7202111 62.3603882 97.2085204 128.8111731 156.2443206 186.6374271 218.324317 249.2699288 279.6008184 310.9970803
60000 47.0708635 92.4599437 138.874287 186.0576007 231.2181381 280.541207 322.9836878 371.3076757 413.4058622 459.6208335
80000 60.7346238 120.3216303 180.471169 241.668982 300.4283548 376.9639188 417.5231669 482.6288981 554.9740194 598.0394434
100000 76.7535915 150.7479245 227.5125656 304.3908046 382.5900043 451.6034296 526.0730786 609.0358776 679.0268121 779.6887277
120000 90.4174626 179.5511355 269.4099593 360.2934453 448.4387573 537.1406039 626.7325734 727.6132992 807.4767327 898.307638
How can I now come up with a function for T (time taken in seconds) as an expression of number of obligors O and number of iterations I
Thanks
I'm not quite sure of the data involved due to the question construction/presentation.
Assuming you're looking for y = f(x). If you load the data into Excel, you can use the methods SLOPE and INTERCEPT on the data ranges to derive an expression of the form
y = mx+c
and thus a linear function.
If you want a quadratic or cubic, you can use LINEST with a column of time data squared/cubed etc. to give you quadratic/cubic parameters, and thus derive an appropriate higher order function.
Spoke to one of the quants here the function is of the from T = KNO, where T is time, K some constant, N iterations, O obligors.
Rearrange for K = T/(NO), plug this into my sample data, take the average of all sample points, use the Std dev for the error
I did this for my data and get:
T = 3.81524E-06 * N * O (with 1.9% error), this is a pretty good approximation.
Create a chart in Excel, add a trendline, and select to have the equation displayed on the chart.
To clarify: You have tabular data below which you want to fit to some function f(O,I)=t?
200 400 600 800 1000 1200 1400 1600 1800 2000
20000 15.627681 30.0968663 44.7592684 60.9037558 75.8267358 90.3718977 105.8749983 121.0030672 135.9191249 150.3331682
40000 31.7202111 62.3603882 97.2085204 128.8111731 156.2443206 186.6374271 218.324317 249.2699288 279.6008184 310.9970803
60000 47.0708635 92.4599437 138.874287 186.0576007 231.2181381 280.541207 322.9836878 371.3076757 413.4058622 459.6208335
80000 60.7346238 120.3216303 180.471169 241.668982 300.4283548 376.9639188 417.5231669 482.6288981 554.9740194 598.0394434
100000 76.7535915 150.7479245 227.5125656 304.3908046 382.5900043 451.6034296 526.0730786 609.0358776 679.0268121 779.6887277
120000 90.4174626 179.5511355 269.4099593 360.2934453 448.4387573 537.1406039 626.7325734 727.6132992 807.4767327 898.307638
A rough guess looks like both O & I are linear. So f is in the form t = aO + bI + c. Plug in a few (O,I,t) and see what a,b,c should be.