I am getting error mentioned in the title and didn't find yet a solution.
X = train[feats].values
y = train['Target'].values
cv = StratifiedKFold(n_splits=3, random_state=2021, shuffle=True)
model = LogisticRegression(solver='liblinear')
scores = []
for train_idx, test_idx in cv.split(X, y):
model.fit(X[train_idx], y[train_idx])
y_pred = model.predict(X[test_idx])
score = mean_absolute_error(y[test_idx], y_pred )
scores.append(score)
print(np.mean(scores), np.std(scores))
fig = plt.figure(figsize=(15,6));
ax1 = fig.add_subplot(121)
ax2 = fig.add_subplot(122)
skplt.metrics.plot_confusion_matrix(y, y_pred, ax = ax1) #error line
skplt.metrics.plot_roc(y, y_pred, ax = ax2)
ValueError: Found input variables with inconsistent numbers of samples: [32561, 10853]
I checked the code, read many threads on this error. Somebody suggested me as a solution to put the cross-validation in a loop, but I don't know how to manage this with code (and also which part of operation to put in a loop, and how to write a condition that should be ending this loop). Please, help me with a specific answer that will help me to easily fix problem with my current level of advancement.
I'm trying to learn more about pyhf and my understanding of what the goals are might be limited. I would love to fit my HEP data outside of ROOT, but I could be imposing expectations on pyhf which are not what the authors intended for it's use.
I'd like to write myself a hello-world example, but I might just not know what I'm doing. My misunderstanding could also be gaps in my statistical knowledge.
With that preface, let me explain what I'm trying to explore.
I have some observed set of events for which I calculate some observable and make a binned histogram of that data. I hypothesize that there are two contributing physics processes, which I call signal and background. I generate some Monte Carlo samples for these processes and the theorized total number of events is close to, but not exactly what I observe.
I would like to:
Fit the data to this two process hypothesis
Get from the fit the optimal values for the number of events for each process
Get the uncertainties on these fitted values
If appropriate, calculate an upper limit on the number of signal events.
My starter code is below, where all I'm doing is an ML fit but I'm not sure where to go. I know it's not set up to do what I want, but I'm getting lost in the examples I find on RTD. I'm sure it's me, this is not a criticism of the documentation.
import pyhf
import numpy as np
import matplotlib.pyplot as plt
nbins = 15
# Generate a background and signal MC sample`
MC_signal_events = np.random.normal(5,1.0,200)
MC_background_events = 10*np.random.random(1000)
signal_data = np.histogram(MC_signal_events,bins=nbins)[0]
bkg_data = np.histogram(MC_background_events,bins=nbins)[0]
# Generate an observed dataset with a slightly different
# number of events
signal_events = np.random.normal(5,1.0,180)
background_events = 10*np.random.random(1050)
observed_events = np.array(signal_events.tolist() + background_events.tolist())
observed_sample = np.histogram(observed_events,bins=nbins)[0]
# Plot these samples, if you like
plt.figure(figsize=(12,4))
plt.subplot(1,3,1)
plt.hist(observed_events,bins=nbins,label='Observations')
plt.legend()
plt.subplot(1,3,2)
plt.hist(MC_signal_events,bins=nbins,label='MC signal')
plt.legend()
plt.subplot(1,3,3)
plt.hist(MC_background_events,bins=nbins,label='MC background')
plt.legend()
# Use a very naive estimate of the background
# uncertainties
bkg_uncerts = np.sqrt(bkg_data)
print("Defining the PDF.......")
pdf = pyhf.simplemodels.hepdata_like(signal_data=signal_data.tolist(), \
bkg_data=bkg_data.tolist(), \
bkg_uncerts=bkg_uncerts.tolist())
print("Fit.......")
data = pyhf.tensorlib.astensor(observed_sample.tolist() + pdf.config.auxdata)
bestfit_pars, twice_nll = pyhf.infer.mle.fit(data, pdf, return_fitted_val=True)
print(bestfit_pars)
print(twice_nll)
plt.show()
Note: this answer is based on pyhf v0.5.2.
Alright, so it looks like you've managed to figure most of the big pieces for sure. However, there's two different ways to do this depending on how you prefer to set things up. In both cases, I assume you want an unconstrained fit and you want to...
fit your signal+background model to observed data
fit your background model to observed data
First, let's discuss uncertainties briefly. At the moment, we default to numpy for the tensor background and scipy for the optimizer. See documentation:
numpy backend
scipy optimizer
However, one unfortunate drawback right now with the scipy optimizer is that it cannot return the uncertainties. What you need to do anywhere in your code before the fit (although we generally recommend as early as possible) is to use the minuit optimizer instead:
pyhf.set_backend('numpy', 'minuit')
This will get you the nice features of being able to get the correlation matrix, the uncertainties on the fitted parameters, and the hessian -- amongst other things. We're working to make this consistent for scipy as well, but this is not ready right now.
All optimizations go through our optimizer API which you can currently view through the mixin here in our documentation. Specifically, the signature is
minimize(
objective,
data,
pdf,
init_pars,
par_bounds,
fixed_vals=None,
return_fitted_val=False,
return_result_obj=False,
do_grad=None,
do_stitch=False,
**kwargs)
There are a lot of options here. Let's just focus on the fact that one of the keyword arguments we can pass through is return_uncertainties which will change the bestfit parameters by adding a column for the fitted parameter uncertainty which you want.
1. Signal+Background
In this case, we want to just use the default model
result, twice_nll = pyhf.infer.mle.fit(
data,
pdf,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
2. Background-Only
In this case, we need to turn off the signal. The way we do this is by setting the parameter of interest (POI) fixed to 0.0. Then we can get the fitted parameters for the background-only model in a similar way, but using fixed_poi_fit instead of an unconstrained fit:
result, twice_nll = pyhf.infer.mle.fixed_poi_fit(
0.0,
data,
pdf,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
Note that this is quite simply a quick way of doing the following unconstrained fit
bkg_params = pdf.config.suggested_init()
fixed_params = pdf.config.suggested_fixed()
bkg_params[pdf.config.poi_index] = 0.0
fixed_params[pdf.config.poi_index] = True
result, twice_nll = pyhf.infer.mle.fit(
data,
pdf,
init_pars=bkg_params,
fixed_params=fixed_params,
return_uncertainties=True,
return_fitted_val=True
)
bestfit_pars, errors = result.T
Hopefully that clarifies things up more!
Giordon's solution should answer all of your question, but I thought I'd also write out the code to basically address everything we can.
I also take the liberty of changing some of your values a bit so that the signal isn't so strong that the observed CLs value isn't far off to the right of the Brazil band (the results aren't wrong obviously, but it probably makes more sense to be talking about using the discovery test statistic at that point then setting limits. :))
Environment
For this example I'm going to setup a clean Python 3 virtual environment and then install the dependencies (here we're going to be using pyhf v0.5.2)
$ python3 -m venv "${HOME}/.venvs/question"
$ . "${HOME}/.venvs/question/bin/activate"
(question) $ cat requirements.txt
pyhf[minuit,contrib]~=0.5.2
black
(question) $ python -m pip install -r requirements.txt
Code
While we can't easily get the best fit value for both the number of signal events as well as the background events we definitely can do inference to get the best fit value for the signal strength.
The following chunk of code (which is long only because of the visualization) should address all of the points of your question.
# answer.py
import numpy as np
import pyhf
import matplotlib.pyplot as plt
import pyhf.contrib.viz.brazil
# Goals:
# - Fit the model to the observed data
# - Infer the best fit signal strength given the model
# - Get the uncertainties on the best fit signal strength
# - Calculate an 95% CL upper limit on the signal strength
def plot_hist(ax, bins, data, bottom=0, color=None, label=None):
bin_width = bins[1] - bins[0]
bin_leftedges = bins[:-1]
bin_centers = [edge + bin_width / 2.0 for edge in bin_leftedges]
ax.bar(
bin_centers, data, bin_width, bottom=bottom, alpha=0.5, color=color, label=label
)
def plot_data(ax, bins, data, label="Data"):
bin_width = bins[1] - bins[0]
bin_leftedges = bins[:-1]
bin_centers = [edge + bin_width / 2.0 for edge in bin_leftedges]
ax.scatter(bin_centers, data, color="black", label=label)
def invert_interval(test_mus, hypo_tests, test_size=0.05):
# This will be taken care of in v0.5.3
cls_obs = np.array([test[0] for test in hypo_tests]).flatten()
cls_exp = [
np.array([test[1][idx] for test in hypo_tests]).flatten() for idx in range(5)
]
crossing_test_stats = {"exp": [], "obs": None}
for cls_exp_sigma in cls_exp:
crossing_test_stats["exp"].append(
np.interp(
test_size, list(reversed(cls_exp_sigma)), list(reversed(test_mus))
)
)
crossing_test_stats["obs"] = np.interp(
test_size, list(reversed(cls_obs)), list(reversed(test_mus))
)
return crossing_test_stats
def main():
np.random.seed(0)
pyhf.set_backend("numpy", "minuit")
observable_range = [0.0, 10.0]
bin_width = 0.5
_bins = np.arange(observable_range[0], observable_range[1] + bin_width, bin_width)
n_bkg = 2000
n_signal = int(np.sqrt(n_bkg))
# Generate simulation
bkg_simulation = 10 * np.random.random(n_bkg)
signal_simulation = np.random.normal(5, 1.0, n_signal)
bkg_sample, _ = np.histogram(bkg_simulation, bins=_bins)
signal_sample, _ = np.histogram(signal_simulation, bins=_bins)
# Generate observations
signal_events = np.random.normal(5, 1.0, int(n_signal * 0.8))
bkg_events = 10 * np.random.random(int(n_bkg + np.sqrt(n_bkg)))
observed_events = np.array(signal_events.tolist() + bkg_events.tolist())
observed_sample, _ = np.histogram(observed_events, bins=_bins)
# Visualize the simulation and observations
fig, ax = plt.subplots()
fig.set_size_inches(7, 5)
plot_hist(ax, _bins, bkg_sample, label="Background")
plot_hist(ax, _bins, signal_sample, bottom=bkg_sample, label="Signal")
plot_data(ax, _bins, observed_sample)
ax.legend(loc="best")
ax.set_ylim(top=np.max(observed_sample) * 1.4)
ax.set_xlabel("Observable")
ax.set_ylabel("Count")
fig.savefig("components.png")
# Build the model
bkg_uncerts = np.sqrt(bkg_sample)
model = pyhf.simplemodels.hepdata_like(
signal_data=signal_sample.tolist(),
bkg_data=bkg_sample.tolist(),
bkg_uncerts=bkg_uncerts.tolist(),
)
data = pyhf.tensorlib.astensor(observed_sample.tolist() + model.config.auxdata)
# Perform inference
fit_result = pyhf.infer.mle.fit(data, model, return_uncertainties=True)
bestfit_pars, par_uncerts = fit_result.T
print(
f"best fit parameters:\
\n * signal strength: {bestfit_pars[0]} +/- {par_uncerts[0]}\
\n * nuisance parameters: {bestfit_pars[1:]}\
\n * nuisance parameter uncertainties: {par_uncerts[1:]}"
)
# Perform hypothesis test scan
_start = 0.0
_stop = 5
_step = 0.1
poi_tests = np.arange(_start, _stop + _step, _step)
print("\nPerforming hypothesis tests\n")
hypo_tests = [
pyhf.infer.hypotest(
mu_test,
data,
model,
return_expected_set=True,
return_test_statistics=True,
qtilde=True,
)
for mu_test in poi_tests
]
# Upper limits on signal strength
results = invert_interval(poi_tests, hypo_tests)
print(f"Observed Limit on µ: {results['obs']:.2f}")
print("-----")
for idx, n_sigma in enumerate(np.arange(-2, 3)):
print(
"Expected {}Limit on µ: {:.3f}".format(
" " if n_sigma == 0 else "({} σ) ".format(n_sigma),
results["exp"][idx],
)
)
# Visualize the "Brazil band"
fig, ax = plt.subplots()
fig.set_size_inches(7, 5)
ax.set_title("Hypothesis Tests")
ax.set_ylabel(r"$\mathrm{CL}_{s}$")
ax.set_xlabel(r"$\mu$")
pyhf.contrib.viz.brazil.plot_results(ax, poi_tests, hypo_tests)
fig.savefig("brazil_band.png")
if __name__ == "__main__":
main()
which when run gives
(question) $ python answer.py
best fit parameters:
* signal strength: 1.5884737977889158 +/- 0.7803435235862329
* nuisance parameters: [0.99020988 1.06040191 0.90488207 1.03531383 1.09093327 1.00942088
1.07789316 1.01125627 1.06202964 0.95780043 0.94990993 1.04893286
1.0560711 0.9758487 0.93692481 1.04683181 1.05785515 0.92381263
0.93812855 0.96751869]
* nuisance parameter uncertainties: [0.06966439 0.07632218 0.0611428 0.07230328 0.07872258 0.06899675
0.07472849 0.07403246 0.07613661 0.08606657 0.08002775 0.08655314
0.07564512 0.07308117 0.06743479 0.07383134 0.07460864 0.06632003
0.06683251 0.06270965]
Performing hypothesis tests
/home/stackoverflow/.venvs/question/lib/python3.7/site-packages/pyhf/infer/calculators.py:229: RuntimeWarning: invalid value encountered in double_scalars
teststat = (qmu - qmu_A) / (2 * self.sqrtqmuA_v)
Observed Limit on µ: 2.89
-----
Expected (-2 σ) Limit on µ: 0.829
Expected (-1 σ) Limit on µ: 1.110
Expected Limit on µ: 1.542
Expected (1 σ) Limit on µ: 2.147
Expected (2 σ) Limit on µ: 2.882
Let us know if you have any further questions!
When I run the following code
from transformers import AutoTokenizer, AutoModelForQuestionAnswering
import torch
tokenizer = AutoTokenizer.from_pretrained("bert-large-uncased-whole-word-masking-finetuned-squad")
model = AutoModelForQuestionAnswering.from_pretrained("bert-large-uncased-whole-word-masking-finetuned-squad")
text = r"""
As checked Dis is not yet on boarded to ARB portal, hence we cannot upload the invoices in portal
"""
questions = [
"Dis asked if it is possible to post the two invoice in ARB.I have not access so I wanted to check if you would be able to do it.",
]
for question in questions:
inputs = tokenizer.encode_plus(question, text, add_special_tokens=True, return_tensors="pt")
input_ids = inputs["input_ids"].tolist()[0]
text_tokens = tokenizer.convert_ids_to_tokens(input_ids)
answer_start_scores, answer_end_scores = model(**inputs)
answer_start = torch.argmax(
answer_start_scores
) # Get the most likely beginning of answer with the argmax of the score
answer_end = torch.argmax(answer_end_scores) + 1 # Get the most likely end of answer with the argmax of the score
answer = tokenizer.convert_tokens_to_string(tokenizer.convert_ids_to_tokens(input_ids[answer_start:answer_end]))
print(f"Question: {question}")
print(f"Answer: {answer}\n")
The answer that I get here is:
Question: Dis asked if it is possible to post the two invoice in ARB.I have not access so I wanted to check if you would be able to do it.
Answer: dis is not yet on boarded to ARB portal
How do I get a score for this answer? Score here is very similar to what is I get when I run Question-Answer pipeline .
I have to take this approach since Question-Answer pipeline when used is giving me Key Error for the below code
from transformers import pipeline
nlp = pipeline("question-answering")
context = r"""
As checked Dis is not yet on boarded to ARB portal, hence we cannot upload the invoices in portal.
"""
print(nlp(question="Dis asked if it is possible to post the two invoice in ARB?", context=context))
This is my attempt to get the score. It appears that I cannot figure out what feature.p_mask. So I could not remove the non-context indexes that contribute to the softmax at the moment.
# ... assuming imports and question and context
model_name="deepset/roberta-base-squad2"
tokenizer = AutoTokenizer.from_pretrained(model_name)
model = AutoModelForQuestionAnswering.from_pretrained(model_name)
inputs = tokenizer(question, context,
add_special_tokens=True,
return_tensors='pt')
input_ids = inputs['input_ids'].tolist()[0]
outputs = model(**inputs)
# used to compute score
start = outputs.start_logits.detach().numpy()
end = outputs.end_logits.detach().numpy()
# from source code
# Ensure padded tokens & question tokens cannot belong to the set of candidate answers.
#?? undesired_tokens = np.abs(np.array(feature.p_mask) - 1) & feature.attention_mask
# Generate mask
undesired_tokens = inputs['attention_mask']
undesired_tokens_mask = undesired_tokens == 0.0
# Make sure non-context indexes in the tensor cannot contribute to the softmax
start_ = np.where(undesired_tokens_mask, -10000.0, start)
end_ = np.where(undesired_tokens_mask, -10000.0, end)
# Normalize logits and spans to retrieve the answer
start_ = np.exp(start_ - np.log(np.sum(np.exp(start_), axis=-1, keepdims=True)))
end_ = np.exp(end_ - np.log(np.sum(np.exp(end_), axis=-1, keepdims=True)))
# Compute the score of each tuple(start, end) to be the real answer
outer = np.matmul(np.expand_dims(start_, -1), np.expand_dims(end_, 1))
# Remove candidate with end < start and end - start > max_answer_len
max_answer_len = 15
candidates = np.tril(np.triu(outer), max_answer_len - 1)
scores_flat = candidates.flatten()
idx_sort = [np.argmax(scores_flat)]
start, end = np.unravel_index(idx_sort, candidates.shape)[1:]
end += 1
score = candidates[0, start, end-1]
start, end, score = start.item(), end.item(), score.item()
print(tokenizer.decode(input_ids[start:end]))
print(score)
See more source code
I have read a lot about the pain of replicate the easy robust option from STATA to R to use robust standard errors. I replicated following approaches: StackExchange and Economic Theory Blog. They work but the problem I face is, if I want to print my results using the stargazer function (this prints the .tex code for Latex files).
Here is the illustration to my problem:
reg1 <-lm(rev~id + source + listed + country , data=data2_rev)
stargazer(reg1)
This prints the R output as .tex code (non-robust SE) If i want to use robust SE, i can do it with the sandwich package as follow:
vcov <- vcovHC(reg1, "HC1")
if I now use stargazer(vcov) only the output of the vcovHC function is printed and not the regression output itself.
With the package lmtest() it is possible to print at least the estimator, but not the observations, R2, adj. R2, Residual, Residual St.Error and the F-Statistics.
lmtest::coeftest(reg1, vcov. = sandwich::vcovHC(reg1, type = 'HC1'))
This gives the following output:
t test of coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.54923 6.85521 -0.3719 0.710611
id 0.39634 0.12376 3.2026 0.001722 **
source 1.48164 4.20183 0.3526 0.724960
country -4.00398 4.00256 -1.0004 0.319041
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
How can I add or get an output with the following parameters as well?
Residual standard error: 17.43 on 127 degrees of freedom
Multiple R-squared: 0.09676, Adjusted R-squared: 0.07543
F-statistic: 4.535 on 3 and 127 DF, p-value: 0.00469
Did anybody face the same problem and can help me out?
How can I use robust standard errors in the lm function and apply the stargazer function?
You already calculated robust standard errors, and there's an easy way to include it in the stargazeroutput:
library("sandwich")
library("plm")
library("stargazer")
data("Produc", package = "plm")
# Regression
model <- plm(log(gsp) ~ log(pcap) + log(pc) + log(emp) + unemp,
data = Produc,
index = c("state","year"),
method="pooling")
# Adjust standard errors
cov1 <- vcovHC(model, type = "HC1")
robust_se <- sqrt(diag(cov1))
# Stargazer output (with and without RSE)
stargazer(model, model, type = "text",
se = list(NULL, robust_se))
Solution found here: https://www.jakeruss.com/cheatsheets/stargazer/#robust-standard-errors-replicating-statas-robust-option
Update I'm not so much into F-Tests. People are discussing those issues, e.g. https://stats.stackexchange.com/questions/93787/f-test-formula-under-robust-standard-error
When you follow http://www3.grips.ac.jp/~yamanota/Lecture_Note_9_Heteroskedasticity
"A heteroskedasticity-robust t statistic can be obtained by dividing an OSL estimator by its robust standard error (for zero null hypotheses). The usual F-statistic, however, is invalid. Instead, we need to use the heteroskedasticity-robust Wald statistic."
and use a Wald statistic here?
This is a fairly simple solution using coeftest:
reg1 <-lm(rev~id + source + listed + country , data=data2_rev)
cl_robust <- coeftest(reg1, vcov = vcovCL, type = "HC1", cluster = ~
country)
se_robust <- cl_robust[, 2]
stargazer(reg1, reg1, cl_robust, se = list(NULL, se_robust, NULL))
Note that I only included cl_robust in the output as a verification that the results are identical.
I am currently runnuing training in matlab on a matrix of logspecrum samples I am constantly dealing with underflow problems.I understood that I need to work with log's in order to deal with underflowing.
I am still strugling with uderflow though , when i calculate the mean (mue) bucause it is negetive i cant work with logs so i need the real values that underflow.
These are equasions i am working with:
In MATLAB code i calulate log_tau in oreder avoid underflow but when calulating mue i need exp(log(tau)) which goes to zero.
I am attaching relevent MATLAB code
**in the code i called the variable alpha is tau ...
for i = 1 : 50
log_c = Logsum(log_alpha,1) - log(N);
c = exp(log_c);
mue = DataMat*alpha./(repmat(exp(Logsum(log_alpha,1)),FrameSize,1));
log_abs_mue = log(abs(mue));
log_SigmaSqr = log((DataMat.^2)*alpha) - repmat(Logsum(log_alpha,1),FrameSize,1) - 2*log_abs_mue;
SigmaSqr = exp(log_SigmaSqr);
for j=1:N
rep_DataMat(:,:,j) = repmat(DataMat(:,j),1,M);
log_gamma(j,:) = log_c - 0.5*(FrameSize*log(2*pi)+sum(log_SigmaSqr)) + sum((rep_DataMat(:,:,j) - mue).^2./(2*SigmaSqr));
end
log_alpha = log_gamma - repmat(Logsum(log_gamma,2),1,M);
alpha = exp(log_alpha);
end
c = exp(log_c);
SigmaSqr = exp(log_SigmaSqr);
does any one see how i can avoid this? or what needs to be fixed in code?
What i did was add this line to the MATLAB code:
mue(isnan(mue))=0; %fix 0/0 problem
and this one:
SigmaSqr(SigmaSqr==0)=1;%fix if mue_k = x_k
not sure if this is the best solution but is seems to work...
any have a better idea?