Vowpal Wabbit - How to get prediction probabilities from contextual bandit model on a test sample - vowpalwabbit

Given a trained contextual bandit model, how can I retrieve a prediction vector on test samples?
For example, let's say I have a train set named "train.dat" containing lines formatted as below
1:-1:0.3 | a b c # <action:cost:probability | features>
2:2:0.3 | a d d
3:-1:0.3 | a b e
....
And I run below command.
vw -d train.dat --cb 30 -f cb.model --save_resume
This produces a file, 'cb.model'. Now, let's say I have a test dataset as below
| a d d
| a b e
I'd like to see probabilities as below
0.2 0.7 0.1
The interpretation of these probabilities would be that action 1 should be picked 20% of the time, action 2 - 70%, and action 3 - 10% of the time.
Is there a way to get something like this?

When you use "--cb K", the prediction is the optimal arm/action based on argmax policy, which is a static policy.
When using "--cb_explore K", the prediction output contains the probability for each arm/action. Depending the policy you pick, the probabilities are calculated differently.

If you send those lines to a daemon running your model, you'd get just that. You send a context, and the reply is a probability distribution across the number of allowed actions, presumably comprising the "recommendation" provided by the model.
Say you have 3 actions, like in your example. Start a contextual bandits daemon:
vowpalwabbit/vw -d train.dat --cb_explore 3 -t --daemon --quiet --port 26542
Then send a context to it:
| a d d
You'll get just what you want as the reply.

In the Workspace Class, initialize the object and then call the method predict(prediction_type: int). Below are the corresponding parameter values
class PredictionType(IntEnum):
SCALAR = pylibvw.vw.pSCALAR
SCALARS = pylibvw.vw.pSCALARS
ACTION_SCORES = pylibvw.vw.pACTION_SCORES
ACTION_PROBS = pylibvw.vw.pACTION_PROBS
MULTICLASS = pylibvw.vw.pMULTICLASS
MULTILABELS = pylibvw.vw.pMULTILABELS
PROB = pylibvw.vw.pPROB
MULTICLASSPROBS = pylibvw.vw.pMULTICLASSPROBS
DECISION_SCORES = pylibvw.vw.pDECISION_SCORES
ACTION_PDF_VALUE = pylibvw.vw.pACTION_PDF_VALUE
PDF = pylibvw.vw.pPDF
ACTIVE_MULTICLASS = pylibvw.vw.pACTIVE_MULTICLASS
NOPRED = pylibvw.vw.pNOPRED

Related

Trying to put together a teaching-example with pyhf

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!

How to calculate shap values for ADABoost model?

I am running 3 different model (Random forest, Gradient Boosting, Ada Boost) and a model ensemble based on these 3 models.
I managed to use SHAP for GB and RF but not for ADA with the following error:
Exception Traceback (most recent call last)
in engine
----> 1 explainer = shap.TreeExplainer(model,data = explain_data.head(1000), model_output= 'probability')
/home/cdsw/.local/lib/python3.6/site-packages/shap/explainers/tree.py in __init__(self, model, data, model_output, feature_perturbation, **deprecated_options)
110 self.feature_perturbation = feature_perturbation
111 self.expected_value = None
--> 112 self.model = TreeEnsemble(model, self.data, self.data_missing)
113
114 if feature_perturbation not in feature_perturbation_codes:
/home/cdsw/.local/lib/python3.6/site-packages/shap/explainers/tree.py in __init__(self, model, data, data_missing)
752 self.tree_output = "probability"
753 else:
--> 754 raise Exception("Model type not yet supported by TreeExplainer: " + str(type(model)))
755
756 # build a dense numpy version of all the tree objects
Exception: Model type not yet supported by TreeExplainer: <class 'sklearn.ensemble._weight_boosting.AdaBoostClassifier'>
I found this link on Git that state
TreeExplainer creates a TreeEnsemble object from whatever model type we are trying to explain, and then works with that downstream. So all you would need to do is and add another if statement in the
TreeEnsemble constructor similar to the one for gradient boosting
But I really don't know how to implement it since I quite new to this.
I had the same problem and what I did, was to modify the file in the git you are commenting.
In my case I use windows so the file is in C:\Users\my_user\AppData\Local\Continuum\anaconda3\Lib\site-packages\shap\explainers but you can do double click over the error message and the file will be opened.
The next step is to add another elif as the answer of the git help says. In my case I did it from the line 404 as following:
1) Modify the source code.
...
self.objective = objective_name_map.get(model.criterion, None)
self.tree_output = "probability"
elif str(type(model)).endswith("sklearn.ensemble.weight_boosting.AdaBoostClassifier'>"): #From this line I have modified the code
scaling = 1.0 / len(model.estimators_) # output is average of trees
self.trees = [Tree(e.tree_, normalize=True, scaling=scaling) for e in model.estimators_]
self.objective = objective_name_map.get(model.base_estimator_.criterion, None) #This line is done to get the decision criteria, for example gini.
self.tree_output = "probability" #This is the last line I added
elif str(type(model)).endswith("sklearn.ensemble.forest.ExtraTreesClassifier'>"): # TODO: add unit test for this case
scaling = 1.0 / len(model.estimators_) # output is average of trees
self.trees = [Tree(e.tree_, normalize=True, scaling=scaling) for e in model.estimators_]
...
Note in the other models, the code of shap needs the attribute 'criterion' that the AdaBoost classifier doesn't have in a direct way. So in this case this attribute is obtained from the "weak" classifiers with the AdaBoost has been trained, that's why I add model.base_estimator_.criterion .
Finally you have to import the library again, train your model and get the shap values. I leave an example:
2) Import again the library and try:
from sklearn import datasets
from sklearn.ensemble import AdaBoostClassifier
import shap
# import some data to play with
iris = datasets.load_iris()
X = iris.data
y = iris.target
ADABoost_model = AdaBoostClassifier()
ADABoost_model.fit(X, y)
shap_values = shap.TreeExplainer(ADABoost_model).shap_values(X)
shap.summary_plot(shap_values, X, plot_type="bar")
Which generates the following:
3) Get your new results:
It seems that the shap package has been updated and still does not contain the AdaBoostClassifier. Based on the previous answer, I've modified the previous answer to work with the shap/explainers/tree.py file in lines 598-610
### Added AdaBoostClassifier based on the outdated StackOverflow response and Github issue here
### https://stackoverflow.com/questions/60433389/how-to-calculate-shap-values-for-adaboost-model/61108156#61108156
### https://github.com/slundberg/shap/issues/335
elif safe_isinstance(model, ["sklearn.ensemble.AdaBoostClassifier", "sklearn.ensemble._weighted_boosting.AdaBoostClassifier"]):
assert hasattr(model, "estimators_"), "Model has no `estimators_`! Have you called `model.fit`?"
self.internal_dtype = model.estimators_[0].tree_.value.dtype.type
self.input_dtype = np.float32
scaling = 1.0 / len(model.estimators_) # output is average of trees
self.trees = [Tree(e.tree_, normalize=True, scaling=scaling) for e in model.estimators_]
self.objective = objective_name_map.get(model.base_estimator_.criterion, None) #This line is done to get the decision criteria, for example gini.
self.tree_output = "probability" #This is the last line added
Also working on testing to add this to the package :)

H2O GLM model: saved MOJO's prediction is very different when running on the same validation data

I built a GLM model using H2O (ver 3.14) in R. Please note that the training data contains integers, and also many NA, which I use MeanImputation to handle them.
glm <- h2o.glm(
training_frame = train.truth,
x=getColNames(train.truth),
y="isFemale",
family = "binomial",
missing_values_handling = "MeanImputation",
seed = 1000000)
I then use a validation data set to look at the perf, and the Precision looks good to me:
h2o.performance(glm, newdata=valid.truth)%>% h2o.confusionMatrix()
Confusion Matrix (vertical: actual; across: predicted) for max f1 # threshold = 0.529384526696015:
0 1 Error Rate
0 41962 300 0.007099 =300/42262
1 863 13460 0.060253 =863/14323
Totals 42825 13760 0.020553 =1163/56585
I then saved the model as a MOJO:
h2o.download_mojo(glm, path="models/mojo", get_genmodel_jar=TRUE)
I exported the validation DF to a CSV file:
dt.valid <- data.table(as.data.frame(valid.truth))
write.table(dt.valid, row.names = F, na="", file="models/test.csv")
I tried to use the saved mojo to do the same prediction by running this on my Linux shell:
java -cp h2o-genmodel.jar hex.genmodel.tools.PredictCsv \
--mojo GLM_model_R_1511161743608_15 \
--decimal --mojo GLM_model_R_1511161743608_15.zip \
--input ../test.csv --output output.csv
However, the result is terrible. All the records were predicted as 0, which is very different from what I got when I ran the model in R.
I have been stuck in this for a day but I couldn't figure out what went wrong. Anyone can shed some light on this?

Error in setting max features parameter in Isolation Forest algorithm using sklearn

I'm trying to train a dataset with 357 features using Isolation Forest sklearn implementation. I can successfully train and get results when the max features variable is set to 1.0 (the default value).
However when max features is set to 2, it gives the following error:
ValueError: Number of features of the model must match the input.
Model n_features is 2 and input n_features is 357
It also gives the same error when the feature count is 1 (int) and not 1.0 (float).
How I understood was that when the feature count is 2 (int), two features should be considered in creating each tree. Is this wrong? How can I change the max features parameter?
The code is as follows:
from sklearn.ensemble.iforest import IsolationForest
def isolation_forest_imp(dataset):
estimators = 10
samples = 100
features = 2
contamination = 0.1
bootstrap = False
random_state = None
verbosity = 0
estimator = IsolationForest(n_estimators=estimators, max_samples=samples, contamination=contamination,
max_features=features,
bootstrap=boostrap, random_state=random_state, verbose=verbosity)
model = estimator.fit(dataset)
In the documentation it states:
max_features : int or float, optional (default=1.0)
The number of features to draw from X to train each base estimator.
- If int, then draw `max_features` features.
- If float, then draw `max_features * X.shape[1]` features.
So, 2 should mean take two features and 1.0 should mean take all of the features, 0.5 take half and so on, from what I understand.
I think this could be a bug, since, taking a look in IsolationForest's fit:
# Isolation Forest inherits from BaseBagging
# and when _fit is called, BaseBagging takes care of the features correctly
super(IsolationForest, self)._fit(X, y, max_samples,
max_depth=max_depth,
sample_weight=sample_weight)
# however, when after _fit the decision_function is called using X - the whole sample - not taking into account the max_features
self.threshold_ = -sp.stats.scoreatpercentile(
-self.decision_function(X), 100. * (1. - self.contamination))
then:
# when the decision function _validate_X_predict is called, with X unmodified,
# it calls the base estimator's (dt) _validate_X_predict with the whole X
X = self.estimators_[0]._validate_X_predict(X, check_input=True)
...
# from tree.py:
def _validate_X_predict(self, X, check_input):
"""Validate X whenever one tries to predict, apply, predict_proba"""
if self.tree_ is None:
raise NotFittedError("Estimator not fitted, "
"call `fit` before exploiting the model.")
if check_input:
X = check_array(X, dtype=DTYPE, accept_sparse="csr")
if issparse(X) and (X.indices.dtype != np.intc or
X.indptr.dtype != np.intc):
raise ValueError("No support for np.int64 index based "
"sparse matrices")
# so, this check fails because X is the original X, not with the max_features applied
n_features = X.shape[1]
if self.n_features_ != n_features:
raise ValueError("Number of features of the model must "
"match the input. Model n_features is %s and "
"input n_features is %s "
% (self.n_features_, n_features))
return X
So, I am not sure on how you can handle this. Maybe figure out the percentage that leads to just the two features you need - even though I am not sure it'll work as expected.
Note: I am using scikit-learn v.0.18
Edit: as #Vivek Kumar commented this is an issue and upgrading to 0.20 should do the trick.

PID control - value of process parameter based on PID result

I'm trying to implement a PID controller following http://en.wikipedia.org/wiki/PID_controller
The mechanism I try to control works as follows:
1. I have an input variable which I can control. Typical values would be 0.5...10.
2. I have an output value which I measure daily. My goal for the output is roughly at the same range.
The two variables have strong correlation - when the process parameter goes up, the output generally goes up, but there's quite a bit of noise.
I'm following the implementation here:
http://code.activestate.com/recipes/577231-discrete-pid-controller/
Now the PID seems like it is correlated with the error term, not the measured level of output. So my guess is that I am not supposed to use it as-is for the process variable, but rather as some correction to the current value? How is that supposed to work exactly?
For example, if we take Kp=1, Ki=Kd=0, The process (input) variable is 4, the current output level is 3 and my target is a value of 2, I get the following:
error = 2-3 = -1
PID = -1
Then I should set the process variable to -1? or 4-1=3?
You need to think in terms of the PID controller correcting a manipulated variable (MV) for errors, and that you need to use an I term to get to an on-target steady-state result. The I term is how the PID retains and applies memory of the prior behavior of the system.
If you are thinking in terms of the output of the controller being changes in the MV, it is more of a 'velocity form' PID, and the memory of prior errors and behavior is integrated and accumulated in the prior MV setting.
From your example, it seems like a manipulated value of -1 is not feasible and that you would like the controller to suggest a value like 3 to get a process output (PV) of 2. For a PID controller to make use of "The process (input) variable is 4,..." (MV in my terms) Ki must be non-zero, and if the system was at steady-state, whatever was accumulated in the integral (sum_e=sum(e)) would precisely equal 4/Ki, so:
Kp= Ki = 1 ; Kd =0
error = SV - PV = 2 - 3 = -1
sum_e = sum_e + error = 4/Ki -1
MV = PID = -1(Kp) + (4/Ki -1)Ki = -1Kp + 4 - 1*Ki = -1 +4 -1 = 2
If you used a slower Ki than 1, it would smooth out the noise more and not adjust the MV so quickly:
Ki = 0.1 ;
MV = PID = -1(Kp) + (4/Ki -1)Ki = -1Kp + 4 - 1*Ki = -1 +4 -0.1 = 2.9
At steady state at target (PV = SV), sum_e * Ki should produce the steady-state MV:
PV = SV
error = SV - PV = 0
Kp * error = 0
MV = 3 = PID = 0 * Kp + Ki * sum_e
A nice way to understand the PID controller is to put units on everything and think of Kp, Ki, Kd as conversions of the process error, accumulated error*timeUnit, and rate-of-change of error/timeUnit into terms of the manipulated variable, and that the controlled system converts the controller's manipulated variable into units of output.

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