Implement truncated responses

[tomicapretto]

This PR adds a function truncated() that can be used in the formula syntax to specify truncated variables.

Example (taken from here)

import bambi as bmb
import numpy as np
import pandas as pd

rng = np.random.default_rng(12345)
slope, intercept, σ, N = 1, 0, 2, 200
x = rng.uniform(-10, 10, N)
y = rng.normal(loc=slope * x + intercept, scale=σ)

def truncate_y(x, y, bounds):
    keep = (y >= bounds[0]) & (y <= bounds[1])
    return (x[keep], y[keep])

bounds = [-5, 5]
xt, yt = truncate_y(x, y, bounds)

df = pd.DataFrame({"x": xt, "y": yt})
priors = {
    "Intercept": bmb.Prior("Normal", mu=0, sigma=1), 
    "x": bmb.Prior("Normal", mu=0, sigma=1),
    "sigma": bmb.Prior("HalfNormal", sigma=1)
}
model = bmb.Model("truncated(y, -5, 5) ~ x", df, priors=priors)

NOTE I still need to implement tests

[codecov-commenter]

Codecov Report

Merging #752 (f4eef3f) into main (ee266d9) will increase coverage by 0.05%. The diff coverage is 94.11%.

@@            Coverage Diff             @@
##             main     #752      +/-   ##
==========================================
+ Coverage   89.84%   89.90%   +0.05%     
==========================================
  Files          45       45              
  Lines        3664     3713      +49     
==========================================
+ Hits         3292     3338      +46     
- Misses        372      375       +3     

Files	Coverage Δ
bambi/terms/response.py	79.68% <100.00%> (+0.32%)	:arrow_up:
bambi/transformations.py	97.87% <100.00%> (+0.45%)	:arrow_up:
bambi/terms/utils.py	91.30% <85.71%> (-2.45%)	:arrow_down:
bambi/backend/terms.py	96.14% <88.23%> (-0.52%)	:arrow_down:

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