Prédire des pentes parallèles
Régression intermédiaire avec statsmodels en Python
Maarten Van den Broeck
Content Developer at DataCamp
Le flux de prédiction
import pandas as pd
import numpy as np
expl_data_length = pd.DataFrame(
{"length_cm": np.arange(5, 61, 5)})
print(expl_data_length)
length_cm
0 5
1 10
2 15
3 20
4 25
5 30
6 35
7 40
8 45
9 50
10 55
11 60
Le flux de prédiction
[A, B, C] x [1, 2] ==> [A1, B1, C1, A2, B2, C2]
from itertools import productproduct(["A", "B", "C"], [1, 2])
length_cm = np.arange(5, 61, 5)species = fish["species"].unique()p = product(length_cm, species)expl_data_both = pd.DataFrame(p, columns=['length_cm', 'species'])print(expl_data_both)
length_cm species
0 5 Bream
1 5 Roach
2 5 Perch
3 5 Pike
4 10 Bream
5 10 Roach
6 10 Perch
...
41 55 Roach
42 55 Perch
43 55 Pike
44 60 Bream
45 60 Roach
46 60 Perch
47 60 Pike
Le flux de prédiction
Prédire mass_g à partir de length_cm seulement
prediction_data_length = expl_data_length.assign(
mass_g = mdl_mass_vs_length.predict(expl_data)
)
Prédire mass_g à partir des deux variables explicatives
prediction_data_both = expl_data_both.assign(
mass_g = mdl_mass_vs_both.predict(expl_data)
)
length_cm mass_g
0 5 -361.7277
1 10 -187.2315
2 15 -12.7353
3 20 161.7610
4 25 336.2572
5 30 510.7534
... # number of rows: 12
length_cm species mass_g
0 5 Bream -459.3991
1 5 Roach -513.9350
2 5 Perch -500.4501
3 5 Pike -876.6133
4 10 Bream -246.5563
5 10 Roach -301.0923
... # number of rows: 48
Visualiser les prédictions
plt.axline(xy1=(0, ic_bream), slope=sl, color="blue")
plt.axline(xy1=(0, ic_perch), slope=sl, color="green")
plt.axline(xy1=(0, ic_pike), slope=sl, color="red")
plt.axline(xy1=(0, ic_roach), slope=sl, color="orange")
sns.scatterplot(x="length_cm",
y="mass_g",
hue="species",
data=fish)
sns.scatterplot(x="length_cm",
y="mass_g",
color="black",
data=prediction_data)
Calculer manuellement des prédictions pour une régression linéaire
coeffs = mdl_mass_vs_length.params
print(coeffs)
intercept, slope = coeffs
explanatory_data = pd.DataFrame(
{"length_cm": np.arange(5, 61, 5)})
prediction_data = explanatory_data.assign(
mass_g = intercept + slope * explanatory_data
)
print(prediction_data)
Intercept -536.223947
length_cm 34.899245
length_cm mass_g
0 5 -361.727721
1 10 -187.231494
2 15 -12.735268
3 20 161.760959
4 25 336.257185
5 30 510.753412
...
9 50 1208.738318
10 55 1383.234545
11 60 1557.730771
Calculer manuellement des prédictions pour une régression multiple
coeffs = mdl_mass_vs_both.params
print(coeffs)
species[Bream] -672.241866
species[Perch] -713.292859
species[Pike] -1089.456053
species[Roach] -726.777799
length_cm 42.568554
ic_bream, ic_perch, ic_pike, ic_roach, slope = coeffs
np.select()
conditions = [
condition_1,
condition_2,
# ...
condition_n
]
choices = [list_of_choices] # same length as conditions
np.select(conditions, choices)
Choisir une ordonnée à l'origine avec np.select()
conditions = [
explanatory_data["species"] == "Bream",
explanatory_data["species"] == "Perch",
explanatory_data["species"] == "Pike",
explanatory_data["species"] == "Roach"
]
choices = [ic_bream, ic_perch, ic_pike, ic_roach]
intercept = np.select(conditions, choices)
print(intercept)
[ -672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46
-672.24 -726.78 -713.29 -1089.46]
L'étape finale de prédiction
prediction_data = explanatory_data.assign( intercept = np.select(conditions, choices), mass_g = intercept + slope * explanatory_data["length_cm"])print(prediction_data)
length_cm species intercept mass_g
0 5 Bream -672.2419 -459.3991
1 5 Roach -726.7778 -513.9350
2 5 Perch -713.2929 -500.4501
3 5 Pike -1089.4561 -876.6133
4 10 Bream -672.2419 -246.5563
5 10 Roach -726.7778 -301.0923
6 10 Perch -713.2929 -287.6073
7 10 Pike -1089.4561 -663.7705
8 15 Bream -672.2419 -33.7136
...
40 55 Bream -672.2419 1669.0286
41 55 Roach -726.7778 1614.4927
42 55 Perch -713.2929 1627.9776
43 55 Pike -1089.4561 1251.8144
44 60 Bream -672.2419 1881.8714
45 60 Roach -726.7778 1827.3354
46 60 Perch -713.2929 1840.8204
47 60 Pike -1089.4561 1464.6572
Comparer à .predict()
mdl_mass_vs_both.predict(explanatory_data)
0 -459.3991
1 -513.9350
2 -500.4501
3 -876.6133
4 -246.5563
5 -301.0923
...
43 1251.8144
44 1881.8714
45 1827.3354
46 1840.8204
47 1464.6572
Passons à la pratique !
Régression intermédiaire avec statsmodels en Python
