Python Regression

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Find a trend in data

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Without diving deep into Scipy, this repo supports fast and accurate data trend analysis.

Get Started

git clone https://github.com/Weilory/python-regression

place regression folder to base level directory.

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Just call regression, it handles all for you.

from regression.regress import regression

x = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
y = [1, 3.2, 5, 7.1, 9, 11, 13, 15.2, 17, 19.1, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39]

expression = regression(x=x, y=y)
print(expression.write)
# y = 2.0 * x**1 + 1.0
my_formula = expression.formula

print(my_formula(2))
# 5.0

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if you know the general equation of the data, it will be faster and more accurate

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Linear

from regression.regress import linear_regression

x = [0, 1, 2, 3, -4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -18, 19]
y = [1, 3.2, 5, 7.1, 9, 11, 13, 15.2, 17, 19.1, 20, 22.5, 24.7, 26.3, 28.9, 31.2, 33.1, 34.3, 36.8, 38.7]

expression = linear_regression(x=x, y=y)
print(expression.write)
# y = 1.939 * x + 1.406
my_formula = expression.formula

print(my_formula(2))
# 5.284625463284197

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Parabola

from regression.regress import parabola_regression

x = [0, 1, 2, 3, -4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -18, 19]
y = [1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, 276, 325, 378, 435, 496, 561, 630, 703, 780]

expression = parabola_regression(x=x, y=y)
print(expression.write)
# y = 2.0 * x**2 + 3.0 * x**1 + 1.0
my_formula = expression.formula

print(my_formula(2))
# 15

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Cubic Parabola

from regression.regress import cubic_regression

x = [0, 1, 2, 3, -4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -18, 19]
y = [1.0, 6.5, 19.0, 41.5, 77.0, 128.5, 199.0, 291.5, 409.0, 554.5, 731.0, 941.5, 1189.0, 1476.5, 1807.0, 2183.5, 2609.0, 3086.5, 3619.0, 4209.5]

expression = cubic_regression(x=x, y=y)
print(expression.write)
# y = 0.5 * x**3 + 2.0 * x**2 + 3.0 * x**1 + 1.0
my_formula = expression.formula

print(my_formula(2))
# 19.0

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Notice that,

• parabola
• cubic
• quadratic they have similarities such that

they are categorised as Exponential Parabola, with a specification to maximum x exponential, for the cubic function exampled above, instead, we can write following:

Exponential Parabola

from regression.regress import exponential_parabola_regression

x = [0, 1, 2, 3, -4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -18, 19]
y = [1.0, 6.5, 19.0, 41.5, 77.0, 128.5, 199.0, 291.5, 409.0, 554.5, 731.0, 941.5, 1189.0, 1476.5, 1807.0, 2183.5, 2609.0, 3086.5, 3619.0, 4209.5]

expression = exponential_parabola_regression(x=x, y=y, x_exp=3)
print(expression.write)
# y = 0.5 * x**3 + 2.0 * x**2 + 3.0 * x**1 + 1.0
my_formula = expression.formula

print(my_formula(2))
# 19.0

this method support any positive integer as max x exponential.

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Truncus

from regression.regress import truncus_regression

x = [0, 1, 2, 3, -4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -18, 19]
y = [1.0, 0.04, 0.012345679012345678, 0.005917159763313609, 0.0034602076124567475, 0.0022675736961451248, 0.0016, 0.0011890606420927466, 0.0009182736455463728, 0.0007304601899196494, 0.000594883997620464, 0.0004938271604938272, 0.00041649312786339027, 0.000355998576005696, 0.0003077870113881194, 0.0002687449610319806, 0.00023668639053254438, 0.00021003990758244065, 0.00018765246762994934, 0.00016866250632484398]

expression = truncus_regression(x=x, y=y)
print(expression.write)
# y ** 1 = 1 / (16.0 * x**2 + 8.0 * x**1 + 1.0)
my_formula = expression.formula

print(my_formula(2))
# 0.012345679012345675

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Notice that,

• truncs general equation

• when we know the power on y, the product, the maximum power on x, we can use Inverse Exponential Parabola
• take the truncs above as an example, where y_exp=1, product=1, 'x_exp=2'

Inverse Exponential Parabola

from regression.regress import inverse_exponential_regression

x = [0, 1, 2, 3, -4, 5, 6, -7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, -18, 19]
y = [1.0, 0.04, 0.012345679012345678, 0.005917159763313609, 0.0034602076124567475, 0.0022675736961451248, 0.0016, 0.0011890606420927466, 0.0009182736455463728, 0.0007304601899196494, 0.000594883997620464, 0.0004938271604938272, 0.00041649312786339027, 0.000355998576005696, 0.0003077870113881194, 0.0002687449610319806, 0.00023668639053254438, 0.00021003990758244065, 0.00018765246762994934, 0.00016866250632484398]

expression = inverse_exponential_regression(x=x, y=y, x_exp=2, y_exp=1, product=1)
print(expression.write)
# y ** 1 = 1 / (16.0 * x**2 + 8.0 * x**1 + 1.0)
my_formula = expression.formula

print(my_formula(2))
# 0.012345679012345675