python 这演示了如何逐步计算python生成器的线性回归，以避免需要加载整个结构

Posted 2021-05-09

'''
This demonstrates how to progressively calculate the linear regression 
of a generator to avoid needing to load the whole structure into memory. 

This is useful when youre trying to analyze infinite streams of data.

by: Cody Kochmann
'''

from operator import mul
from generators import started

@started
def sum_of_squares():
    out = 0
    while 1:
        out += (yield out)**2

@started
def cor():
    out = 0
    while 1:
        out += mul(*(yield out))

@started
def sd():
    _len = 0
    _sum = 0
    _input = 0
    _sum_of_squares = sum_of_squares().send
    while 1:
        _input = yield _sum**2 - _len*_sum_of_squares(_input)
        _len += 1
        _sum += _input

@started
def linear_slope():
    # (sum(x)*sum(y) - len(x)*cor(x,y)) / sd(x)
    x_sum = 0
    y_sum = 0
    x_len = 0
    slope = 0
    _cor = cor().send
    _sd = sd().send
    while 1:
        x,y = yield slope
        x_sum += x
        y_sum += y
        x_len += 1
        _deviation = _sd(x)
        slope = (x_sum*y_sum - x_len*_cor((x,y)))/(_deviation if _deviation!=0 else 1)

@started
def y_intercept():
    # (cor(x,y)*sum(x) - sum_of_squares(x)*sum(y)) / sd(x)
    _cor = cor().send
    x_sum = 0
    y_sum = 0
    _sum_of_squares = sum_of_squares().send
    _sd = sd().send
    intercept = 0
    while 1:
        x,y = yield intercept
        _deviation = _sd(x)
        x_sum += x
        y_sum += y
        intercept = (_cor((x,y))*x_sum - _sum_of_squares(x)*y_sum)/(_deviation if _deviation!=0 else 1)

_sum_of_squares = sum_of_squares().send
_cor = cor().send
_sd = sd().send
_linear_slope = linear_slope().send
_y_intercept = y_intercept().send

for x,y in zip(range(10), range(3,13)):
    #print(_sum_of_squares(x))
    #print(_cor((x,y)))
    #print(_sd(x))
    print(_linear_slope((x,y)))
    print(_y_intercept((x,y)))