我有数据，我希望形式的几个线性相关
y_i = a_i + b_i * t_i, i = 1 .. N
其中N是先验未知。问题的简短版本是：给定一个合适的
N？在下面的可重现示例中，我的数据(t,y)包含相应的参数p1（级别p1_1，p1_2）和p2（级别{{1} }，p2_1，p2_2）。
因此，数据看起来像p2_3，其中最多 2 * 3个不同的最佳拟合线，而线性拟合则最多 2 * 2 * 3非零系数。
我遇到了以下问题：假设我有三个方程式
(t, y, p1, p2)y1 = 5 + 3*t (for p1=p1_1, p2=p2_1)
y2 = 3 + t (for p1=p1_2, p2=p2_2)
y3 = 1 – t (for p1=p1_2, p2=p2_3)
期望的结果：程序应该建议4个方程y1，纠正y4，y5和y6，希望有一个很好的理由（哪一个？）忽略y6。
(Intercept) 5
t 3 => y1 = 5 + 3t
p1p1_2 -2 => y2 = 3 + 3t?
p2p2_2 .
p2p2_3 -2 => y3 = 1 + 3t?
t:p1p1_2 -2 => y4 = 3 + t (or y4 = 1 + t?)
t:p2p2_2 .
t:p2p2_3 -2 => y5 = 1 - t
p1p1_2:p2p2_2 .
p1p1_2:p2p2_3 -0.1 => y6 = 0.9 – t?
t:p1p1_2:p2p2_2 .
t:p1p1_2:p2p2_3 .
期望的结果：程序应该建议3个方程y1，y3和y6
我是否忽视了一些明显的东西？
第三列是包含噪音的虚拟因素。为简单起见，不考虑此列
(Intercept) 5
t 3 => y1 = 5 + 3t
p1p1_2 -4 => y2 = 1 + 3t?
p2p2_2 2 => y3 = 3 + 3t
p2p2_3 .
t:p1p1_2 -4 => y5 = 1 - x (or y4 = 3 - t?)
t:p2p2_2 2 => y6 = 3 + t?
t:p2p2_3 .
p1p1_2:p2p2_2 .
p1p1_2:p2p2_3 .
t:p1p1_2:p2p2_2 .
t:p1p1_2:p2p2_3 .
相关帖子：
# Create testdata
sigma <- 0.5
t <- seq(0,10, length.out = 1000) # large sample of x values
# Create 3 linear equations of the form y_i = a*t_i + b
a <- c(3, 1, -1) # slope
b <- c(5, 3, 1) # offset
# create t_i, y_ti (theory) and y_i (including noise)
d <- list()
y <- list()
y_t <- list()
for (i in 1:3) {
set.seed(33*i)
d[[i]] <- sort(sample(t, 50, replace = F))
set.seed(33*i)
noise <- rnorm(10, 0, sigma)
y[[i]] <- a[i]*d[[i]] + b[i] + noise
y_t[[i]] <- a[i]*d[[i]] + b[i]
}
# Final data set
df1 <- data.frame(t=d[[1]], y=y[[1]], p1=rep("p1_1"), p2=rep("p2_1"),
p3=sample(c("p3_1", "p3_2", "p3_3"), length(d[[1]]), replace = T))
df2 <- data.frame(t=d[[2]], y=y[[2]], p1=rep("p1_2"), p2=rep("p2_2"),
p3=sample(c("p3_1", "p3_2", "p3_3"), length(d[[1]]), replace = T))
df3 <- data.frame(t=d[[3]], y=y[[3]], p1=rep("p1_2"), p2=rep("p2_3"),
p3=sample(c("p3_1", "p3_2", "p3_3"), length(d[[1]]), replace = T))
mydata <- rbind(df1, df2, df3)
mydata$p1 <- factor(mydata$p1)
mydata$p2 <- factor(mydata$p2)
mydata$p3 <- factor(mydata$p3)
mydata <- mydata[sample(nrow(mydata)), ]
# What the raw data looks like:
plot(x = mydata$t, y = mydata$y)
cols <- rainbow(length(levels(mydata$p1))*length(levels(mydata$p2))*length(levels(mydata$p3)))
rm(.Random.seed, envir=.GlobalEnv)
cols <- sample(cols) # most likely similar colors are not next to each other;-)
# Fit using lm disabled - just uncomment and comment the part below
# fit <- lm(y ~ t * p1 * p2, data = mydata)
# coef <- as.matrix(fit$coefficients)
# mydata$pred <- predict(fit)
# Fit using glmnet
set.seed(42)
fit_type <- c("lambda.min", "lambda.1se")[1]
x <- model.matrix(y ~ t * p1 * p2, data = mydata)[,-1]
fit <- glmnet::cv.glmnet(x, mydata$y, intercept = TRUE, nfolds = 10, alpha = 1)
coef <- glmnet::coef.cv.glmnet(fit, newx = x, s = fit_type)
mydata$pred <- predict(fit, newx = x, s = fit_type)
# plots
plot(d[[1]], y_t[[1]], type = "l", lty = 3, col = "black", main = "Raw data",
xlim = c(0, 10), ylim = c(min(mydata$y), max(mydata$y)), xlab = "t", ylab = "y")
lines(d[[2]], y_t[[2]], col = "black", lty = 3)
lines(d[[3]], y_t[[3]], col = "black", lty = 3)
# The following for loops are fixed right now. In the end this should be automated using
# the input from the fit (and the knowledge how to extract N and the lines above).
pn <- 0
for (p1 in 1:length(levels(mydata$p1))) {
for (p2 in 1:length(levels(mydata$p2))) {
pn <- pn + 1
tmp <- mydata[mydata$p1 == levels(mydata$p1)[p1] & mydata$p2 == levels(mydata$p2)[p2], ]
points(x = tmp$t, y = tmp$y, col = cols[pn]) # original data
points(x = tmp$t, y = tmp$pred, col = cols[pn], pch = 3) # estimated data from predict
if (length(tmp$pred) > 0) {
abline(lm(tmp$pred ~ tmp$t), col = cols[pn])
}
}
}
和a）。答案 0 :(得分：2)
我认为你误解了回归结果。如果等式包含术语p1_m和p2_n，则它还必须包含互动t:p1_m和t:p2_n。它不可能是一个而不是另一个。在样本数据中有三对系数：
> unique(mydata[,3:4])
# p1 p2
# 96 p1_2 p2_2
# 1 p1_1 p2_1
# 135 p1_2 p2_3
查看lm结果，我们将方程式重建为：
y = 5 + 3t + p1p1_2 + (t:p1p1_2)*t + p2p2_2 + (t:p2p2_2)*t = 3 + t; y = 5 + 3t + p1p1_1 + (t:p1p1_1)*t + p2p2_1 + (t:p2p2_1)*t = 5 + 3t; y = 5 + 3t + p1p1_2 + (t:p1p1_2)*t + p2p2_3 + (t:p2p2_3)*t = 1 - t。这些匹配您在开头指定的等式，因此没有歧义。