1

考虑这段代码：

# Load libraries
library(RCurl)
library(TraMineR)
library(PST)

# Get data
x <- getURL("https://gist.githubusercontent.com/aronlindberg/08228977353bf6dc2edb3ec121f54a29/raw/c2539d06771317c5f4c8d3a2052a73fc485a09c6/challenge_level.csv")
data <- read.csv(text = x)

# Load and transform data
data <- read.table("thread_level.csv", sep = ",", header = F, stringsAsFactors = F)

# Create sequence object
data.seq <- seqdef(data[2:nrow(data),2:ncol(data)], missing = NA, right= NA, nr = "*")

# Make a tree
S1 <- pstree(data.seq, ymin = 0.05, L = 6, lik = FALSE, with.missing = TRUE)

# Look at contexts
cmine(S1, pmin = 0, state = "N3", l = 3)

然后，我可以通过以下方式计算两个特定“关联规则”的提升值的重要性阈值：

# Calculate lift threshold for N2-QU->N3
ngood_idea <- sum(data.seq == "N3")
nn <- nrow(data.seq)*ncol(data.seq)
p_good_idea <- ngood_idea/nn

x <- seqdef("N2-QU")
p_context <- predict(S1, x, decomp = F, output = "prob")
p_not_context_good_idea <- (1-p_context)*(1-(p_good_idea))
p_context_good_idea <- p_context*p_good_idea
N2_QU_N3_threshold <- 1+1.645*sqrt(((1/nn)*(p_not_context_good_idea/p_context_good_idea)))

# Calculate lift threshold for N2-QU->N1
nbad_idea <- sum(data.seq == "N1")
nn <- nrow(data.seq)*ncol(data.seq)
p_bad_idea <- nbad_idea/nn

p_not_context_bad_idea <- (1-p_context)*(1-(p_bad_idea))
p_context_bad_idea <- p_context*p_bad_idea
N2_QU_N1_threshold <- 1+1.645*sqrt(((1/nn)*(p_not_context_bad_idea/p_context_bad_idea)))

# Print lift thresholds
N2_QU_N3_threshold
N2_QU_N1_threshold

但是，如果我想比较两个提升值，看看它们是否存在显着差异（类似于我如何比较两个回归系数以查看它们是否显着不同），该怎么办？其他）？我怎样才能做到这一点？

4

1 回答 1

1

利用这个等式：

$Z = \frac{\beta_1-\beta_2}{\sqrt{(SE\beta_1)^2+(SE\beta_2)^2}}$

$SE\beta$的标准误在哪里$\beta$。

该方程由 Clogg 等人 (1995) 提供

资料来源：https ://stats.stackexchange.com/questions/93540/testing-equality-of-coefficients-from-two-different-regressions

我们可以类推，使用提升作为系数，并根据 Lenca 等人 (2008, p. 619) 计算每个提升的方差

# Calculate conditional probability for I3
cp_good <- query(S1, context = "N2-QU", output= "prob")@.Data[attr(query(S1, context = "N2-QU", output= "prob")@.Data, "dimnames")[[2]]=="I3"]
cp_good <- unlist(cp_good)

# Calculate conditional probability for I1
cp_bad <- query(S1, context = "N2-QU", output= "prob")@.Data[attr(query(S1, context = "N2-QU", output= "prob")@.Data, "dimnames")[[2]]=="I1"]
cp_bad <- unlist(cp_bad)

# Calculate lift for I3
ngood_idea <- sum(data.seq == "I3")
nn <- nrow(data.seq)*ncol(data.seq)
p_good_idea <- ngood_idea/nn

good_lift <- cp_good/p_good_idea

# Calculate lift for I1
nbad_idea <- sum(data.seq == "I1")
nn <- nrow(data.seq)*ncol(data.seq)
p_bad_idea <- nbad_idea/nn

bad_lift <- cp_bad/p_bad_idea

# Calculate z_diff
p_context <- predict(S1, x, decomp = F, output = "prob")

p_not_context_good_idea <- (1-p_context)*(1-(p_good_idea))
p_context_good_idea <- p_context*p_good_idea

p_not_context_bad_idea <- (1-p_context)*(1-(p_bad_idea))
p_context_bad_idea <- p_context*p_bad_idea

var_good_idea <- ((1/nn)*(p_not_context_good_idea/p_context_good_idea))
var_bad_idea <- ((1/nn)*(p_not_context_bad_idea/p_context_bad_idea))

z_diff <- (good_lift-bad_lift)/sqrt(var_good_idea+var_bad_idea)
z_diff

差值的 z 值为0.2556881。

参考

Clogg, CC, Petkova, E. 和 Haritou, A. (1995)。比较模型之间回归系数的统计方法。美国社会学杂志, 100 (5), 1261-1293.]

Lenca, P.、Meyer, P.、Vaillant, B. 和 Lallich, S. 2008。“关于为关联规则选择兴趣度度量：面向用户的描述和多标准决策辅助”，欧洲运筹学杂志 (184:2 ），第 610-626 页（doi：10.1016/j.ejor.2006.10.059）。

于 2017-03-02T16:22:42.133 回答