使用ggplot2生成“模糊”RD图

Question

我的问题类似于this，但那里的答案对我不起作用。基本上，我正在尝试使用“模糊”设计生成回归不连续图，该设计使用治疗组和对照组的所有数据，但仅绘制治疗组和对照组“范围”内的回归线。

下面，我模拟了一些数据并生成了带有基本图形的模糊RD图。我希望用ggplot2复制这个情节。请注意，最重要的部分是浅蓝色回归线适合所有蓝点，而桃色回归线适合所有红点，尽管只绘制在个人意图的范围内接受治疗。这是我在ggplot中难以复制的部分。

我想转移到ggplot，因为我想使用faceting在参与者嵌套的各个单元中生成相同的图。在下面的代码中，我使用geom_smooth显示了一个非示例。当组内没有模糊时，它可以正常工作，但否则会失败。如果我可以将geom_smooth仅限于特定范围，我想我会被设定。任何和所有的帮助表示赞赏。

模拟数据

library(MASS)
mu <- c(0, 0)
sigma <- matrix(c(1, 0.7, 0.7, 1), ncol = 2)

set.seed(100)
d <- as.data.frame(mvrnorm(1e3, mu, sigma))

# Create treatment variable
d$treat <- ifelse(d$V1 <= 0, 1, 0)

# Introduce fuzziness
d$treat[d$treat == 1][sample(100)] <- 0
d$treat[d$treat == 0][sample(100)] <- 1

# Treatment effect
d$V2[d$treat == 1] <- d$V2[d$treat == 1] + 0.5

# Add grouping factor
d$group <- gl(9, 1e3/9)

使用基数

生成回归不连续图

library(RColorBrewer)
pal <- brewer.pal(5, "RdBu")

color <- d$treat
color[color == 0] <- pal[1]
color[color == 1] <- pal[5]

plot(V2 ~ V1, 
    data = d, 
    col = color,
    bty = "n")
abline(v = 0, col = "gray", lwd = 3, lty = 2)

# Fit model
m <- lm(V2 ~ V1 + treat, data = d)

# predicted achievement for treatment group
pred_treat <- predict(m, 
            newdata = data.frame(V1 = seq(-3, 0, 0.1), 
                                 treat = 1))
# predicted achievement for control group
pred_no_treat <- predict(m, 
            newdata = data.frame(V1 = seq(0, 4, 0.1), 
                                 treat = 0))

# Add predicted achievement lines
lines(seq(-3, 0, 0.1), pred_treat, col = pal[4], lwd = 3)
lines(seq(0, 4, 0.1), pred_no_treat, col = pal[2], lwd = 3)

# Add legend
legend("bottomright", 
    legend = c("Treatment", "Control"),
    lty = 1,
    lwd = 2,
    col = c(pal[4], pal[2]),
    box.lwd = 0)

非ggplot

的例子

d$treat <- factor(d$treat, labels = c("Control", "Treatment"))

library(ggplot2)
ggplot(d, aes(V1, V2, group = treat)) + 
    geom_point(aes(color = treat)) +
    geom_smooth(method = "lm", aes(color = treat)) +
    facet_wrap(~group)

请注意，回归线超出了第1组和第2组的治疗范围。

Answer 1

使用geom_smooth创建行可能会更优雅，但可以与geom_segment一起攻击。如果你愿意，可以在绘图调用之外使用数据框。

ggplot(d, aes(x = V1, y = V2, color = factor(treat, labels = c('Control', 'Treatment')))) + 
    geom_point(shape = 21) + 
    scale_color_brewer(NULL, type = 'qual', palette = 6) + 
    geom_vline(aes(xintercept = 0), color = 'grey', size = 1, linetype = 'dashed') + 
    geom_segment(data = data.frame(t(predict(m, data.frame(V1 = c(-3, 0), treat = 1)))), 
                 aes(x = -3, xend = 0, y = X1, yend = X2), color = pal[4], size = 1) + 
    geom_segment(data = data.frame(t(predict(m, data.frame(V1 = c(0, 4), treat = 0)))), 
                 aes(x = 0, xend = 4, y = X1, yend = X2), color = pal[2], size = 1)

另一个选项是geom_path：

df <- data.frame(V1 = c(-3, 0, 0, 4), treat = c(1, 1, 0, 0))
df <- cbind(df, V2 = predict(m, df))

ggplot(d, aes(x = V1, y = V2, color = factor(treat, labels = c('Control', 'Treatment')))) + 
    geom_point(shape = 21) + 
    geom_vline(aes(xintercept = 0), color = 'grey', size = 1, linetype = 'dashed') + 
    scale_color_brewer(NULL, type = 'qual', palette = 6) + 
    geom_path(data = df, size = 1)

对于使用构面进行编辑，如果我理解了您想要的内容，则可以使用lapply计算每个组的模型并预测每个组。在这里，我重新组合dplyr::bind_rows而不是do.call(rbind, ...)，.id参数从列表元素名称中插入组号，尽管还有其他方法可以做同样的事情。

df <- data.frame(V1 = c(-3, 0, 0, 4), treat = c('Treatment', 'Treatment', 'Control', 'Control'))
m_list <- lapply(split(d, d$group), function(x){lm(V2 ~ V1 + treat, data = x)})
df <- dplyr::bind_rows(lapply(m_list, function(x){cbind(df, V2 = predict(x, df))}), .id = 'group')

ggplot(d, aes(x = V1, y = V2, color = treat)) + 
    geom_point(shape = 21) + 
    geom_vline(aes(xintercept = 0), color = 'grey', size = 1, linetype = 'dashed') + 
    geom_path(data = df, size = 1) + 
    scale_color_brewer(NULL, type = 'qual', palette = 6) + 
    facet_wrap(~group)