绘制lm预测值

Question

我是新的可视化回归结果，需要帮助获得显示线性模型回归预测值的图表。

使用dput获得的数据结构：

`> dput(head(dat,4))
structure(list(X = c(809L, 3L, 1L, 2L), cntry = structure(c(1L, 
1L, 1L, 1L), .Label = c("AT", "BE", "CH", "CZ", "DE", "DK", "ES", 
"FI", "GB", "GR", "HU", "IE", "IL", "NL", "NO", "PL", "PT", "SE", 
"SI", "EE", "IS", "LU", "SK", "TR", "UA", "BG", "CY", "FR", "RU", 
"HR", "LV", "RO", "LT", "AL", "IT", "XK"), class = "factor"), 
ipshabt = c(4L, 2L, 3L, 2L), ipsuces = c(3L, 2L, 3L, 1L), 
imprich = c(3L, 3L, 3L, 3L), iprspot = c(3L, 3L, 4L, 4L), 
impsafe = c(3L, 3L, 2L, 4L), ipstrgv = c(2L, 2L, 2L, 3L), 
ipfrule = c(3L, 2L, 1L, 6L), ipbhprp = c(3L, 2L, 4L, 4L), 
ipmodst = c(3L, 3L, 2L, 5L), imptrad = c(2L, 2L, 1L, 6L), 
ipeqopt = c(1L, 2L, 1L, 1L), ipudrst = c(1L, 2L, 3L, 3L), 
impenv = c(3L, 2L, 2L, 1L), iphlppl = c(1L, 2L, 1L, 4L), 
iplylfr = c(2L, 2L, 2L, 3L), ipcrtiv = c(2L, 2L, 2L, 1L), 
impfree = c(2L, 3L, 1L, 1L), impdiff = c(2L, 3L, 3L, 1L), 
ipadvnt = c(6L, 4L, 3L, 1L), ipgdtim = c(2L, 2L, 1L, 1L), 
impfun = c(6L, 5L, 1L, 3L), gndr = c(2L, 2L, 1L, 1L), agea = c(69L, 
63L, 54L, 50L), hincfel = c(1L, 2L, 1L, 3L), educ = c(1L, 
2L, 3L, 3L), year = c(2002L, 2002L, 2002L, 2002L), Achievement = c(3.5, 
2, 3, 1.5), Power = c(3, 3, 3.5, 3.5), Security = c(2.5, 
2.5, 2, 3.5), Conformity = c(3, 2, 2.5, 5), Tradition = c(2.5, 
2.5, 1.5, 5.5), Universalism = c(1.66666666666667, 2, 2, 
1.66666666666667), Benevolence = c(1.5, 2, 1.5, 3.5), SelfDirection = c(2, 2.5, 1.5, 1), Stimulation = c(4, 3.5, 3, 1), Hedonism = c(4, 
3.5, 1, 2), SelfEnh = c(3.25, 2.5, 3.25, 2.5), SelfTran = c(1.6, 
2, 1.8, 2.4), Cons = c(2.66666666666667, 2.33333333333333, 
2, 4.66666666666667), Open = c(3.33333333333333, 3.16666666666667, 
1.83333333333333, 1.33333333333333), SelfTranNet = c(-1.65, 
-0.5, -1.45, -0.1), OpenNet = c(0.666666666666667, 0.833333333333333, 
-0.166666666666667, -3.33333333333333), east = c(0, 0, 0, 
0), eastyear = c(0, 0, 0, 0), income = c(1L, 2L, 1L, 3L), 
year2002 = c(1, 1, 1, 1), eastyear2002 = c(0, 0, 0, 0), year2004 = c(0, 
0, 0, 0), eastyear2004 = c(0, 0, 0, 0), year2006 = c(0, 0, 
0, 0), eastyear2006 = c(0, 0, 0, 0), year2008 = c(0, 0, 0, 
0), eastyear2008 = c(0, 0, 0, 0), year2010 = c(0, 0, 0, 0
), eastyear2010 = c(0, 0, 0, 0), year2012 = c(0, 0, 0, 0), 
eastyear2012 = c(0, 0, 0, 0), year2014 = c(0, 0, 0, 0), eastyear2014 = c(0, 
0, 0, 0), year2016 = c(0, 0, 0, 0), eastyear2016 = c(0, 0, 
0, 0)), .Names = c("X", "cntry", "ipshabt", "ipsuces", "imprich", 
"iprspot", "impsafe", "ipstrgv", "ipfrule", "ipbhprp", "ipmodst", 
"imptrad", "ipeqopt", "ipudrst", "impenv", "iphlppl", "iplylfr", 
"ipcrtiv", "impfree", "impdiff", "ipadvnt", "ipgdtim", "impfun", 
"gndr", "agea", "hincfel", "educ", "year", "Achievement", "Power", 
"Security", "Conformity", "Tradition", "Universalism", "Benevolence", 
"SelfDirection", "Stimulation", "Hedonism", "SelfEnh", "SelfTran", 
"Cons", "Open", "SelfTranNet", "OpenNet", "east", "eastyear", 
"income", "year2002", "eastyear2002", "year2004", "eastyear2004", 
"year2006", "eastyear2006", "year2008", "eastyear2008", "year2010", 
"eastyear2010", "year2012", "eastyear2012", "year2014", "eastyear2014", 
"year2016", "eastyear2016"), row.names = c(NA, 4L), class =     "data.frame")`

我的线性回归模型： > modelAchievement <- lm(Achievement~east+year+year2002+eastyear2002+year2004+eastyear2004+year2006+eastyear2006+year2008+eastyear2008+year2010+eastyear2010+year2012+eastyear2012+year2014+eastyear2014+year2016+eastyear2016+agea+gndr+income+educ, data = dat)

现在，我希望获得两个预测的因变量线，即＆＃34;成就＆＃34;，在与y轴相同的情节＆＃34;成就＆＃34;和x轴＆＃34;年＆＃34;。第一行：如果是虚拟变量＆＃34; east＆＃34; = 1;第二行：如果是虚拟变量＆＃34; east＆＃34; = 0。

我不知道如何继续并试图使用ggplot(modelAchievement, aes(y = Achievement, x = year))，但它给出了一个空图。

任何建议都将不胜感激。

链接到完整数据： data

Answer 1

根据您的公式，您似乎与east的每个级别进行year互动，我们可以更紧凑地表达

fit <- lm(Achievement ~ east * factor(year) + agea + gndr + income + educ, data = dat)

要计算east和year的不同值的预测结果，我们首先必须定义其他4个变量的值，agea，gndr，{{1 }和income。我将这些值设置为它们的样本均值，尽管您可以使用任何您想要的值。

educ

然后，我们会将此数据框与另一个包含library(dplyr) new_dat <- summarise_at(dat, vars(agea, gndr, income, educ), mean) # agea gndr income educ # 1 47.88262 1.536708 2.031206 3.16173 和east的所有组合的数据框合并。

year

然后我们使用new_dat <- cbind(expand.grid(year = seq(2002, 2016, 2), east = 0:1), new_dat) new_dat # year east agea gndr income educ # 1 2002 0 47.88262 1.536708 2.031206 3.16173 # 2 2004 0 47.88262 1.536708 2.031206 3.16173 # 3 2006 0 47.88262 1.536708 2.031206 3.16173 # 4 2008 0 47.88262 1.536708 2.031206 3.16173 # 5 2010 0 47.88262 1.536708 2.031206 3.16173 # 6 2012 0 47.88262 1.536708 2.031206 3.16173 # 7 2014 0 47.88262 1.536708 2.031206 3.16173 # 8 2016 0 47.88262 1.536708 2.031206 3.16173 # 9 2002 1 47.88262 1.536708 2.031206 3.16173 # 10 2004 1 47.88262 1.536708 2.031206 3.16173 # 11 2006 1 47.88262 1.536708 2.031206 3.16173 # 12 2008 1 47.88262 1.536708 2.031206 3.16173 # 13 2010 1 47.88262 1.536708 2.031206 3.16173 # 14 2012 1 47.88262 1.536708 2.031206 3.16173 # 15 2014 1 47.88262 1.536708 2.031206 3.16173 # 16 2016 1 47.88262 1.536708 2.031206 3.16173 来计算此新数据集的预测结果：

predict

现在我们可以绘制

new_dat$predicted <- predict(fit, new_dat)