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绘制lm预测值

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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 .

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,我想在y轴“成就”和x轴“年”的同一图上得到两个预测的因变量线,即“成就” . 第一行:如果虚拟变量“east”= 1;第二行:如果虚拟变量“east”= 0 .

我不知道如何继续并尝试使用 ggplot(modelAchievement, aes(y = Achievement, x = year)) ,但它给出了一个空的情节 .

任何建议将不胜感激 .

链接到完整数据:data

1 回答

  • 3

    根据您的公式,您似乎正在与 year 的每个级别进行交互 east ,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 eastyear 的不同值的预测结果,我们首先必须为其他4个变量 ageagndrincomeeduc 定义值 . 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 .

    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
    

    然后,我们将此数据框架与另一个具有 eastyear 的所有组合的数据框合并 .

    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
    

    然后,我们使用 predict 来计算此新数据集的预测结果:

    new_dat$predicted <- predict(fit, new_dat)
    

    现在我们可以策划

    library(ggplot2)
    ggplot(new_dat, aes(x = year, y = predicted, colour = factor(east), group = east)) +
      geom_line()
    

    enter image description here

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