R 回归，数据集中所有变量的列表
有吗 在R中有没有一种简单或不那么简单的方法来创建 数据集中所有变量的列表 我想循环我的变量，回归到另一个变量上，这样我就可以在关于自变量的假设下预测每个变量的值 在斯塔塔，这看起来像R 回归，数据集中所有变量的列表,r,list,regression,stata,R,List,Regression,Stata,有吗 在R中有没有一种简单或不那么简单的方法来创建 数据集中所有变量的列表 我想循环我的变量，回归到另一个变量上，这样我就可以在关于自变量的假设下预测每个变量的值 在斯塔塔，这看起来像 foreach v of var a b c { qui regress `v' i di "`v' {col 2O}" _b[`i'] * 42 } 42 is an arbitrary number 例如，如果我有三个变量a、b和c，还有另一个独立变量 变量i，我想回归a i，回归
foreach v of var a b c {
qui regress `v' i
di "`v' {col 2O}" _b[`i'] * 42
}
42 is an arbitrary number
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"DMFlag", "HhOut", "IndOut", "ind_wt", "IMonth", "IYear", "DMSex",
"DM014", "DM016", "DM510", "DM1115", "DM1619", "WorkSta", "act1_1",
"act1_2", "act1_3", "act1_4", "act1_5", "act1_6", "act1_7", "act1_8",
"act1_9", "act1_10", "act1_11", "act1_12", "act1_13", "act1_14",
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"act1_81", "act1_82", "act1_83", "act1_84", "act1_85", "act1_86",
"act1_87", "act1_88", "act1_89", "act1_90", "act1_91", "act1_92",
"act1_93", "act1_94", "act1_95", "act1_96", "act1_97", "act1_98",
"act1_99", "act1_100", "act1_101", "act1_102", "act1_103", "act1_104",
"act1_105", "act1_106", "act1_107", "act1_108", "act1_109", "act1_110",
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"act1_117", "act1_118", "act1_119", "act1_120", "act1_121", "act1_122",
"act1_123", "act1_124", "act1_125", "act1_126", "act1_127", "act1_128",
"act1_129", "act1_130", "act1_131", "act1_132", "act1_133", "act1_134",
"act1_135", "act1_136", "act1_137", "act1_138", "act1_139", "act1_140",
"act1_141", "act1_142", "act1_143", "act1_144"), row.names = c(NA,
5L), class = "data.frame")
下面是一个不雅观但有效的方法。首先，由于您的数据看起来比理解此方法的要点所需的数据要混乱一些：我创建了一个玩具示例数据集： 设定种子的重复性 第123集 创建我们的变量 我 >电话： >lmformula=as.formulapaste0j，~i，data=dataset > >残差： >最小1季度中值3季度最大值 > -1.6085 -0.5056 -0.2152 0.6932 2.0118 > >系数： >估计标准误差t值Pr>t >截距0.1720 0.1534 1.121 0.272 >i 4.8660 0.1589 30.629-- >签名。代码：0'***'0.001'***'0.01'*'0.05'.'0.1''1 > >剩余标准误差：28个自由度上的0.8393 >倍数R平方：0.971，调整后的R平方：0.97 >F-统计量：1和28 DF上为938.1，p值：<2.2e-16 > >电话： >lmformula=as.formulapaste0j，~i，data=dataset > >残差： >最小1季度中值3季度最大值 > -2.39301 -0.56909 0.03468 0.51764 2.08387 > >系数： >估计标准误差t值Pr>t >截距0.0313 0.1596 0.196 0.846 >i-1.8540 0.1653-11.2187.18e-12*** > -- >签名。代码：0'***'0.001'***'0.01'*'0.05'.'0.1''1 > >剩余标准误差：28个自由度上的0.8731 >倍数R平方：0.818，调整后的R平方：0.8115 >F-统计量：1和28 DF上的125.8，p-值：7.177e-12 > >电话： >lmformula=as.formulapaste0j，~i，data=dataset > >残差： >最小1季度中值3季度最大值 > -1.4989 -0.6435 -0.1436 0.5917 2.2613 > >系数： >估计标准误差t值Pr>| t| >截距-0.09205 0.16855-0.546 0.589 >i 0.03914 0.17454 0.224 0.824 > >剩余标准误差：28个自由度上的0.9221 >多重R平方：0.001793，调整后的R平方：-0.03386 >F统计：1和28 DF上的0.05029，p值：0.8242 你的预测呢 lapplymodels，functionmodel{ predictmodel，newdata=data.framei=42 } > [[1]] > 1 > 204.5426 > > [[2]] > 1 > -77.8355 > > [[3]] > 1 > 1.551869 由v0.2.1于2019-02-08创建 您应该注意到，这种方法与如何为解释变量模拟Stata宏的相关问题非常相似。主要的区别，也是为什么我觉得你的问题应该有一个单独的答案，是因为你有多个DV，你需要多次运行lm，这在某种程度上改变了方法。在这个问题上，在一个lm调用中使用as.formula和paste就足够了，即lmas.formulapasteincome~，pasterregistors，collapse=+，data=Duncan，这里我们需要在循环或*apply调用中使用这种方法，即lapplyDVnames，functionj{lmas.formulapaste0j，~i，data=dataset}。然后我们还需要使用*apply函数来进行所需的预测