Running several simple Regression in R

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野的像风
野的像风 2021-01-28 10:38

So I have a data set that has 188 rows and 65 columns relating to World development indicators and Birth statistics. I am trying to do a purposeful selection method to create a

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  •  醉梦人生
    2021-01-28 10:58

    First, I don't recommend you doing this unless you know what you are doing. Else read about things like selection bias, false discovery rate, etc.

    In the following, I am using the iris dataset, and regress the first three columns on the fourth one. You can easily change this to data you have.

    Using the broom package isn't mandatory. If you don't want that, remove tidy`` command in thelapply` function.

    library(broom)
    
    list_out <- lapply(colnames(iris)[1:3], function(i)
                 tidy(lm(as.formula(paste("Petal.Width ~", i)), data = iris)))
    
    # [[1]]
    # term   estimate  std.error statistic      p.value
    # 1  (Intercept) -3.2002150 0.25688579 -12.45773 8.141394e-25
    # 2 Sepal.Length  0.7529176 0.04353017  17.29645 2.325498e-37
    # 
    # [[2]]
    # term   estimate std.error statistic      p.value
    # 1 (Intercept)  3.1568723 0.4130820  7.642242 2.474053e-12
    # 2 Sepal.Width -0.6402766 0.1337683 -4.786461 4.073229e-06
    # 
    # [[3]]
    # term   estimate   std.error statistic      p.value
    # 1  (Intercept) -0.3630755 0.039761990 -9.131221 4.699798e-16
    # 2 Petal.Length  0.4157554 0.009582436 43.387237 4.675004e-86
    

    Put them into a data.frame

    do.call(rbind, list_out)
    
    #          term   estimate   std.error  statistic      p.value
    # 1  (Intercept) -3.2002150 0.256885790 -12.457735 8.141394e-25
    # 2 Sepal.Length  0.7529176 0.043530170  17.296454 2.325498e-37
    # 3  (Intercept)  3.1568723 0.413081984   7.642242 2.474053e-12
    # 4  Sepal.Width -0.6402766 0.133768277  -4.786461 4.073229e-06
    # 5  (Intercept) -0.3630755 0.039761990  -9.131221 4.699798e-16
    # 6 Petal.Length  0.4157554 0.009582436  43.387237 4.675004e-86
    

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