8 thoughts on “StatQuest: Linear Discriminant Analysis (LDA), clearly explained

  1. Hello friend,
    Thank you for your helpful video. Could you please provide me the code to run this analysis in R?
    Best regards

  2. Hi Joshua! Thanks a lot for your very helpful video!!

    I have a little question for you…
    I was thinking about how to find the variables that mainly contribute to the formation of the new axes. How do you concretely do that?
    How would you correlate LD1 (coefficients of linear discriminants) with the variables?

    Thanks in advance,

    • Madeleine,
      I use R, so here’s how to do it in R. First do the LDA…

      library(MASS) ## Load the “MASS” package (which contains the lda() function)
      data(iris) ## load an example dataset
      head(iris, 3) ## look at the first 3 rows..
      Sepal.Length Sepal.Width Petal.Length Petal.Width Species
      1 5.1 3.5 1.4 0.2 setosa
      2 4.9 3.0 1.4 0.2 setosa
      3 4.7 3.2 1.3 0.2 setosa

      lda.results <- lda(formula = Species ~ ., data = iris) # now do LDA
      lda.results$scaling # now look at the the linear combination of coefficients
      LD1 LD2
      Sepal.Length 0.8293776 0.02410215
      Sepal.Width 1.5344731 2.16452123
      Petal.Length -2.2012117 -0.93192121
      Petal.Width -2.8104603 2.83918785

      … roughly speaking, the absolute value of the "scaling" values will tell you which variables were the most important for each linear discriminant.

      If you want to know how much variation each linear discriminate accounts for…


      • Hi Josh,

        thanks for your answer!
        the higher the absolute value is the most the variable contributes to the groups/categories separation?

        Do you think that one can “filtering variables” using the ones with highest “scaling” values and recompute lda? …or lda can not really be used as variable selection?

        sorry, I’m wandering a bit off.

        thanks in advance

  3. Yes, I think you can use LDA iteratively to filter out variables that are not helpful. Just like when you do a regression with a ton of variables and then leave some out and see if the sums of squares are still pretty much in your favor.

  4. Hello Josh, I am traying to run the code for my own data but when I tried to my data frame for building a graph, when I put Y as Y = data.lda.values$x[,2] there is an error:
    subscript out of bounds

    so in data.lda.values$x there is only 1 lda. Why I only get 1 lda?

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