Correlation graph with coefficients and significance together

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I am trying to represent a correlation matrix that contains the correlation coefficients and significance in the same graph.I have read in another post that with the corrplot it would not be possible to put these two information in the same cell of the graph. It would be possible to place the correlation coefficients at the top and the significance in the lower half?

Correlação 

library(readxl)
seed <- read_excel("seed.xlsx")
seed
cor(seed, method = "pearson")
M <- cor(seed, method = "pearson")
library(corrplot)
library(RColorBrewer)
corrplot(M, type = "upper", method = "color", order = "original", outline = "white", 
         tl.col="black", tl.srt=45, addCoef.col = "black",
         col = colorRampPalette(c("blue", "yellow", "red"))(20))
library(Hmisc)
m <- rcorr(as.matrix(seed))
m
P <- cor.mtest(seed, conf.level = .95)
corrplot(M, p.mat = P$p, insig = "label_sig", sig.level = c(.001, .01, .05),
         pch.cex = 2, pch.col = "white", type = "upper", method = "color", 
         order = "original", outline = "white", tl.col="black", tl.srt=45,diag = FALSE,
         addCoef.col = "black", col = colorRampPalette(c("blue", "yellow", "red"))(20)
r
  • 2

    Please attach the file Seed.xlsx or better still, provide the data via dput(head(Seed, 20)). I did some research, and I couldn’t find any package ready that would provide both information the way you suggest. Why do you want to present both values literally? Don’t think the correlation value is more important than the p-value in this case (assuming that all variables have n equal)

  • > dput(head(Seed, 20)) Structure(list(TT = c(151.65, 169.51, 173.91, 158.95, 159.05,146.85, 157.21, 146.01, 178.45, 165.75, 185.95, 143.81, 182.25, 175.51, 217.95, 273.75, 217.61, 206.01, 213.75), RT = c(145.33, 120.15, 135.31, 148.12, 127.45, 112.05, 130.21, 113.65, 136.71, 126.65, 153.61, 113.55, 148.21, 166.85, 188.45, 234.75, 195.31, 199.32, 193.11), NRI = c(4.91, 4.51, 5.41, 5.35, 6.15, 6.61, 4.51, 5.95, 5.55, 4.51, 5.11, 5.51, 5.95, 6.91, 7.11, 6.55, 8.21, 6.75, 8.11), continues

  • NSI = c(182.51, 95.45, 124.21, 199.91, 117.41, 166.35, 189.65, 171.75, 203.35, 166.25, 125.45, 128.91, 143.51, 187.41, 211.81, 169.01, 216.61, 77.85, 262.41), WTS = c(3.46, 3.38, 4.07, 3.89, 3.33, 3.78, 2.88, 3.34, 3.02, 3.68, 3.55, 4.17, 3.66, 3.11, 3.09, 3.41, 3.29, 3.18, 2.91), SY = c(79.81, 19.97, 60.92, 92.74, 23.92, 55.43, 57.44, 53.29, 56.62, 58.08, 53.83, 59.55, 55.26, 75.13, 73.18, 90.69, 78.48, 107.89, 81.46)), Row.Names = c(NA, -19L), class = c("tbl_df", "tbl", "frame."))

  • 6 quantitative variables with 19 repetitions. Thank you!

  • this result is better if placed in the body of the question, anyway, I did not find the specific answer to your question

  • Thank you very much William! I thank you for your attention. Although I am a beginner of R, migrating from SAS, to take advantage of the graphical parts of R, in logic it would be possible to divide the graph.... I will take a break and resume this endeavor. My many thanks, hug.

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I did a similar search of this for some time, and I remade it again, but could not find such a result. So I thought it wouldn’t be an interesting visualization, because no one has done it before and it’s something simple.

That way, I suggest you wear it something thus.

But you can also program what you want. Just use a for for column variables, and other for rows and store the function result cor.test in an object (ex.: corre). In the same loop, in the upper triangular part of the matrix you place the result of corre$estimate (correlation), and in the lower triangular the result of corre$p.value. Then you can print the result of this matriz/data.frame combining the functions kable and kableExtra (to assemble the layout as you wish).

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