geom_tile fill in confusion matrix correspond to value - r

Question

I am trying to get the colours of a confusion matrix to correspond to the percent value in the middle of each matrix.
I have tried adjusting the geom_tile section fill to various options of Freq, or percentage, but with no luck.
valid_actualFunc <- as.factor(c(conf$ObsFunc))
valid_predFunc <- as.factor(c(conf$PredFunc))
cfmFunc <- confusionMatrix(valid_actualFunc, valid_predFunc)
ggplotConfusionMatrix <- function(m){
mytitle <- paste("Accuracy", percent_format()(m$overall[1]),
"Kappa", percent_format()(m$overall[2]))
data_c <- mutate(group_by(as.data.frame(m$table), Prediction ),
percentage=percent(Freq/sum(Freq)))
p <-
ggplot(data = data_c,
aes(x = Reference, y = Prediction)) +
geom_tile(aes(fill = Freq/sum(Freq)), colour = "white") +
scale_fill_gradient(low = "white", high = "red", na.value="white") +
geom_text(aes(x = Reference, y = Prediction, label = percentage)) +
theme(axis.text.x=element_text(angle = -90, hjust = 0),
axis.ticks=element_blank(), legend.position="none") +
ggtitle(mytitle)+
scale_y_discrete(limits = rev(levels(as.factor(valid_predFunc))))
return(p)
}
conf2Func=ggplotConfusionMatrix(cfmFunc)
conf2Func
Currently the fill is not equal to the value in the middle, i.e. a tile with 89% is lighter than one with 70%
As per the comment the return is
dput(head(cfmFunc))
list(positive = NULL, table = structure(c(2331L, 102L, 262L,
52L, 290L, 1986L, 178L, 89L, 495L, 74L, 2966L, 52L, 189L, 58L,
92L, 800L), .Dim = c(4L, 4L), .Dimnames = list(Prediction = c("Algae",
"Hard Coral", "Other", "Other Inv"), Reference = c("Algae", "Hard Coral",
"Other", "Other Inv")), class = "table"), overall = c(Accuracy =
0.807008785942492,
Kappa = 0.730790156424558, AccuracyLower = 0.799141307917932,
AccuracyUpper = 0.814697342402988, AccuracyNull = 0.358126996805112,
AccuracyPValue = 0, McnemarPValue = 6.95780670112837e-62), byClass =
structure(c(0.848562067710229,
0.780967361384192, 0.826874825759688, 0.702370500438982,
0.866006328243225,
0.968687274187073, 0.917249961113703, 0.978258420637603,
0.705295007564297,
0.894594594594595, 0.847913093196112, 0.805639476334341,
0.938012218745343,
0.928553104155977, 0.904725375882172, 0.962429347223761,
0.705295007564297,
0.894594594594595, 0.847913093196112, 0.80563947633434, 0.848562067710229,
0.780967361384192, 0.826874825759688, 0.702370500438982,
0.770323859881031,
0.833928196514802, 0.837261820748059, 0.75046904315197, 0.274261182108626,
0.253893769968051, 0.358126996805112, 0.113718051118211,
0.232727635782748,
0.198282747603834, 0.296126198083067, 0.0798722044728434,
0.329972044728434,
0.221645367412141, 0.349241214057508, 0.0991413738019169,
0.857284197976727,
0.874827317785633, 0.872062393436696, 0.840314460538292), .Dim = c(4L,
11L), .Dimnames = list(c("Class: Algae", "Class: Hard Coral",
"Class: Other", "Class: Other Inv"), c("Sensitivity", "Specificity",
"Pos Pred Value", "Neg Pred Value", "Precision", "Recall", "F1",
"Prevalence", "Detection Rate", "Detection Prevalence", "Balanced
Accuracy"
))), mode = "sens_spec", dots = list())

Answer 1

If you check the structure of your dataset to be plotted str(data_c), you will see that percentage is a character vector, and needs to be converted to numeric to be used as continuous input to the fill gradient.
data_c$percentage.numeric <- as.numeric(gsub("%", "", data_c$percentage))
You can use percentage.numeric for aes fill and percentage for aes label.
ggplot(data = data_c,
aes(x = Reference, y = Prediction)) +
geom_tile(aes(fill = percentage.numeric), colour = "white") +
scale_fill_gradient(low = "white", high = "red", na.value="white") +
geom_text(aes(x = Reference, y = Prediction, label = percentage)) +
theme(axis.text.x=element_text(angle = -90, hjust = 0),
axis.ticks=element_blank(), legend.position="none") +
ggtitle(mytitle)
Note scale_y_discrete(limits = rev(levels(as.factor(valid_predFunc)))) produces an error in your example
Error in as.factor(valid_predFunc) : object 'valid_predFunc' not found