使用heatmaply创建不需要转换的原始值/数据的热图？

Question

使用heatmaply::heatmaply()创建热图时，对原始数据执行一些转换。这一点在下面的示例中很明显，其中提供了一些虹膜数据集，但是在热图上徘徊显示了一些负值(这些值不在提供给heatmaply()的原始数据中)，而旁边的颜色条也显示了负值。

https://i.stack.imgur.com/Mt0vGm.png

library(heatmaply)
mat <- iris[1:20, 1:4] %>% as.matrix
p <- heatmaply(mat, 
                   dendrogram = "none",
                   xlab = "", ylab = "", 
                   main = "",
                   scale = "column",
                   margins = c(60,100,40,20),
                   grid_color = "white",
                   grid_width = 0.00001,
                   titleX = FALSE,
                   hide_colorbar = FALSE,
                   branches_lwd = 0.1,
                   label_names = c("Country", "Feature:", "Value"),
                   fontsize_row = 5, fontsize_col = 5,
                   labCol = colnames(mat),
                   labRow = rownames(mat),
                   heatmap_layers = theme(axis.line=element_blank()),
                   colors = rev(c("000000", heat.colors(30)[1:28]))
)

p

对iris[1:20, 1:4]的不同规模

https://i.stack.imgur.com/nuMcn.png

问题

如何使heatmaply::heatmaply()完全基于提供的原始值(不进行转换)生成基于热图的heatmaply::heatmaply()？

注:开放使用其他软件包/功能/建议

Answer 1

删除scale参数-默认为“无”。或者改为“无”。

heatmaply(mat, 
          dendrogram = "none",
          xlab = "", ylab = "", 
          main = "",
          # scale = "column",  # <-- remove this line. Or change to "none"
          margins = c(60,100,40,20), ...

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