"Too many" of the same value causes an error

[mmahmoudian]

Thanks for the awesome package. The following is the issue I am facing:

When giving the vioplot too many of the same value (and only that value) it produces an error.

These work fine:

> vioplot(x = c(0))
> vioplot(x = c(rep.int(0, 100)))

Adding one order of magnitude and it breaks:

vioplot(x = c(rep.int(0, 1000)))
vioplot(x = c(rep.int(1, 1000)))

The error I get is:

Error in cut.default(x, breaks = breaks) : 'breaks' are not unique

To clarify, if there is even one more number with another value, it will work as intended:

vioplot(x = c(rep.int(c(1), 10000), 2))

---

My session's info:

R version 4.0.3 (2020-10-10)
Platform: x86_64-pc-linux-gnu (64-bit)
Running under: Manjaro Linux

Matrix products: default
BLAS:   /usr/lib/libopenblasp-r0.3.12.so
LAPACK: /usr/lib/liblapack.so.3.9.0

locale:
 [1] LC_CTYPE=en_US.UTF-8       LC_NUMERIC=C              
 [3] LC_TIME=en_US.UTF-8        LC_COLLATE=en_US.UTF-8    
 [5] LC_MONETARY=fi_FI.UTF-8    LC_MESSAGES=en_US.UTF-8   
 [7] LC_PAPER=fi_FI.UTF-8       LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C            
[11] LC_MEASUREMENT=fi_FI.UTF-8 LC_IDENTIFICATION=C       

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods  
[7] base     

other attached packages:
[1] vioplot_0.3.5   zoo_1.8-8       sm_2.2-5.6     
[4] beeswarm_0.2.3  varhandle_2.0.5 RCurl_1.98-1.2 
[7] XML_3.99-0.5   

loaded via a namespace (and not attached):
 [1] rstudioapi_0.13 knitr_1.30      magrittr_1.5   
 [4] hms_0.5.3       lattice_0.20-41 R6_2.5.0       
 [7] rlang_0.4.8     highr_0.8       tcltk_4.0.3    
[10] tools_4.0.3     grid_4.0.3      xfun_0.19      
[13] ellipsis_0.3.1  yaml_2.2.1      tibble_3.0.4   
[16] lifecycle_0.2.0 crayon_1.3.4    readr_1.4.0    
[19] vctrs_0.3.4     bitops_1.0-6    compiler_4.0.3 
[22] pillar_1.4.6    pkgconfig_2.0.3

[TomKellyGenetics]

Thanks for reporting. I've done some quick testing and it seems to be thresholded at >100. Unclear why this is.

> library("vioplot")
Loading required package: sm
Package 'sm', version 2.2-5.6: type help(sm) for summary information
Loading required package: zoo

Attaching package: ‘zoo’

The following objects are masked from ‘package:base’:

    as.Date, as.Date.numeric

> vioplot(x = c(0))
> vioplot(x = c(rep.int(0, 100)))
> vioplot(x = c(rep.int(0, 500)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 250)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 150)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 100)))
> vioplot(x = c(rep.int(0, 125)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 110)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 101)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 100)))

Is there a particular reason you wish to do this? I can't imagine a use case where plotting all of the same value as a violin would be informative. If you plot multiple groups this does not occur.

> vioplot(x = c(rep.int(0, 100), 1))
> vioplot(x = c(rep.int(0, 100)), y = c(rep.int(0, 100)))
> vioplot(x = c(rep.int(0, 101)), y = c(rep.int(0, 101)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
> vioplot(x = c(rep.int(0, 100)), y = c(rep.int(0, 100), 1))
> vioplot(x = c(rep.int(0, 101)), y = c(rep.int(0, 101)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique```

If I can identify the issue we could replace the error by a point making the average (well mean, median, and mode would be the same in this case).

[TomKellyGenetics]

This appears to be an issue with a dependency but we could test for the exception in the package.

> options(error = function() {
+     calls <- sys.calls()
+     if (length(calls) >= 2L) {
+         sink(stderr())
+         on.exit(sink(NULL))
+         cat("Backtrace:\n")
+         calls <- rev(calls[-length(calls)])
+         for (i in seq_along(calls)) {
+             cat(i, ": ", deparse(calls[[i]], nlines = 1L), "\n", sep = "")
+         }
+     }
+     if (!interactive()) {
+         q(status = 1)
+     }
+ })
> vioplot(x = c(rep.int(0, 101)), y = c(rep.int(0, 101)))
Error in cut.default(x, breaks = breaks) : 'breaks' are not unique
Backtrace:
1: stop("'breaks' are not unique")
2: cut.default(x, breaks = breaks)
3: cut(x, breaks = breaks)
4: binning.1d(x, y, breaks = breaks, nbins = nbins)
5: binning(x, y, nbins = opt$nbins)
6: h.select(x = x, y = NA, weights = weights, ...)
7: sm.density(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
8: do.call("sm.density", c(list(data, xlim = est.xlim), args))
9: vioplot.default(x = c(rep.int(0, 101)), y = c(rep.int(0, 101)))
10: vioplot(x = c(rep.int(0, 101)), y = c(rep.int(0, 101)))```

[mmahmoudian]

So this is how I stumbled upon the bug:
I was plotting some RNA-seq read counts for case vs. control for certain genes. I also wanted to add gaps in between each gene (so two violins, gap, two violins, gap, ... two violins). For this, I create the plot as an empty canvas and in every iteration I add two violins and one gap, therefore the violins are plotted one by one and for genes that have value of 0 expression for one group, the vioplot throws the error.

The way I bypassed the error in my code was to add e^-10 to the first lowest number, but of course you cannot do that and instead you can check if there are more than 100 items in the vector, you just trim it to 100 and then plot it.

[TomKellyGenetics]

This is corrected in vioplot 0.3.6 on CRAN.