How to deal with log of zero in R in image.plot?

Question

I have a matrix, and the entries are all probabilities. Most of the entries have very low probabilities. Some have zeros. I need to do log of the matrix. However, since there are zeros in the matrix, R generates -inf for those zero entries. My goal is to feed this log(matrix) into the image.plot(). When I feed this into the image.plot, I kept getting this error:

Error in seq.default(minz + binwidth/2, maxz - binwidth/2, by = binwidth) : 
  invalid (to - from)/by in seq(.)

Is there any solution in here that can help me get around this?

Here is what the matrix looks like:

                  0          1          2          3         4         5         6
 [1,] -0.0007854138 -8.9132811 -10.011893 -10.705041 -9.606428 -9.318746      -Inf
 [2,] -0.3402118357 -1.6137090  -2.742625  -4.215836 -5.721434 -7.121522 -9.606428
 [3,] -0.2912175507 -2.0296478  -3.521929  -4.275321 -4.426519 -4.187369 -3.715705
 [4,] -1.5244380532 -0.7048802  -2.001368  -3.405243 -3.713864 -3.143919 -3.781412
 [5,] -0.7572491288 -0.7487709  -3.981208  -5.110329 -5.228577 -5.095569 -5.293395
 [6,] -0.0007629648       -Inf  -8.759130  -7.613998 -9.606428      -Inf      -Inf
 [7,] -0.0020658381 -7.4861648  -7.526987  -7.094123 -9.318746      -Inf      -Inf
 [8,] -0.0295715883 -6.7160566  -7.208533  -6.610696 -6.485533 -6.813220 -6.387552
 [9,] -0.0032128722 -6.7160566  -7.613998  -7.871827 -7.760602 -8.759130 -8.759130
[10,] -0.4869248130 -1.3225132  -2.518576  -3.768698 -5.140520 -6.183252 -7.208533
               7          8          9
 [1,]       -Inf -10.705041 -10.011893
 [2,]       -Inf       -Inf  -7.149693
 [3,]  -4.965248  -5.968842  -6.428374
 [4,]  -4.696227  -5.091913  -4.669559
 [5,]  -5.163777  -5.468599  -6.577906
 [6,]       -Inf       -Inf       -Inf
 [7,]       -Inf       -Inf       -Inf
 [8,]  -6.627503  -6.456545  -6.400976
 [9,] -10.011893 -10.011893       -Inf
[10,]  -8.402456  -7.814669  -6.546158

Here is the structure :

structure(c(0.999214894571557, 0.71161956034096, 0.747353073126963, 
0.217743382682817, 0.468954688200987, 0.999237326155227, 0.997936294302378, 
0.970861372812921, 0.996792283535218, 0.614513234634365, 0.000134589502018843, 
0.199147599820547, 0.13138178555406, 0.49416778824585, 0.472947510094213, 
0, 0.000560789591745177, 0.00121130551816958, 0.00121130551816958, 
0.266464782413638, 4.48631673396142e-05, 0.0644010767160162, 
0.0295423956931359, 0.135150291610588, 0.0186630776132795, 0.00015702108568865, 
0.00053835800807537, 0.000740242261103634, 0.000493494840735756, 
0.0805742485419471, 2.24315836698071e-05, 0.0147599820547331, 
0.0139075818752804, 0.0331987438313145, 0.00603409600717811, 
0.000493494840735756, 0.000829968595782862, 0.00134589502018843, 
0.000381336922386721, 0.0230820995962315, 6.72947510094213e-05, 
0.00327501121579183, 0.0119560340960072, 0.0243831314490803, 
0.00536114849708389, 6.72947510094213e-05, 8.97263346792284e-05, 
0.00152534768954688, 0.000426200089726335, 0.00585464333781965, 
8.97263346792284e-05, 0.000807537012113055, 0.0151861821444594, 
0.0431135038133692, 0.00612382234185734, 0, 0, 0.00109914759982055, 
0.00015702108568865, 0.00206370569762225, 0, 6.72947510094213e-05, 
0.0243382682817407, 0.022790489008524, 0.00502467474203679, 0, 
0, 0.00168236877523553, 0.00015702108568865, 0.000740242261103634, 
0, 0, 0.00697622252131, 0.00912965455361149, 0.00572005383580081, 
0, 0, 0.00132346343651862, 4.48631673396142e-05, 0.000224315836698071, 
2.24315836698071e-05, 0, 0.00255720053835801, 0.00614625392552714, 
0.00421713772992373, 0, 0, 0.0015702108568865, 4.48631673396142e-05, 
0.000403768506056528, 4.48631673396142e-05, 0.000785105428443248, 
0.00161507402422611, 0.00937640197397936, 0.00139075818752804, 
0, 0, 0.00165993719156572, 0, 0.00143562135486765), .Dim = c(10L, 
10L), .Dimnames = list(NULL, c("0", "1", "2", "3", "4", "5", 
"6", "7", "8", "9")))

Answer 1

If these zeros are caused by a physical measurement which should yield a positive-definite results but fails to do so for technical reasons, it might be reasonable to substitute 1/2 of the lower limit of detection for the zeros.

 M2 <- M
 print( min(M[M!=0]), digits=16)
#[1] 2.24315836698071e-05
 M2[M2==0] <- 0.5*min(M[M!=0])
 image(M2)
 image(log(M2))