The Wilcoxon Signed-Rank (Mann-Whitney U) test two-sample test and the
Kruskal-Wallis test for comparing $k > 2$ samples, are both based on
ranks. That is, the numerical values you input are reduced to ranks.
For a single sample, ranking goes like this (output from R):
x = c(1, 4, 7, 3, 11, 6); x
[1] 1 4 7 3 11 6
rank(x)
[1] 1 3 5 2 6 4
If there are ties in the data, then ranking is not quite so straightforward:
y = c(1, 4, 6, 2, 4, 6, 11); y
[1] 1 4 6 2 4 6 11
rank(y)
[1] 1.0 3.5 5.5 2.0 3.5 5.5 7.0
Various texts and software packages treat ties differently.
In order to find a p-value for the test, all of them begin
with with distribution theory based on ranks from data without ties.
Some have approximate distribution theory for the less-tidy ranks
that result from ties. (When more than one sample is involved,
not only duplicated values within a sample count as ties, but
duplicated value anywhere among the groups also count.)
If there are only a few ties in a dataset, then printed warnings
about ties can often be ignored. If there are many ties, then
either no p-value will be given, or the p-value provided may be
essentially useless.
An unofficial trick can be used to assess how serious ties are
in any one analysis. You can artificially jitter the data with
just enough random noise to break ties and see if the p-value
changes by enough to matter. (Best to do this a couple of times
with different jittering.)
Example:
x = c(1, 2, 2, 4, 5, 3, 0); y = c(4, 6, 3, 8, 11, 11)
wilcox.test(x, y)
Wilcoxon rank sum test with continuity correction
data: x and y
W = 4, p-value = 0.01778
alternative hypothesis: true location shift is not equal to 0
Warning message:
In wilcox.test.default(x, y) : cannot compute exact p-value with ties
.
# Jittering
jx = runif(length(x), -.01, .01); jy = runif(length(y), -.01, .01)
wilcox.test(x+jx, y+jy)
Wilcoxon rank sum test
data: x + jx and y + jy
W = 4, p-value = 0.01399
alternative hypothesis: true location shift is not equal to 0
The original and jittered $x$s look like this:
x; rank(x)
[1] 1 2 2 4 5 3 0
[1] 2.0 3.5 3.5 6.0 7.0 5.0 1.0
round(x+jx, 4); rank(x+jx)
[1] 0.9958 1.9900 2.0035 3.9945 5.0093 2.9995 -0.0095
[1] 2 3 4 6 7 5 1
Another run with different random jittering also gave P-value about 0.014.
So if we are working at the 5% level, is seems safe to say that the
original version of the test (with warnings about ties) gave a useful result.
If there is any doubt or if the result is for review or publication,
then you must do a formal test on the data. My first choice would be a simulated permutation test. One elementary
review of permutation tests is Eudey, et al.. (Sect. 3 is
most relevant to this discussion.) Also, perhaps see this Q&A. Permutation tests do
not use ranks and can handle ties without difficulty.
Finally, I don't know exactly how your groups with all 0's should be treated because I don't know what process is producing the 0's. Clearly
not with rank-based tests. You are correct that two groups with all 0's
can't be distinguished from one another by hypothesis testing. If one
group has a few 0's and another has mostly 0's, a binomial test (0's vs non-0's) or permutation test
should help to judge whether they came from different populations.
