How to analyze these data? The biological data is listed as following: 
   V1    V2    V3    V4    V5    V6
0.064 0.014 0.016 0.012 0.013 0.023
0.056 0.000 0.000 0.008 0.010 0.000
0.042 0.014 0.024 0.008 0.017 0.023
0.031 0.014 0.016 0.008 0.013 0.023
0.068 0.000 0.008 0.004 0.020 0.000
0.081 0.000 0.000 0.004 0.010 0.000
0.060 0.014 0.016 0.006 0.010 0.023

or you can download the data from http://www.mediafire.com/?6yp9l9m47jv433a.
A<- dat[,1] 
B<- dat[,2:6]

I want to compare the difference between the first column to other columns of the data.Because only dat[,2] and dat[,6] not subject to normal distribute,I used wilcox.test instead of t.test function to caculate in R. But the warning messages rised up,such as "In wilcox.test.default(A, B[, 1]) : cannot compute exact p-value with ties". Could you give me some suggestions? Thank you.
wilcox.test(A,B[,1])

Wilcoxon rank sum test with continuity correction

data: A and B[, 1] W = 49, p-value = 0.00184 alternative hypothesis: true location shift is not equal to 0

Warning message: In wilcox.test.default(A, B[, 1]) : cannot compute exact p-value with ties

 A: Sometimes a formal statistical test is overkill.  Row by row, the entries in the first column are the largest.  Draw a picture to make this apparent: side-by-side boxplots or dotplots would work nicely.
Although this is a post-hoc comparison, if the initial intent had been to compare the first column against the rest for a shift in distribution, the most extreme characterizations would be that either all maxima or all minima occur in the first column (a two-sided test).  The chance of this occurring by chance, if all columns contained values drawn at random from a common distribution, would be $2 (\frac{1}{6})^7$ = about 0.0007%.
In fact, the first two contains the largest 7 of the 42 values.  Again, ex post facto, the chance of such an extreme ordering occurring equals $\frac{2}{42 \choose 7}$ = about 0.000007%.
These results indicate that any reasonably powerful test you choose to conduct will conclude there's a highly significant difference.
In any event, You don't need a p-value; you need to characterize how large the difference is (the right way to do this depends on what the data mean) and you need to seek an explanation for the difference.
A: For most of your variables (e.g. V2), some observations have identical values, hence the warning message thrown by R: unique ranks cannot be computed for all observations, and there are ties, precluding the computation of an exact p-value. For your variable named V2, there are in fact only two distinct values (out of 7), so I am very puzzled by the approach you took to analysis your data. With such a high number of tied data, I would not trust any Wilcoxon test. Moreover, in most non-parametric tests we assume that the sampled populations are symmetric and have the same dispersion or shape, which is hardly verifiable in your case.
Thus, I think a permutation test would be more appropriate in your case, see e.g. permTS (perm), pperm (exactRankTests), or the coin package.
A: Thank you very much, chl, whuber and Gaetan Lion. But do you think is there any problem that if I change to caculate the differene among the data using Kruskal-Wallis test instead of comparing the difference between the first column with other columns?

kruskal.test(as.list(Data))

    Kruskal-Wallis rank sum test

data:  as.list(Data) 
Kruskal-Wallis chi-squared = 19.9149, df = 5, p-value = 0.001297

kruskal.test(as.list(Data[,2:6]))

    Kruskal-Wallis rank sum test

data:  as.list(Data[, 2:6]) 
Kruskal-Wallis chi-squared = 3.8242, df = 4, p-value = 0.4303
The result also shows the 1st column has great difference between the other columns. Is it that right?
