Interpreting Fisher's test and power analysis in R Currently I want to check whether there is a significant difference between the data from two clusters with n1 = n2 = 6.
    cluster1 cluster2
av         0        1
cem        3        4
cl         2        4
cm         1        1
md         2        4
mtt        0        3
pf         1        1
pul        0        0
r          1        1
va         1        2
vl         1        5
vm         0        5
vp         0        0

As frequencies aren't bigger than 5 I decided to use Fisher's exact test, but I'm somewhat lost. Something that I noticed is how I put the data into Fisher's exact test. 
For example fisher.test(df$cluster1, df$cluster2) delivers a p-value of 0.1192 for a two-sided Fisher's test while fisher.test(df) delivers a p-value of 0.7792. What is the correct usage? Secondly, I wonder how to conduct a compromise power analysis for Fisher's test. As my sample size is rather small using alpha = .05 is not a good idea. However, for calculating a power analysis I need the proportions but iirc the proportion is the ratio of the category of interest and the sample size.
Further, I wonder why I get a p-value of 0.7792 if it is the right p-value. As you can see I made a barplot for the distribution of thalamic nuclei in the two clusters. In my eyes, the barplot suggest that there is a significant difference between the two clusters, however, the p-value suggests there isn't.
Am I using the Fisher's test wrong? Is the p-value correct and my interpretation is simply wrong? And are there some other methods to analyse the distribution of my variables?

 A: Two tests seem appropriate. Counts in Cluster 2 tend to
be larger than counts in Cluster 1. I assume row counts are
determined independently.
Sign test:  Of the 13 row categories, Cluster 2 had higher
counts than Cluster 1 for 8 categories, and and there is no
difference for 5 categories. No row category had a smaller count
in Cluster 2.
The sign test ignores the 5 categories with no difference.
In 8 tosses of a fair coin the probability of getting all heads
or all
tails is $(1/2)^7 = 0.0078 < 0.05$ so there is a significant
difference at the 5% level. Minitab output:
Sign test of median =  0.00000 versus ≠ 0.00000

      N  Below  Equal  Above       P  Median
Dif  13      0      5      8  0.0078   1.000

Wilcoxon signed-rank test: Look at the 13 differences for
Cluster 2 - Cluster 1. This test takes into account the sizes
of the positive differences. The P-value is $0.014 < 0.05,$
again significant.
Wilcoxon Signed Rank Test 

Test of median = 0.000000 versus median ≠ 0.000000

         N for   Wilcoxon         Estimated
      N   Test  Statistic      P     Median
Dif  13      8       36.0  0.014      1.000

Note: Neither Fisher's exact test nor (even with higher counts) 
a chi-squared test would be appropriate. The counts in
Cluster 2 are consistently higher than the counts in Cluster 1
for the same category. But both of these tests are looking
for inconsistency between the two clusters. 
Here are Binomial counts, consistently about three times as large in B as in A:
set.seed(1234)
A = rbinom(13, 25, .3);  B = rbinom(13, 25, .9)
DTA = rbind(A,B);  DTA
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
A    5    8    8    8   10    8    3    6    8     8     9     8     6
B   20   23   21   23   24   24   24   23   23    24    25    24    21

I suppose you would have been happy with such a strong result in your
experiment if you had more data. However, the hypothesis of the chi-squared
test for homogeneity is (appropriately) not rejected.
chisq.test(DTA)

        Pearson's Chi-squared test

data:  DTA
X-squared = 3.9673, df = 12, p-value = 0.984

