Perform multiple fisher.test for groups in data frame

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I would like to perform multiple fisher.test() for each tax and each column (ABCB1 and ABL1 in the example below) in the data frame below. The contingency tables should be extracted from the rows like shown below.

EDIT:

The contingency table should be calculated as shown below in the example:

   data_frame(c(42,1),c(20,3))

contingency example:

             ABCB1      NotABCB1
tax1Present      42     1
tax1NotPresent   20     3



42 is 43-1
1 is the cell value Present:tax1Present
20 is 23-3
3 is the cell value in NotPresent: tax1NotPresent

data:

structure(list(group = c("tax1Present", "tax1NotPresent", "tax2Present", 
"tax2NotPresent", "tax3Present", "tax3NotPresent", "tax4Present", 
"tax4NotPresent", "tax5Present", "tax5NotPresent"), ABCB1 = c(1L, 
3L, 4L, 5L, 3L, 6L, 6L, 3, 2, 6L), ABL1 = c(18L, 14, 12L, 
9L, 1L, 5L, 0L, 0L, 7L, 0L), Present = c(43L, 43, 23L, 23, 
9L, 9, 7L, 7, 20, 20L),NotPresent = c(23, 23, 18, 18, 
7L, 7L, 10L, 10L, 10, 10L), tax = c("tax1", "tax1", "tax2", 
"tax2", "tax3", "tax3", "tax4", "tax4", "tax5", "tax5")), row.names = c(NA, 
10L), class = "data.frame")


> df
            group ABCB1 ABL1 Present NotPresent  tax
1     tax1Present     1   18    43   23          tax1
2  tax1NotPresent     3   14    43   23          tax1
3     tax2Present     4   12    23   18         tax2
4  tax2NotPresent     5    9    23   18          tax2
5     tax3Present     3    1    9    7         tax3
6  tax3NotPresent     6    5    9    7        tax3
7     tax4Present     6    0    7    10         tax4
8  tax4NotPresent     3    0    7    10         tax4
9     tax5Present     2   7    20   10          tax5
10 tax5NotPresent     6    0    20   10          tax5
tidyverse dataframe r
1 Answers

Try using apply:

Data:

df <- structure(list(group = c("tax1Present", "tax1NotPresent", "tax2Present", 
"tax2NotPresent", "tax3Present", "tax3NotPresent", "tax4Present", 
"tax4NotPresent", "tax5Present", "tax5NotPresent"), ABCB1 = c(1L, 
3L, 4L, 5L, 3L, 6L, 6L, 12L, 13L, 6L), ABL1 = c(18, 14, 12, 9, 
1, 5, 0, 0, 7, 0), Present = c(43, 43, 23, 23, 9, 9, 7, 7, 20, 
20), NotPresent = c(23, 23, 18, 18, 7, 7, 1, 13, 10, 10), tax = c("tax1", 
"tax1", "tax2", "tax2", "tax3", "tax3", "tax4", "tax4", "tax5", 
"tax5")), row.names = c(NA, 10L), class = "data.frame")

# set the columns to use
columns <- c("ABCB1", "ABL1")

dat_test <- sapply( column, function(colx) 
  lapply( unique(df$tax), function(x) 
    fisher.test( data.frame( 
      a=c(( df[ which(df$tax %in% x)[1] ,"Present"] - 
      df[ which(df$tax %in% x)[1], colx] ), df[ which(df$tax %in% x)[1], colx]), 
      b=c(( df[ which(df$tax %in% x)[2],"NotPresent"] - 
      df[ which(df$tax %in% x)[2], colx] ), df[ which(df$tax %in% x)[2], colx]) ))
 ) )

# set rownames
rownames(dat_test) <- unique( df$tax )

dat_test
     ABCB1  ABL1  
tax1 List,7 List,7
tax2 List,7 List,7
tax3 List,7 List,7
tax4 List,7 List,7
tax5 List,7 List,7

TEST:

#p-values == manual calculation [and exactly the same
#values as with previous df$Total]:
     ABCB1      ABL1      
tax1 0.1179487  0.1971581 
tax2 0.4709802  1         
tax3 0.06013986 0.03496503
tax4 1          1         
tax5 1          0.06371942

Also see: https://stats.stackexchange.com/questions/332224/2x2-fisher-exact-test-contingency-tables

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