Writing a function to filter and summarize data into proportion table

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I want to create a large proportion table that involves filtering out certain values based on one column and outputting the proportion of values equal to 0 and those greater than 0 in table. Here's an example of the data frame (df):

     ID   a   b   c   d   e   f   g
1     1   1   2   3   0   4   5   A 
2     2   0   0   1   0   2   0   A
3     3   1   5   2   1   0   0   B
4     4   5   1   2   0   1   1   B
5     5   2   0   1   0   0   0   C
...

From this, I want to come up with the proportion that b=0 or b>0 IF column a>0. For your reference, I can get this information with the following code:

prop.table(table(df$b[df$a>0]!=0))*100

However, I want to do the same with columns c and d as well as e and f (same sort of pattern so that you're filtering out when c=0 and when e=0 to get those >0 and =0 proportions for d and f, respectively). Additionally, I would love to have this output all into a single table. Might look something like this:

      b.perc   d.perc   f.perc
TRUE   75.00    20.00    66.67
FALSE  25.00    80.00    33.33

Any help is appreciated. Also, I would like to calculate the TRUE percentages across groups listed in column G, giving me an output like this:

      b.perc   d.perc   f.perc
A     100.00    0.00     50.00
B     100.00   50.00    100.00
C     0.00      0.00      0.00
summarize filter r
1 Answers

We subset the alternate columns, use each set as inputs to mapply, get the table and prop.table based on the condition mentioned in the OP's post

out <- round(mapply(function(x, y) prop.table(table(x[y > 0] != 0)) * 100,
          df[c(FALSE, TRUE)], df[c(TRUE, FALSE)]), 2)
colnames(out) <- paste0(colnames(out), ".perc")
out
#      b.perc d.perc f.perc
#FALSE     25     80  33.33
#TRUE      75     20  66.67

If we are just interested in the TRUE percentage, then we can do this with colMeans as well

colMeans((df[c(FALSE, TRUE)] * NA^!(df[c(TRUE, FALSE)] > 0)) != 0, na.rm = TRUE)
#       b         d         f 
#0.7500000 0.2000000 0.6666667

data

df <- structure(list(a = c(1L, 0L, 1L, 5L, 2L), b = c(2L, 0L, 5L, 1L, 
0L), c = c(3L, 1L, 2L, 2L, 1L), d = c(0L, 0L, 1L, 0L, 0L), e = c(4L, 
2L, 0L, 1L, 0L), f = c(5L, 0L, 0L, 1L, 0L)), class = "data.frame",
row.names = c("1", 
"2", "3", "4", "5"))
user_37322710
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