Repeat condition over columns - r

Question

I have a data frame of 14 columns and thousands of rows. I want to count or select rows where value in column 1 is 0 and more than 0 in the other 13 columns, then count those were value is 0 in second column and more than 0 in the other 13 columns and so on for all 14 columns.
Any hint on how to do that ?
Many thanks

Answer 1

Try this. The first line is to replicate the data and the second line shows the counting result based on your logical expression
df <- data.frame(replicate(14, sample(0:5, 1000, replace = T)))
result <- sapply(1:14, function(i) {sum(df[,i]==0 & apply(df[-i]>0, 1, all))})
names(result) <- paste0("Col_", 1:14)
result
Col_1 Col_2 Col_3 Col_4 Col_5 Col_6 Col_7 Col_8 Col_9 Col_10 Col_11 Col_12 Col_13 Col_14
12 12 19 15 18 20 19 13 19 15 12 17 15 18

Answer 2

Are you aware of the function apply? If you write a function that reads a vector of length 14 and and outputs T or F depending on whether the vector satisfies the requirement, then you can use apply to apply this function to all rows of the data.frame, yielding a vector of thousands of Ts and Fs that can be used for selecting or counting (the latter by simply putting the vector into the sum function).
Example:
cow <- function(colnr, x){#colnr is number of column you want zero, x is vector of length 14
all(x[-colnr] > 0) & x[colnr] == 0)
}
horse <- function(colnr){#produces sequence of Trues and Falses telling you which columns satisfy the condition
apply(yourdataframe, 1, cow)
}
#example output:
horse(1)
#while we're at it: create a vector of length 14 containing the number of rows satisfying the 14 conditions:
sapply(seq(1:14), horse)
The 1 in apply is because you want to apply to rows, not columns. The function sapply is like apply but then applying a function to each element of a vector rather than each row of a dataframe.
Update: this answer is the same as zyunaidi's which appeared while I was typing.

Answer 3

Using the sample data from zyurnaidi you can do this.
Find all 0 values in your data.frame using which with array indices set on TRUE, then remove duplicated rows (0 in other columns) and count the occurence per column:
set.seed(1234)
df <- data.frame(replicate(14, sample(0:5, 1000, replace = T)))
a <- which(df == 0, arr.ind = T)
table(a[ !(duplicated(a[, 1]) | duplicated(a[, 1], fromLast=T)), 2])
1 2 3 4 5 6 7 8 9 10 11 12 13 14
18 26 19 14 11 20 21 10 24 21 15 11 22 11