Return rows where consecutive values meet criterion

Question

I have the following dataframe df . I would like to return a vector result that indicates which rows meet the following criterion: at least 2 consecutive values in that row are lower than -1.7.

set.seed(123)

df <- data.frame(V1=rnorm(10,-1.5,.5),
                 V2=rnorm(10,-1.5,.5),
                 V3=rnorm(10,-1.5,.5),
                 V4=rnorm(10,-1.5,.5),
                 V5=rnorm(10,-1.5,.5),
                 V6=rnorm(10,-1.5,.5),
                 V7=rnorm(10,-1.5,.5),
                 V8=rnorm(10,-1.5,.5),
                 V9=rnorm(10,-1.5,.5),
                 V10=rnorm(10,-1.5,.5))
rownames(df) <- c(seq(1976,1985,1))

The result would be a vector:

result <- c(1977,1979,1980,1982,1983,1985)

Answer 1

One option is to loop through the rows with apply , create a logical condition with rle , check if there are any TRUE elements that have lengths more than 1, extract the names

names(which(apply(df, 1, function(x) with(rle(x < - 1.7), any(lengths[values] > 1)))))
#[1] "1977" "1979" "1980" "1982" "1983" "1985"

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Or a better approach is to vectorize it by placing two logical matrices (ie remove the first column of the dataset, check whether it is less than -1.7, similarly remove the last column and do the same), Reduce it to a single logical matrix by checking whether the corresponding elements are TRUE , get the rowSums , if the value is greater than 0, we extract the row names

names(which(rowSums(Reduce(`&`, list(df[-ncol(df)] < -1.7, df[-1] < -1.7))) > 0))
#[1] "1977" "1979" "1980" "1982" "1983" "1985"

Answer 2

A fun option using which with arr.ind = TRUE

temp <- which(df < -1.7, arr.ind = TRUE)
rownames(df)[aggregate(col~row, temp, function(x) any(diff(x) == 1))[, 2]]

#[1] "1977" "1979" "1980" "1982" "1983" "1985"

We first get all row and column positions where value is less than -1.7. Using aggregate we group col for every row and check if there is at least one consecutive value in a row and for values which return TRUE subset its rownames .

Answer 3

A solution which uses the lagged sum to get the sum of each pair of numbers in a vector. If the lagged sum gets 2, then it means at least 2 consecutive values in that row meet the condition.

rownames(df)[apply(df < -1.7, 1, function(x) any(x[-nrow(df)] + x[-1] == 2))]

# [1] "1977" "1979" "1980" "1982" "1983" "1985"