向量列表中向量元素的比率

Question

I have a data frame which has a column:

> head(df$lengths,5)
[[1]]
[1] "28"

[[2]]
[1] "33"

[[3]]
[1] "47" "37" "42" "41"

[[4]]
[1] "41" "39" "64" "54"

[[5]]
[1] "45" "22" "23"

I would like to operate on the elements in the vectors, to obtain the ratios of the element(i) to the element(ik) in each vector. Where a ratio cannot be obtained because element(ik) has invalid index, the result should be NA. The desired output is like this, where I specified k=1:

[[1]]
[1] NA

[[2]]
[1] NA

[[3]]
[1] NA (37/47) (42/37) (41/42)

[[4]]
[1] NA (39/41) (64/39) (54/64)

[[5]]
[1] NA (22/45) (23/22)

as for k=2:

[[1]]
[1] NA

[[2]]
[1] NA

[[3]]
[1] NA NA (42/47) (41/37)

[[4]]
[1] NA NA (64/41) (54/39)

[[5]]
[1] NA NA (23/45)

I have little clue on how to approach this, I would think to perform some loops, but in R, it seems complicated. Please advice.

Answer 1

We loop through the list elements ( lapply(.. ), if the length of the list element is 1, we return 'NA' or else divide the next value by the current value and concatenate with NA . We convert to numeric as the original list elements were character class.

 lapply(df$lengths, function(x) if(length(x)==1) NA
            else c(NA, as.numeric(x[-1])/as.numeric(x[-length(x)])))

Update

We could use the lag/lead function in dplyr/data.table for k values greater than 1.

library(dplyr)
k <- 2
lapply(df$lengths, function(x) {x <- as.numeric(x)
               if(length(x)==1) NA 
                 else c(rep(NA,k), na.omit(lead(x,k)))/na.omit(lag(x,k))})
#[[1]]
#[1] NA

#[[2]]
#[1] NA

#[[3]]
#[1]       NA       NA 0.893617 1.108108

#[[4]]
#[1]       NA       NA 1.560976 1.384615

#[[5]]
#[1]        NA        NA 0.5111111

Or without using any packages, we can do with head/tail functions

lapply(lst, function(x) {x <- as.numeric(x)
               if(length(x)==1) NA 
               else c(rep(NA, k), tail(x, -k)/head(x,-k))})