How to construct this binary variable in R?

Question

The aim is check if value at index i is 1 and then make the previous six entries as 1.

x <- c(0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1)

## Required output 
y <- c(1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1)

## Attempt 
for(j in seq_along(x)){
if(x[j] == 1){
   for(i in (j-6):j)
   x[i] = 1
   }}

Could you help solve this or better approach ?

Thanks.

Roland · Accepted Answer · 2014-11-13 12:43:11Z

5

A fully vectorized solution using filter:

as.integer( #turn logical value into numbers
  as.logical( #coerce to logical --> 0 becomes FALSE, everything else TRUE
   rev( #reverse order
    filter( #linear filtering
      c(rep(0, 6), #pad with zeros in the beginning to avoid NAs
        rev(x)), #revers order of input vector
          c(rep(1, 7)), sides=1 #y_i = x_i * 1 + x_(i-1) * 1 +...+ x_(i-6) * 1 
  )[-(1:6)]))) #remove NA values

#[1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

answered Nov 13, 2014 at 12:43

Roland

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4 Comments

David Arenburg Over a year ago

(+1) Nice vectorization.

Anusha Over a year ago

Isnt filter used for moving average type of calculations ? Could you explain how the filter function is working in this context ? Thanks.

Roland Over a year ago

filter calculates a moving sum (for past values or for a window on both sides). If the moving sum isn't zero I turn it into a 1.

Anusha Over a year ago

+1 Brilliant. I did think of rollapply but not of taking moving sum.

David Arenburg · Accepted Answer · 2014-11-13 13:11:57Z

5

You could try the following options (though don't forget to initialize x when trying each option as I'm overriding it)

indx <- mapply(function(x, y) x:y, which(x == 1) - 6 , which(x == 1))
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

Or even simpler

indx <- sapply(which(x == 1) - 6, function(x) x:(x + 6))
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

Or

indx <- apply(cbind(which(x == 1) - 6 , which(x == 1)), 1, function(x) x[1]:x[2])
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

Or

indx <- seq_len(6)
indx <- sapply(which(x == 1), function(x) x - indx)
x[indx[indx > 0]] <- 1
x
## [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

edited Nov 13, 2014 at 13:11

answered Nov 13, 2014 at 12:40

David Arenburg

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1 Comment

Anusha Over a year ago

+1. I was trying to use apply function on index but didnt think of extracting the indexes using apply and then assigning all those to 1. Thanks for this logic.

rnso · Accepted Answer · 2014-11-13 12:32:10Z

3

Using 'for' loop:

ddf = data.frame(x,y=0)
for(i in 1:nrow(ddf)){
    if(ddf[i,'x']==1){
        j = i-5
        if(j<1) j=1
        ddf[j:i,'y'] = 1
    }
}
ddf
   x y
1  0 1
2  0 1
3  0 1
4  1 1
5  0 0
6  0 0
7  0 0
8  0 0
9  0 0
10 0 0
11 0 0
12 0 1
13 0 1
14 0 1
15 0 1
16 0 1
17 1 1
18 0 1
19 1 1

y = ddf$y
y
 [1] 1 1 1 1 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

answered Nov 13, 2014 at 12:32

rnso

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5 Comments

Anusha Over a year ago

Thanks for the reply. Could you explain why my loop was returning the y vector as all 1's. Why do I need a dataframe here ?

rnso Over a year ago

You were not correcting if j-6 was less than 1. Also data.frame are easier to work with.

Anusha Over a year ago

+1. Would it be possible to put this into a function and then use apply ? Just curious how the sapply or lapply would be defined for this for loop.

rnso Over a year ago

See answers by David and CathG

Anusha Over a year ago

Yes but there they are extracting indexes using apply and then assigning those to 1. I was asking how to convert the for loop here into lapply.

Cath · Accepted Answer · 2014-11-13 14:32:33Z

2

y<-x
y[unlist(sapply(which(x==1),
                function(val){
                   val:(max(val-6,1))
                 }
                )
        )
  ]<-1

> y
 [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

Explanation :

I first look for indices of x=1 with which(x==1). Then, for each of the indices I get the indices from the one with x=1 to the 6th before that with sapply(...) then I unlist the result to only have a vector of indices for which y must be 1. I then assigned 1 to the corresponding y values.

another writing, in 2 steps :

y<-x
ind<-unlist(sapply(which(x==1),function(val){val:(max(val-6,1))}))
y[ind]<-1

> y
 [1] 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1

edited Nov 13, 2014 at 14:32

answered Nov 13, 2014 at 12:33

Cath

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4 Comments

Anusha Over a year ago

Thanks for the reply. Could you explain how you defined the function. I was looking for an *apply approach. How is that assignment <- 1 part working ?

Cath Over a year ago

@Anusha, I edited my answer to add an explanation, tell me if that's ok now or if you need me to detail it a bit more.

Anusha Over a year ago

I understood that part. I am curious about {val:(max(val-6,1))}))]<-1 , the bracketting and the logic.

Cath Over a year ago

@Anusha I "re-organize" my answer so you can better see what goes with what : the function applied to each element of which(x==1) calculates the indices of y values that should be one but it is the whole unlisted result of sapply function (in this case 4 3 2 1 17 16 15 14 13 12 11 19 18 17 16 15 14 13) that really goes between the brackets. In the end, it's like writing y[c(4,3,2,1,17,16,15,14,13,12,11,19,18,17,16,15,14,13)]<-1. I hope I understood right you comment and that this answers it

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