r - Replace row values in dataframe that meet special conditions -

i need replace 0's 1's, groups of values meet following conditions; "1 0 1" or "0 1" (if @ beginning) or "1 0" (if @ end). given example dataframe:

df <- data.frame(a = c(1,0,1,0,1,1,1,0,1,1,1),                  b = c(1,1,1,0,1,1,1,0,1,1,1),                  c = c(1,0,1,1,1,0,1,0,1,1,1),                  d = c(1,1,1,0,1,1,1,1,1,1,1),                  e = c(1,0,1,0,1,1,1,1,1,1,1),                  f = c(1,1,1,1,1,1,1,1,1,0,1)) df 

it need return :

df.result <- data.frame(a = c(1,1,1,0,1,1,1,0,1,1,1),                         b = c(1,1,1,0,1,1,1,0,1,1,1),                         c = c(1,1,1,1,1,1,1,0,1,1,1),                         d = c(1,1,1,0,1,1,1,1,1,1,1),                         e = c(1,1,1,0,1,1,1,1,1,1,1),                         f = c(1,1,1,1,1,1,1,1,1,1,1)) df.result 

notice relevant 0's have changed 1's. essentially, i'm trying replace 0's occur alone in row.

any idea how achieve in r?

thanks in advance.

here solution rle():

foo <- function(x) {   r <- rle(x)   r$values[r$values==0 & r$lengths==1] <- 1   inverse.rle(r) } foo(c(0,1,0,0,1,0,1))  # testing working horse: # [1] 1 1 0 0 1 1 1 

now apply function on each row , give result desired form. apply() coerces first argument matrix:

t(apply(df, 1, foo)) # > t(apply(df,1,foo)) #      [,1] [,2] [,3] [,4] [,5] [,6] # [1,]    1    1    1    1    1    1 # [2,]    1    1    1    1    1    1 # [3,]    1    1    1    1    1    1 # [4,]    0    0    1    0    0    1 # [5,]    1    1    1    1    1    1 # [6,]    1    1    1    1    1    1 # [7,]    1    1    1    1    1    1 # [8,]    0    0    0    1    1    1 # [9,]    1    1    1    1    1    1 # [10,]   1    1    1    1    1    1 # [11,]   1    1    1    1    1    1 

if want dataframe result can do:

df.result <- df df.result[,] <- t(apply(df,1,foo))