Filling data frame with previous row value

user_2813055 0
Viewed 7240

I have a data frame that has 2 columns.

column1 has random numbers in column2 is a place holding column for what i want column3 to look like

  random    temp
0.502423373 1
0.687594055 0
0.741883739 0
0.445364032 0
0.50626137  0.5
0.516364981 0
...

I want to fill column3 so it takes the last non-zero number (1 or .5 in this example) and continuously fills the following rows with that value until it hits a row with a different number. then it repeats the process for the entire column. 

random     temp state
0.502423373 1   1
0.687594055 0   1
0.741883739 0   1
0.445364032 0   1
0.50626137  0.5 0.5
0.516364981 0   0.5
0.807804708 0   0.5
0.247948445 0   0.5
0.46573337  0   0.5
0.103705154 0   0.5
0.079625868 1   1
0.938928944 0   1
0.677713019 0   1
0.112231619 0   1
0.165907178 0   1
0.836195267 0   1
0.387712998 1   1
0.147737077 0   1
0.439281543 0.5 0.5
0.089013503 0   0.5
0.84174743  0   0.5
0.931738707 0   0.5
0.807955172 1   1

thanks for any and all help

calculated-columns dataframe r
7 Answers

Perhaps you can make use of na.locf from the "zoo" package after setting values of "0" to NA. Assuming your data.frame is called "mydf":

mydf$state <- mydf$temp
mydf$state[mydf$state == 0] <- NA

library(zoo)
mydf$state <- na.locf(mydf$state)
#      random temp state
# 1 0.5024234  1.0   1.0
# 2 0.6875941  0.0   1.0
# 3 0.7418837  0.0   1.0
# 4 0.4453640  0.0   1.0
# 5 0.5062614  0.5   0.5
# 6 0.5163650  0.0   0.5

If there were NA values in your original data.frame in the "temp" column, and you wanted to keep them as NA in the newly generated "state" column too, that's easy to take care of. Just add one more line to reintroduce the NA values:

mydf$state[is.na(mydf$temp)] <- NA
user_12706950

Here is an interesting way with the Reduce function.

temp = c(1,0,0,0,.5,0,0,0,0,0,1,0,0,0,0,0,1,0,0.5,0,0,0,1)
fill_zero = function(x,y) if(y==0) x else y
state = Reduce(fill_zero, temp, accumulate=TRUE)

If you're worried about speed, you can try Rcpp.

library(Rcpp)
cppFunction('
  NumericVector fill_zeros( NumericVector x ) {
    for( int i=1; i<x.size(); i++ )
     if( x[i]==0 ) x[i] = x[i-1];
    return x;
  }
')
state = fill_zeros(temp)
user_14428210

A loop along the following lines should do the trick for you - 

for(i in seq(nrow(df)))
{
  if (df[i,"v1"] == 0) df[i,"v1"] <- df[i-1,"v1"]
}

Output - 

> df
   v1 somedata
1   1       33
2   2       24
3   1       36
4   0       49
5   2       89
6   2       48
7   0        4
8   1       98
9   1       60
10  2       76
> 
> for(i in seq(nrow(df)))
+ {
+   if (df[i,"v1"] == 0) df[i,"v1"] <- df[i-1,"v1"]
+ }
> df
   v1 somedata
1   1       33
2   2       24
3   1       36
4   1       49
5   2       89
6   2       48
7   2        4
8   1       98
9   1       60
10  2       76
user_24734680

I suggest using the run length encoding functions, it's a natural way for dealing with steaks in a data set. Using @Kevin's example vector:

temp = c(1,0,0,0,.5,0,0,0,0,0,1,0,0,0,0,0,1,0,0.5,0,0,0,1)
y <- rle(temp)
#str(y)
#List of 2
# $ lengths: int [1:11] 1 3 1 5 1 5 1 1 1 3 ...
# $ values : num [1:11] 1 0 0.5 0 1 0 1 0 0.5 0 ...
# - attr(*, "class")= chr "rle"


for( i in seq(y$values)[-1] ) {
   if(y$values[i] == 0) {
      y$lengths[i-1] = y$lengths[i] + y$lengths[i-1]
      y$lengths[i] = 0
   }
}

#str(y)
#List of 2
# $ lengths: num [1:11] 4 0 6 0 6 0 2 0 4 0 ...
# $ values : num [1:11] 1 0 0.5 0 1 0 1 0 0.5 0 ...
# - attr(*, "class")= chr "rle"

inverse.rle(y)
#  [1] 1.0 1.0 1.0 1.0 0.5 0.5 0.5 0.5 0.5 0.5 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 0.5
# [20] 0.5 0.5 0.5 1.0
user_9867930

Simply use a loop with a global variable ,

globalvariable used here is m, r is a dataframe with two columns A and B.

r$B = c(1,NA, NA, NA, 3, NA,6)


m=1

for( i in 1:nrow(r) ){

  if(is.na(r$B[i])==FALSE ){

    m <<- i # please note the assign sign ,  " <<- "
    next()

  } else {

    r$B[i] = r$B[m]

  }

}

After Execution : r$B = 1 1 1 1 3 3 6

user_111283400
Related