Replace 0's with previous non-zero value per ID (lag)

问题

How can I replace all 0's with the last non-zero value per ID in R?

Example:

Input:

df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2),
         Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0))

Output:

df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2),
         Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0),
         res = c(0,10,30,30,30,50,80,0,0,57,57))

Is there an easy way with lag function?

回答1:

Here's a tidyverse approach:

library(tidyverse)
df %>% 
  group_by(ID) %>% 
  mutate(x = replace(Var1, cumsum(Var1 !=0) > 0 & Var1 == 0, NA)) %>% 
  fill(x)
# # A tibble: 11 x 4
# # Groups:   ID [2]
# ID  Var1   res     x
# <dbl> <dbl> <dbl> <dbl>
# 1    1.    0.    0.    0.
# 2    1.   10.   10.   10.
# 3    1.   30.   30.   30.
# 4    1.    0.   30.   30.
# 5    1.    0.   30.   30.
# 6    1.   50.   50.   50.
# 7    1.   80.   80.   80.
# 8    2.    0.    0.    0.
# 9    2.    0.    0.    0.
# 10    2.   57.   57.   57.
# 11    2.    0.   57.   57.

In the mutate step, we replace 0's with NA except for those that are at the beginning of each ID-run because in those cases we have no values to replace the NAs afterwards.

If you have multiple columns to adjust, you can use:

df %>% 
  group_by(ID) %>% 
  mutate_at(vars(starts_with("Var")), 
            funs(replace(., cumsum(. !=0) > 0 & . == 0, NA))) %>% 
  fill(starts_with("Var"))

where df could be:

df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2),
                 Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0),
                 Var2 = c(4,0, 30, 0, 0,50,0,16, 0, 57, 0))

回答2:

Without using any package, just using the loops:

df <- data.frame(ID = c(1,1,1,1,1,1,1,2,2,2,2),
                 Var1 = c(0,10, 30, 0, 0,50,80,0, 0, 57, 0)) 

for(i in 1:nrow(df)){
  if(i!=1){    
    if(df$ID[i-1]==df$ID[i] && df$Var1[i]==0){  # if value is zero and value of current and previous rows ID are same
      if(df$Var1[i-1]!=0){           # If previous value is not zero then store it 
        df$res[i]=df$Var1[i-1]       # Use previous value of var1
        a=0
        a=df$Var1[i-1]
      }else{
        df$res[i]=a     # Use previous value var1
        a=0
      }
    }else{
      df$res[i]=df$Var1[i]  # Use the current value of var1
    }

  }else{
    df$res[i]=df$Var1[i]    # Set the first point as it is
  }
}

Output:

> df
      ID Var1 res
   1   1    0   0
   2   1   10  10
   3   1   30  30
   4   1    0  30
   5   1    0  30
   6   1   50  50
   7   1   80  80
   8   2    0   0
   9   2    0   0
   10  2   57  57
   11  2    0  57

来源：https://stackoverflow.com/questions/50966647/replace-0s-with-previous-non-zero-value-per-id-lag

标签

replace

lag

zero