CODEWITHSUNDEEP
rdataframereplacetidyversezoo
Hydro
Hydro

Reputation: 1117

how to replace zero with incremental value in in R data frame?

I have a dataset where I want to replace each zero with incrementing values: the first zero becomes 0.001, the second 0.002, the third 0.003, and so on. Each time a zero appears, it should increase by 0.001 from the previous replacement.

set.seed(123)
DF <- data.frame(Year = 1981:2000, Value = sample(0:5, 20, replace = T))

Upvotes: 3

Views: 86

Answers (5)

Andre Wildberg
Andre Wildberg

Reputation: 19191

val <- DF$Value == 0

transform(DF, Value = replace(Value, val, (cumsum(val) * 0.001)[val]))
   Year Value
1  1981 2.000
2  1982 5.000
3  1983 2.000
4  1984 1.000
5  1985 1.000
6  1986 5.000
7  1987 2.000
8  1988 4.000
9  1989 3.000
10 1990 5.000
11 1991 5.000
12 1992 0.001
13 1993 1.000
14 1994 2.000
15 1995 4.000
16 1996 2.000
17 1997 2.000
18 1998 0.002
19 1999 3.000
20 2000 0.003

Upvotes: 1

Friede
Friede

Reputation: 7979

We can do

# DF$Value = 
with(DF, { i = Value==0; replace(Value, i, .001 * cumsum(i)[i]) })

giving

 [1] 2.000 5.000 2.000 1.000 1.000 5.000 2.000 4.000 3.000
[10] 5.000 5.000 0.001 1.000 2.000 4.000 2.000 2.000 0.002
[19] 3.000 0.003

Upvotes: 1

margusl
margusl

Reputation: 17544

set.seed(123)
DF <- data.frame(Year = 1981:2000, Value = sample(0:5, 20, replace = T))

idx <- which(DF$Value == 0)
DF$Value[idx] <- seq_along(idx) * .001
DF
#>    Year Value
#> 1  1981 2.000
#> 2  1982 5.000
#> 3  1983 2.000
#> 4  1984 1.000
#> 5  1985 1.000
#> 6  1986 5.000
#> 7  1987 2.000
#> 8  1988 4.000
#> 9  1989 3.000
#> 10 1990 5.000
#> 11 1991 5.000
#> 12 1992 0.001
#> 13 1993 1.000
#> 14 1994 2.000
#> 15 1995 4.000
#> 16 1996 2.000
#> 17 1997 2.000
#> 18 1998 0.002
#> 19 1999 3.000
#> 20 2000 0.003

Created on 2024-11-12 with reprex v2.1.1

Upvotes: 5

one
one

Reputation: 3902

in dplyr:

DF%>%
  mutate(Value=if_else(Value==0,cumsum(Value==0)*0.001,Value))

equivalently in base R:

DF$Value <- with(DF,ifelse(Value==0,cumsum(Value==0)*0.001,Value))

Upvotes: 5

Jilber Urbina
Jilber Urbina

Reputation: 61214

library(dplyr)
DF %>% 
  filter(Value != 0) %>% 
  rbind(.,
    DF %>% 
      filter(Value == 0) %>% 
      mutate(Value = cumsum(Value + .001))) %>% 
  arrange(Year)

   Year Value
1  1981 2.000
2  1982 5.000
3  1983 2.000
4  1984 1.000
5  1985 1.000
6  1986 5.000
7  1987 2.000
8  1988 4.000
9  1989 3.000
10 1990 5.000
11 1991 5.000
12 1992 0.001
13 1993 1.000
14 1994 2.000
15 1995 4.000
16 1996 2.000
17 1997 2.000
18 1998 0.002
19 1999 3.000
20 2000 0.003

Upvotes: 1

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