Replace "ND" (Non Detect) values in data frame to zero

tidyverse
1

I am attempting to write code to find all "ND" values in a large data frame named "Master" and reassign them as 0. I tried this:

library(tidyverse)
library(readr)
#VSS

Master1 <- Master %>%
  
  mutate(VSS = ifelse(str_detect(
         string=VSS,
         pattern="ND"
         ),
         parse_number(gsub(
           pattern = "ND",
           replacement ="", x = 0, parse_number(VSS)
  ))))

but I wind up with this error message:

# Error in ifelse(str_detect(string = VSS, pattern = "ND"), parse_number(gsub(pattern = "ND", : argument "no" is missing, with no default

If there is a better method to search all columns in the dataframe at once and replace them, that would be preferred.

Data Example: (Master)
TDS TSS VSS TKN DOC TOC DON Chla
10 2.2 3.5 ND 10 2.2 3.5 ND
14 2.5 3.75 0.01 14 2.5 3.75 0.01
ND N/A 5 0.2 ND N/A 5 0.2
10 2.2 3.5 ND 10 2.2 3.5 ND
14 2.5 3.75 0.01 14 2.5 3.75 0.01
ND N/A 5 0.2 ND N/A 5 0.2
14 2.5 3.75 0.01 14 2.5 3.75 0.01
ND N/A 5 0.2 ND N/A 5 0.2
10 2.2 3.5 ND 10 2.2 3.5 ND
14 2.5 3.75 0.01 14 2.5 3.75 0.01
ND N/A 5 0.2 ND N/A 5 0.2

*Please note that I want to maintain N/As, which I will change to blanks later

2

Is this what you wan to do?

library(dplyr)

master <- data.frame(
  stringsAsFactors = FALSE,
               TDS = c("10","14","ND","10","14",
                       "ND","14","ND","10","14","ND"),
               TSS = c("2.2","2.5","N/A","2.2",
                       "2.5","N/A","2.5","N/A","2.2","2.5","N/A"),
               VSS = c(3.5, 3.75, 5, 3.5, 3.75, 5, 3.75, 5, 3.5, 3.75, 5),
               TKN = c("ND","0.01","0.2","ND",
                       "0.01","0.2","0.01","0.2","ND","0.01","0.2"),
               DOC = c("10","14","ND","10","14",
                       "ND","14","ND","10","14","ND"),
               TOC = c("2.2","2.5","N/A","2.2",
                       "2.5","N/A","2.5","N/A","2.2","2.5","N/A"),
               DON = c(3.5, 3.75, 5, 3.5, 3.75, 5, 3.75, 5, 3.5, 3.75, 5),
              Chla = c("ND","0.01","0.2","ND",
                       "0.01","0.2","0.01","0.2","ND","0.01","0.2")
)

master %>% 
    mutate_if(is.character, ~ if_else(. == "ND", "0", .)) %>% 
    mutate_all(as.numeric) # Your variables are saved as text, so this extra step would be a good idea

#>    TDS TSS  VSS  TKN DOC TOC  DON Chla
#> 1   10 2.2 3.50 0.00  10 2.2 3.50 0.00
#> 2   14 2.5 3.75 0.01  14 2.5 3.75 0.01
#> 3    0  NA 5.00 0.20   0  NA 5.00 0.20
#> 4   10 2.2 3.50 0.00  10 2.2 3.50 0.00
#> 5   14 2.5 3.75 0.01  14 2.5 3.75 0.01
#> 6    0  NA 5.00 0.20   0  NA 5.00 0.20
#> 7   14 2.5 3.75 0.01  14 2.5 3.75 0.01
#> 8    0  NA 5.00 0.20   0  NA 5.00 0.20
#> 9   10 2.2 3.50 0.00  10 2.2 3.50 0.00
#> 10  14 2.5 3.75 0.01  14 2.5 3.75 0.01
#> 11   0  NA 5.00 0.20   0  NA 5.00 0.20

Created on 2020-03-19 by the reprex package (v0.3.0.9001)

Have in mind that NA (Not Available) is the way R handles the "blanks", if you replace that for let's say this " ", you are going to get a character variable and you are not going to be able to use it for numerical calculations.

3

Yes, thank you! I appreciate the heads up on the NAs too, I will leave them as is

4

I have actually run into a problem here. This code only works if I change everything to numeric (mutate_all(as.numeric)), although I have to run the code through a filter that eliminates inequalities by replacing them with half of their value. If I filter the code first, NAs are created by coercion and NDs are lost. If I run this code first, the inequalities are lost when I change everything to numeric. I believe you helped me with my filter code as well, but I will include it for reference:

library(tidyverse)
library(readr)
#Pheophytin

Master2 <- Master %>%

mutate(Pheophytin = ifelse(str_detect(
string=Pheophytin,
pattern="<"
),
parse_number(gsub(
pattern = "<",
replacement ="", x = Pheophytin
))/2, parse_number(Pheophytin)
))

5

Never mind, I fixed it!

6

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