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Juan Luis
Asked: 2020-03-16 04:41:01 +0800 CST 2020-03-16 04:41:01 +0800 CST

How to convert given numeric values ​​to NA values ​​efficiently in R?

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I have 2 dataframes about 2 surveys (National Household and Adult Health Survey), in which many of the variables give "Don't know/No answer" as an answer option, assigning a number or a pair of numbers to said answer.

This response (NS/NC):

  • in most occasions/variables/question they are assigned the numbers 8 and 9 to indicate if "does not know" or "does not answer", respectively.
  • In a variable it corresponds only to 8, since it only gives the option to indicate "does not know". Similarly for 9.
  • In another variable they are 98 and 99
  • in others the 998 and 999
  • There are other variables that use these values ​​(8,9,998...) to indicate other options that do not correspond to "don't know" or "no answer".

So what I want is to convert all these NS/NC into NA values ​​to later treat them in a more comfortable way with the rest of the missing values. I am doing it as follows:

# Variables de la Encuesta Adultos con respuesta de NS/NC
vars_na_89_A <- c("enf_cron","SM_estres","tipo_act_fis","frec_act_fis","fuma")

vars_na_8_A <- c("sit_lab")
vars_na_9_A <- c("clase_pr")
vars_na_9899_A <- c("niv_est")
vars_na_998999_A <- c("altura","peso")

# Variables de la Encuesta Hogares con respuesta de NS/NC
vars_na_89_H <- c("ruido","malos","agua","limpieza", "cont_indus","cont_otras",
                  "escasez_verde","molest_animal" ,"delincuencia")
vars_na_9899_H <- c("n_dormitorios", "ingreso")
vars_na_998999_H <-c("m2")

ifelse(datos_adultos[vars_na_89_A]!=8 | datos_adultos[vars_na_89_A!=9],datos_adultos,NA)
ifelse(datos_adultos[vars_na_8_A]!=8,datos_adultos,NA)
.
.
.

In this way I have to create an ifelse for each type of NS/NC of each of the two surveys. My question is: How can I get the same result but without so many lines of code?

r

2 Answers

  1. It is difficult to give you a general answer, since I do not know what all the total columns that your df contains are, so I cannot give you a reproducible example with your data, I recommend the following code which only works if you know exactly which ones are the columns that contain NA based on the patterns you mention assuming that the columns: vars_na_89_A,vars_na_8_A,vars_na_9_A,vars_na_9899_A,vars_na_998999_A,vars_na_89_H,vars_na_9899_H

    keys 89, 8 , 9 ,9899,998999,,89,9899 do not have a different meaning, for example, for vars_na_8_A 9 it has a different meaning than "NA"

    library(map) #De esta libraría utilizaremos la función purrr que es muy similar apply
library(dplyr) 
library(stringr) # esta librería utilizaremos str_detect para identificar si la 
                 # columna tiene los patrones que deseamos

   df_modificar<-df %>% 
          select(vars_na_89_A,vars_na_8_A,vars_na_9_A,vars_na_9899_A,vars_na_998999_A,
              vars_na_89_H,vars_na_9899_H)
#No agrego la columna vars_na_m2_H por que no se si m2 quiere decir NA
#Para cambiarlo sobre todas las columnas y no realizarlo todo en una sola linea
#utilizo purrr

#Patrones que queremos modificar
 patrones<-paste(c(8,9,98,99,998,999),collapse = '|')
#Lo que vamos realizar es buscar los patrones sobre todas las columnas
 df_limpio<-map_df(df_modificar,function(x) ifelse(str_detect(x,pattern= 
                   patrones),"NA",x))
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  2. Best Answer

    This is a proof of concept of a possible way to reduce some code, although I don't think you gain much. The idea is to define a list of replacements, where you indicate the columns and values ​​that you want to replace with NA. Assuming one data.framelike this:

    set.seed(2020)
df <- as.data.frame(matrix(sample(c(1:9,98,99,998,999), 100, replace = TRUE), ncol=10))
df

    V1  V2  V3  V4  V5  V6 V7  V8  V9 V10
1  998   1   2   4 999 999  9 999  98  98
2  998 999 999   5 998   9  3   2   1   1
3    7   8   8   4   8 999  8   5   6 998
4    6   8   8 998 999  99  2   6   9   1
5    8  98   4  99 999   9  1   4 998   1
6    1   2   2   6   2   5  6   3   6   8
7    1   6 998   2 999  99  7   7   8 998
8    4 999   7  98   1  98  8   4   7  98
9   98   2   4   6   6   6  5   4  98   2
10   6   3   2  99   3   2  8   2  98 999

    And a replacement list, where we define different criteria:

    lista_reemplazo <- list(
  list(cols=c("V1", "V2"), na_vals=c(9,8)),
  list(cols=c("V3"), na_vals=c(998)),
  list(cols=c("V4", "V5", "V6"), na_vals=c(998, 999)),
  list(cols=c("V7", "V8"), na_vals=c(9)),
  list(cols=c("V9", "V10"), na_vals=c(8, 9, 998, 999))
)

    For example, the first criteria is that in V1and V2the values ​​8 and 9 are replaced by NA, the following is to iterate to process each criteria:

    for (reemplazo in lista_reemplazo) {
  for (col in reemplazo$cols) {
    df[df[, col] %in% reemplazo$na_vals, col] <- NA
  }
}

df

    V1  V2  V3 V4 V5 V6 V7  V8 V9 V10
1  998   1   2  4 NA NA NA 999 98  98
2  998 999 999  5 NA  9  3   2  1   1
3    7  NA   8  4  8 NA  8   5  6  NA
4    6  NA   8 NA NA 99  2   6 NA   1
5   NA  98   4 99 NA  9  1   4 NA   1
6    1   2   2  6  2  5  6   3  6  NA
7    1   6  NA  2 NA 99  7   7 NA  NA
8    4 999   7 98  1 98  8   4  7  98
9   98   2   4  6  6  6  5   4 98   2
10   6   3   2 99  3  2  8   2 98  NA
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