Replacing a value in df (all variables) with na [duplicate] - r

Question

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How do I replace NA values with zeros in an R dataframe?
(29 answers)
Conditional replacement of values in a data.frame
(5 answers)
Closed 3 years ago.
I am trying to change portions of my data frame in multiple variables from 8 and 9 to NA
Also, does anyone know a quick way to reverse code a vector? (likert scale where 1 is strongly agree, I want the most weight to be at 5)
Any help would be appreciated. Cheers.
naniar::replace_with_na_all(data = amer, condition = ~.x == -8)
data %>% mutate_all(.funs = function(x) replace(var, which(var == -9 | var == -8), NA))
df %>% mutate_each(funs(replace(., .>7, NA))
dep. evidently

Answer 1

Please see the comment to understand how to make your question reproducible for future posts. It's always a good idea to include sample data; if you can't share your data, provide code to generate representative mock data or use one of the built-in datasets.
As to your question, you can use mutate_all in the following way
library(dplyr)
data %>% mutate_all(~ifelse(.x %in% c(-8, -9), NA, .x))
Or you can use replace
data %>% mutate_all(~replace(.x, which(.x %in% c(-8, -9)), NA))
Reproducible example
Let's take mtcars as sample data. To replace all 3 and 4 entries across all columns with NA we can do
mtcars %>% mutate_all(~ifelse(.x %in% c(3, 4), NA, .x))
# mpg cyl disp hp drat wt qsec vs am gear carb
#1 21.0 6 160.0 110 3.90 2.620 16.46 0 1 NA NA
#2 21.0 6 160.0 110 3.90 2.875 17.02 0 1 NA NA
#3 22.8 NA 108.0 93 3.85 2.320 18.61 1 1 NA 1
#4 21.4 6 258.0 110 3.08 3.215 19.44 1 0 NA 1
#5 18.7 8 360.0 175 3.15 3.440 17.02 0 0 NA 2
#6 18.1 6 225.0 105 2.76 3.460 20.22 1 0 NA 1
#7 14.3 8 360.0 245 3.21 3.570 15.84 0 0 NA NA
#8 24.4 NA 146.7 62 3.69 3.190 20.00 1 0 NA 2
#9 22.8 NA 140.8 95 3.92 3.150 22.90 1 0 NA 2
#10 19.2 6 167.6 123 3.92 3.440 18.30 1 0 NA NA
#11 17.8 6 167.6 123 3.92 3.440 18.90 1 0 NA NA
#12 16.4 8 275.8 180 3.07 4.070 17.40 0 0 NA NA
#13 17.3 8 275.8 180 3.07 3.730 17.60 0 0 NA NA
#14 15.2 8 275.8 180 3.07 3.780 18.00 0 0 NA NA
#15 10.4 8 472.0 205 2.93 5.250 17.98 0 0 NA NA
#16 10.4 8 460.0 215 NA 5.424 17.82 0 0 NA NA
#17 14.7 8 440.0 230 3.23 5.345 17.42 0 0 NA NA
#18 32.4 NA 78.7 66 4.08 2.200 19.47 1 1 NA 1
#19 30.4 NA 75.7 52 4.93 1.615 18.52 1 1 NA 2
#20 33.9 NA 71.1 65 4.22 1.835 19.90 1 1 NA 1
#21 21.5 NA 120.1 97 3.70 2.465 20.01 1 0 NA 1
#22 15.5 8 318.0 150 2.76 3.520 16.87 0 0 NA 2
#23 15.2 8 304.0 150 3.15 3.435 17.30 0 0 NA 2
#24 13.3 8 350.0 245 3.73 3.840 15.41 0 0 NA NA
#25 19.2 8 400.0 175 3.08 3.845 17.05 0 0 NA 2
#26 27.3 NA 79.0 66 4.08 1.935 18.90 1 1 NA 1
#27 26.0 NA 120.3 91 4.43 2.140 16.70 0 1 5 2
#28 30.4 NA 95.1 113 3.77 1.513 16.90 1 1 5 2
#29 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 NA
#30 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
#31 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
#32 21.4 NA 121.0 109 4.11 2.780 18.60 1 1 NA 2
Using replace as
mtcars %>% mutate_all(~replace(.x, which(.x %in% c(3, 4)), NA))
gives the same result.