I have to check the names of all variables in a data.frame and if match found, need to replace the NA values in that variable with Median, else for others replace NAs with mean.
The data.frame cyl_spec has 11 variables and I have to replace NA as below:
I can certainly do it by picking the variables one at a time but I was trying the following code :
attach(cyl_spec)
var <- colnames(cyl_spec)
for(val in var)
{
if(val == 'viscosity'){viscosity[is.na(viscosity == T)] <- median(viscosity, na.rm = T)}
else if(val == 'wax'){wax[is.na(wax == T)] <- median(wax, na.rm = T)}
else {val[is.na(val == T)] <- mean(val, na.rm = T)}
}
detach(cyl_spec)
Somehow the code is not doing anything and I am still getting the same no of NA in the variable using this command :
sum(is.na(cyl_spec$viscosity)
Also, when I run this code I get the following warning message :
Warning messages:
1: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
2: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
3: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
4: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
5: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
6: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
7: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
8: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
9: In mean.default(val, na.rm = T) :
argument is not numeric or logical: returning NA
Could someone please help me with finding the solution for this, am stuck! Thanks in advance!!
You do not need a loop to do this. Moreover, the correct syntax to test for na values is is.na(var)
, not is.na(var == TRUE)
. Finally, if you want to avoid typing the name of your dataframe, you would need to use some function that does it (like with
or the dplyr
functions). Here, R is looking for an object named viscosity
that is nowhere to be found because it is the name of a column inside cyl_spec
(same for the other variable names).
cyl_spec$viscosity[is.na(cyl_spec$viscosity)] <- median(cyl_spec$viscosity, na.rm = T)
cyl_spec$wax[is.na(cyl_spec$wax)] <- median(cyl_spec$wax, na.rm = T)
cyl_spec$val[is.na(cyl_spec$val)] <- mean(cyl_spec$val, na.rm = T)
If all you need is to deal with this data.frame and only those three variables, I strongly recommend you stick to this base-r solution. If, however, you are looking to do this on a data frame with more variables and you want to automate it, you could look into the dplyr::mutate_each
. Here is an example with simulated data.
We create a data.frame with 7 variables and assign some NA values.
library(dplyr)
set.seed(10)
df <- data.frame(n=runif(100),
m=runif(100),
d=runif(100),
o=runif(100),
e=runif(100),
f=runif(100),
g=runif(100))
df <- mutate_each(df,funs(ifelse(.>.8,NA,.)))
head(df)
n m d o e f g
1 0.50747820 0.34434350 0.2230884 0.347860110 NA NA NA
2 0.30676851 0.06132255 0.5358950 0.007992606 0.6855115 NA 0.7478783
3 0.42690767 0.36897981 0.6625291 0.401344915 0.6296311 NA 0.7225419
4 0.69310208 0.40759356 NA 0.588350693 0.7508252 0.29063776 0.5457709
5 0.08513597 NA 0.1491831 NA NA 0.07203601 0.2641231
6 0.22543662 NA 0.6700994 0.708542599 0.3600703 0.55888842 0.3057243
Now, we apply to each variable a function to infer NA values from either mean or median :
df <- df %>%
## Which variables are to be recoded with mean? here, n and m
mutate_each(funs(ifelse(is.na(.),mean(.,na.rm = TRUE),.)),n,m) %>%
## Which variables are to be recoded with median? here, d,o,e,f,g
mutate_each(funs(ifelse(is.na(.),median(.,na.rm = TRUE),.)),d,o,e,f,g)
head(df)
n m d o e f g
1 0.50747820 0.34434350 0.2230884 0.347860110 0.3602354 0.39956699 0.4499041
2 0.30676851 0.06132255 0.5358950 0.007992606 0.6855115 0.39956699 0.7478783
3 0.42690767 0.36897981 0.6625291 0.401344915 0.6296311 0.39956699 0.7225419
4 0.69310208 0.40759356 0.4407363 0.588350693 0.7508252 0.29063776 0.5457709
5 0.08513597 0.40892568 0.1491831 0.378731867 0.3602354 0.07203601 0.2641231
6 0.22543662 0.40892568 0.6700994 0.708542599 0.3600703 0.55888842 0.3057243