if statement - How to use ifelse inside the for loop in R -

i have check names of variables in data.frame , if match found, need replace na values in variable median, else others replace nas mean.

the data.frame cyl_spec has 11 variables , have replace na below:

1. viscosity: impute median
2. wax: impute median
3. others: impute mean

i can picking variables 1 @ time trying 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 code not doing , still getting same no of na in variable using command :

sum(is.na(cyl_spec$viscosity)  

also, when run code following warning message :

warning messages: 1: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 2: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 3: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 4: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 5: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 6: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 7: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 8: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 9: in mean.default(val, na.rm = t) :   argument not numeric or logical: returning na 

could please me finding solution this, stuck! in advance!!

you not need loop this. moreover, correct syntax test na values is.na(var), not is.na(var == true). finally, if want avoid typing name of dataframe, need use function (like with or dplyr functions). here, r looking object named viscosity found because name of column inside cyl_spec (same 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 need deal data.frame , 3 variables, recommend stick base-r solution. if, however, looking on data frame more variables , want automate it, dplyr::mutate_each. here example simulated data.

we create data.frame 7 variables , assign 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, apply each variable function infer na values either mean or median :

df <- df %>% ## variables recoded mean? here, n , m   mutate_each(funs(ifelse(is.na(.),mean(.,na.rm = true),.)),n,m) %>%  ## variables recoded 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