converting all columns (factor to numeric) without affecting rownames/colnames

Question

For my sample dataset, I use the following code to convert the data from factor to numeric:

sample = as.data.frame(lapply(sample, function(x) as.numeric(as.character(x))))

in order to then replace all NA values with 0s using this code:

sample[is.na(sample)] = 0

however, when I convert from factor to numeric, the column names change and the rownames disappear. why does this happen and how can I prevent it from happening when converting all columns to numeric?

dput(sample)
structure(list(`2015-10-08 00:05:00` = structure(c(NA, NA, NA, 
NA, 2L, NA), .Names = c("72", "79", "82", "83", "116", "120"), .Label = c(" 1", 
" 2", " 3", " 5", "2015-10-08 00:05:00"), class = "factor"), 
    `2015-10-08 00:12:00` = structure(c(NA, 1L, NA, NA, NA, NA
    ), .Names = c("72", "79", "82", "83", "116", "120"), .Label = c(" 1", 
    " 2", " 3", "2015-10-08 00:12:00"), class = "factor"), `2015-10-08 00:34:00` = structure(c(NA, 
    NA, NA, NA, 1L, NA), .Names = c("72", "79", "82", "83", "116", 
    "120"), .Label = c(" 1", " 2", " 3", " 4", "2015-10-08 00:34:00"
    ), class = "factor"), `2015-10-08 00:40:00` = structure(c(NA_integer_, 
    NA_integer_, NA_integer_, NA_integer_, NA_integer_, NA_integer_
    ), .Names = c("72", "79", "82", "83", "116", "120"), .Label = c(" 1", 
    " 2", "2015-10-08 00:40:00"), class = "factor"), `2015-10-08 01:32:00` = structure(c(NA, 
    NA, 1L, 1L, 3L, NA), .Names = c("72", "79", "82", "83", "116", 
    "120"), .Label = c(" 1", " 2", " 3", " 4", " 6", " 8", "2015-10-08 01:32:00"
    ), class = "factor"), `2015-10-08 01:52:00` = structure(c(1L, 
    NA, NA, NA, NA, NA), .Names = c("72", "79", "82", "83", "116", 
    "120"), .Label = c(" 1", " 2", " 3", "2015-10-08 01:52:00"
    ), class = "factor")), .Names = c("2015-10-08 00:05:00", 
"2015-10-08 00:12:00", "2015-10-08 00:34:00", "2015-10-08 00:40:00", 
"2015-10-08 01:32:00", "2015-10-08 01:52:00"), row.names = c("72", 
"79", "82", "83", "116", "120"), class = "data.frame")

Answer 1

You can use data.frame instead of as.data.frame . check.names=F tell the function to keep the column names. Use row.names to inherit the row names.

BTW, try not to use sample as a variable name in R as it is a reserved word of R.

d1 = data.frame(lapply(d1, function(x) as.numeric(as.character(x))),
                   check.names=F, row.names = rownames(d1))
d1[is.na(d1)] = 0