[R] replacing all NA's in a dataframe with zeros...

[Peter Dalgaard]

Gavin Simpson wrote:
> On Wed, 2007-03-14 at 20:16 -0700, Steven McKinney wrote:
>   
>> Since you can index a matrix or dataframe with
>> a matrix of logicals, you can use is.na()
>> to index all the NA locations and replace them
>> all with 0 in one command.
>>
>>     
>
> A quicker solution, that, IIRC,  was posted to the list by Peter
> Dalgaard several years ago is:
>
> sapply(mydata.df, function(x) {x[is.na(x)] <- 0; x}))
>   
I hope your memory fails you, because it doesn't actually work.....

> sapply(test.df, function(x) {x[is.na(x)] <- 0; x})
     x1 x2 x3
[1,]  0  1  1
[2,]  2  2  0
[3,]  3  3  0
[4,]  0  4  4

is a matrix, not a data frame.

Instead:

> test.df[] <- lapply(test.df, function(x) {x[is.na(x)] <- 0; x})
> test.df
  x1 x2 x3
1  0  1  1
2  2  2  0
3  3  3  0
4  0  4  4

Speedwise, sapply() is doing lapply() internally, and the assignment
overhead should be small, so I'd expect similar timings.

-- 
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