[R] searching and replacing in a data frame.

[Joshua Wiley]

Hi,

A slight modification corrects this:

ttt <- data.frame(A = c(Inf, 0, 0), B = c(1, 2, 3))
apply(ttt, 2, function(x) {x[is.infinite(x)] <- 0; x})

and please do use spaces in your code.  It is much more legible.  Most
notably spaces are after commas.

Cheers,

Josh

On Sun, Jul 17, 2011 at 10:01 PM, Ashim Kapoor <ashimkapoor at gmail.com> wrote:
> Dear David,
>
> Many thanks for your reply.
>
> On Fri, Jul 15, 2011 at 5:24 PM, David Winsemius <dwinsemius at comcast.net>wrote:
>
>>
>> On Jul 15, 2011, at 5:20 AM, Ashim Kapoor wrote:
>>
>>  Dear R helpers,
>>>
>>>
>>> Please have a look at the following : -
>>>
>>> Note : My goal is to find and replace all Inf's in a data array with 0.
>>>
>>>  t<-data.frame(A=c(Inf,0,0),B=**c(1,2,3))
>>>> t
>>>>
>>>   A B
>>> 1 Inf 1
>>> 2   0 2
>>> 3   0 3
>>>
>>>  str(t)
>>>>
>>> 'data.frame':    3 obs. of  2 variables:
>>> $ A: num  Inf 0 0
>>> $ B: num  1 2 3
>>>
>>>> t[which(t==Inf,arr.ind=T)]
>>>>
>>> [1] Inf
>>>
>>
>> Several problems here.
>> `t` is a perfectly good function name so using it as an object name is
>> confusing.
>>
>>
>>
>>  t[which(t==Inf,arr.ind=T)]<-0
>>>>
>>> Error in `[<-.data.frame`(`*tmp*`, which(t == Inf, arr.ind = T), value =
>>> 0)
>>> :
>>>  only logical matrix subscripts are allowed in replacement
>>>
>>> Query : Why does the search work but the replace not work ?
>>>
>>
>> Because you gave a numeric matrix as an argument to "data.frame.[<-" and it
>> wanted a different mode. I think it would have worked if `t` were a matrix.
>>
>>
>>
>>
>>> Many thanks for your time and efforts.
>>>
>>
>> Two methods that would accomplish the task:
>>
>> ttt<-data.frame(A=c(Inf,0,0),**B=c(1,2,3))
>>
>> ttt[is.infinite(as.matrix(ttt)**)] <- 0
>>
>> Or:
>>
>> apply(ttt, 1:2, function(x) x[is.infinite(x)] <- 0 )
>>
>>>
>>>
>> ttt
>    A B
> 1 Inf 1
> 2   0 2
> 3   0 3
>> apply(ttt,c(1,2),function(x) x[is.infinite(x)]<-0)
>     A B
> [1,] 0 0
> [2,] 0 0
> [3,] 0 0
>> apply(ttt,c(1,2),function(x) x[is.infinite(x)]<-0)
>     A B
> [1,] 0 0
> [2,] 0 0
> [3,] 0 0
>
> Why do I have ALL zeroes ?

well x is altered, but x is not what gets returned from apply.  My
alteration just makes it so x gets returned rather than the result of
the subassignment.

>
> The other method DOES work.
>
> Thank you.
> Ashim
>
>        [[alternative HTML version deleted]]
>
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>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
https://joshuawiley.com/