Message-ID: <49CD245D.9020209@statistik.tu-dortmund.de>
Date: 2009-03-27T19:09:17Z
From: Uwe Ligges
Subject: how to get all iterations if I meet NaN?
In-Reply-To: <54d792b30903271036s51159ad8y5d12fbfef4b0ad53@mail.gmail.com>
huiming song wrote:
> hi, everybody, please help me with this question:
>
> If I want to do iteration for 1000 times, however, for the 500th iteration,
> there is NaN appears. Then the iteration will stop. If I don't want the stop
> and want the all the 1000 iterations be done. What shall I do?
>
>
> suppose I have x[1:1000] and z[1:1000],I want to do some calculation for all
> x[1] to x[1000].
>
> z=rep(0,1000)
> for (i in 1:1000){
> z[i]=sin(1/x[i])
> }
>
> if x[900] is 0, in the above code it will not stop when NaN appears. Suppose
> when sin(1/x[900]) is NaN appears and the iteration will now fulfill the
> rest 100 iterations. How can I write a code to let all the 1000 iterations
> be done?
See ?try
Uwe Ligges
> Thanks!
>
> [[alternative HTML version deleted]]
>
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