Thank you.
I understand what do you mean. Here is my thought:
My original method to deal with a trajectory is:
1.trajin the trajectory --> 2.strip the uncessary part --> 3.matrix mwcovar name bbprotein.matrix out bbprotein.matrix (get the correlation matrix)--> 4. analyze matrix bbprotein.matrix out bbprotein.mode vecs 400
The best way is to average the correlation matrix generated after the 3rd step and use the average matrix as input of 4th step. However, it seems "analyze matrix" instruction can not accept external matrix (I also asked this in the maillist and no reply). So I can only average the trajectory and put the average trajectory as input of the 1st step. Because calculating correlation is linear process, matrix obtained by average trajectory is the same.
Is my process correct? I feel really puzzled about that.
cyk5056
2009-11-18
发件人： Carlos Simmerling
发送时间： 2009-11-18 20:10:59
收件人： AMBER Mailing List
抄送：
主题： Re: [AMBER] Can I get an average trajectory?
i'm not sure how you would do this, but I don't think it's really what you
want. averaging over trajectories that are different will probably give you
unphysical structures. what you normally want to do is average the analysis
data, not the structures.
if you want to "eliminate stochastic properties", why not average over a
period of time in your trajectory and see if it looks like what you want?
this will help decide if it's worth the effort to average over multiple
trajectories.
On Tue, Nov 17, 2009 at 11:46 PM, cyk5056 <cyk5056.163.com> wrote:
> Hi Amber users,
> I am doing MD and have obtained several binary trajectory files. To
> eliminate the stochastic properties, I want to average these trajectory(all
> 0-500ps) into one 0-500ps trajectory. How can I do that? I think maybe it is
> not easy to average them in binary form. Maybe I can use Matlab to average
> them if they are in ascii form. So how to convert them into ascii form?
> Either of these two approaches will work, I think.
> BTW: I have tried "average" instruction and it will return a average
> structure, not a trajectory
>
> Thank you very much!
>
>
> Best,
>
>
>
>
>
> cyk5056
> 2009-11-18
>
> _______________________________________________
> AMBER mailing list
> AMBER.ambermd.org
> http://lists.ambermd.org/mailman/listinfo/amber
>
>
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Received on Wed Nov 18 2009 - 08:30:03 PST
