On 07/18/2011 05:32 PM, Daniel Roe wrote:
> Hi,
>
> Assuming your mdcrd file has no box coords and is not corrupted you
> can calculate the number of lines that each frame of your trajectory
> should have based on this formula:
>
> #LinesPerFrame = ((#atoms * 3) / 10)
>
> If there is any fraction after the division, round up. If your
> trajectory has box coordinates add 1. From this, you can calculate the
> number of frames from the total number of lines in your trajectory:
>
> #Frames = (#LinesInTrajectory - 1) / #LinesPerFrame
>
> The "-1" in the first term is because every trajectory has a title
> line. Hope this helps,
Dan,
thanks for your answer. First of all, I should clarify that I am looking
for the most efficient programmatic approach to get this number directly
from the mdcrd file (independent from its format).
Your solution won't work for binary file formats, right?
ptraj seems to be the right tool to get the number I am after, but I do
not know how to do it best... consider the following ptraj input:
trajin md_heatup.mdcrd
trajout f netcdf
It runs and prints to stdout:
PTRAJ: trajin md_heatup.mdcrd
Checking coordinates: md_heatup.mdcrd
NETCDF file:
PTRAJ: trajout f netcdf
md_heatup.mdcrd: 100 frames.
There it is: 100.
I could parse the output, but in order to make this efficient, I would
need some non-time-consuming ptraj action to be run, because this number
is only printed if there is a `trajout` statement -- usually connected
to some time-consuming I/O operation. Is there a ptraj command executing
very quickly even for big files? Going this way would be a (dirty) option...
There should be a better way. I noticed this number being printed by
ptraj very quickly even for input mdcrd files of several gigabytes. So,
is this number stored in the mdcrd file header (of at least NetCDF files)?
Thanks!
>
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Received on Mon Jul 18 2011 - 09:00:05 PDT
