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I'm working now with xAct in Mathematica obtaining huge list (around 50 000 terms) of contractions of a metric tensor with vector fields. I know that some elements of these list are multiplied for certain factor, let us call this for simplicity $f[x]$, and those elements should be deleted for consistency. So for example I will have

 List1={f[x]y,f[x]z,z^2,y,y^2,f[x]^2};

So a tried the simplest solution: divide the list by $f[x]$ and check the Denominator

 Redc=List1/(f[x]);
 Numc=Denominator[Redc];

Thus, if the Denominator is different from 1, I will keep that entry and delete the rest (from the original list). Unfortunately I could't find the solution. Perhaps It will be quite easy, but I'm not really expert on Mathematica. I appreciate any help. Thanks

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  • $\begingroup$ what about DeleteCases? As in {f[x] y, f[x] z, z^2, y} // DeleteCases[f[x] _]. This won't delete the f[x]^2 instance, but it is easily adapted for those cases if that's what you needed $\endgroup$ – glS Nov 2 '16 at 20:14
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I actually think that Position[] works best here, because it is more versatile than Cases[], and will also delete powers. If this still misses some instances of the function, such as Log[f[x]] you may need to look at FreeQ, this assumes that this is not the case.

list={y fx[x], z fx[x], z^2, y, y^2, fx[x]^2};
Delete[list, Partition[Position[list, fx[x]][[All, 1]], 1]]

{z^2, y, y^2}

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  • $\begingroup$ Jesus!!..thank you so much @Feyre the solution works perfectly! :) $\endgroup$ – Alejandro Guarnizo Nov 2 '16 at 23:42
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Cases[ List1, _?(FreeQ[f])]

or

DeleteCases[ List1,_?(!FreeQ[f][#]&)]

{z^2, y, y^2}

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