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I have a list of numerical functions (interpolation functions) that I'd like to get the values of each one at different points. For example: the functions are {y1[t],y2[t]} and the points are {t1, t2}; I'd like to get the result {y1[t1], y2[t2]}.

I used Diagonal to get the answer:

Diagonal[{y1[t],y2[t]}/. t-> {t1,t2}]

I am wondering if there is a better way to avoid unnecessary replacements.

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MapThread does the trick:

In[1]:= MapThread[#[#2] &, {{y1, y2}, {t1, t2}}]
Out[1]= {y1[t1], y2[t2]}
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    $\begingroup$ If you don't mind using deprecated/undocumented functions, you could do MapThread[Compose, {{y1, y2}, {t1, t2}}] instead. $\endgroup$ – Carl Woll Aug 28 '17 at 16:10
  • $\begingroup$ And {y1,y2} can be obtained as {y1[t], y2[t]}[[All, 0]] using the OP's input. Or Head /@ {y1[t], y2[t]}. $\endgroup$ – Szabolcs Aug 28 '17 at 16:28
  • $\begingroup$ Thank you. Both methods work great. I ended up using Mapthread with Compose. $\endgroup$ – Amir Jalali Aug 28 '17 at 16:31
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MapThread is probably your best shot, and in particular this version most closely mirrors your original approach, which did not require extracting the heads or pure-function versions of the $y_i$:

MapThread[#1 /. t -> #2 &, {{y1[t], y2[t]}, {t1, t2}}]

You could do this directly using Thread:

Thread[f[{y1[t], y2[t]}, {t1, t2}]] /. f -> (#1 /. t -> #2 &)

By defining an auxially variable $t_i=\{t_1,t_2\}$, you could also do this using MapIndexed:

ti = {t1, t2};
MapIndexed[#1 /. t -> ti[[First@#2]] &, {y1[t], y2[t]}]
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  • $\begingroup$ Thank you. very nice application of MapIndexed. $\endgroup$ – Amir Jalali Aug 28 '17 at 17:15

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