I have a list and a If-statement:
a = {-3, 0, 5};
If[a[[1]] > 1 || a[[2]] > 1 || a[[3]] > 1 , 1, 0]
Output: 1
Is there a way to write the condition shorter? If the list is longer, the condition will get very long.
Or probably more general: Is there a way to create a function which would take the elements of a list and connect them with an operation?
If you have a very long list and want to check if any of the elements is larger than one, the fastest way is probably a vectorized operation such as
If[Times @@ UnitStep[1 - a] == 0, 1, 0]
For the particular question:
If[Or @@ (# > 1& /@ a), 1, 0]
but there is a problem you should be aware of: Or returns immediately after a True clause, so if your list is $10^{10}$ long, the version I wrote will take time proportional to $10^{10}$ even if the first comparison returns True.
For general functions, the answer is similar.
a = {2, 0, 2}; b = {0, 2, 2}; should output 1 because the last element of both lists is greater than 1. But a = {2, 0, 2}; b = {0, 2, 0}; should output 0, because there are no element pairs which are greater than 1. In the end, I want to check for different conditions in the two lists. Like: If a[[1]] > 1 and b[[1]] == 1 OR a[[2]] > 1 and b[[2]] == 1 OR ... and so on.
$\endgroup$
If[a[[1]] > 1 && b[[1]] == 1 || a[[2]] > 1 && b[[2]] == 1 || a[[3]] > 1 && b[[3]] == 1, 1, 0]
$\endgroup$
If[AnyTrue[a, # > 1 &], 1, 0]orIf[Or @@ Thread[a > 1], 1, 0]is your after? $\endgroup$Count[a, _?(# > 1 &)] // Unitize$\endgroup$