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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?

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    $\begingroup$ If[AnyTrue[a, # > 1 &], 1, 0] or If[Or @@ Thread[a > 1], 1, 0] is your after? $\endgroup$ – yode Apr 18 '17 at 2:10
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    $\begingroup$ Count[a, _?(# > 1 &)] // Unitize $\endgroup$ – Bob Hanlon Apr 18 '17 at 6:24
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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]
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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.

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  • $\begingroup$ Is there a way to apply this on two lists, not by checking the whole lists separate, but by checking if the first, second,... last element of both lists is greater than 1? For example: 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. $\endgroup$ – drcyberz Apr 18 '17 at 11:27
  • $\begingroup$ Can you help me regarding the post above? I want to check for two different conditions in two long lists. If[a[[1]] > 1 && b[[1]] == 1 || a[[2]] > 1 && b[[2]] == 1 || a[[3]] > 1 && b[[3]] == 1, 1, 0] $\endgroup$ – drcyberz Apr 19 '17 at 21:45

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