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How to check is there any negative term in a large list?


Issue with very large lists in MathematicaHow to check if all the members of list lies in specific rangeQuery Dataset to check if row contains any from a set of valuesDeleting any list that contains a negative numberDefining a function that detects square matricesHow to use Contains functions on matrices?Ordering real numeric quantitiescount the pairs in a set of DataSpeed up Flatten[] of a large nested listHow to check expression depends on symbol in a particular way













7












$begingroup$


I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.






list-manipulation expression-test







share|improve this question









New contributor




a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is slightly pensive♦
    15 hours ago






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    15 hours ago










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    14 hours ago















7












$begingroup$


I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.






list-manipulation expression-test







share|improve this question









New contributor




a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$







  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is slightly pensive♦
    15 hours ago






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    15 hours ago










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    14 hours ago













7












7








7


2



$begingroup$


I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.






list-manipulation expression-test







share|improve this question









New contributor




a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I want to check if a data set of size $10^10$ contains any non-positive elements. Positive[Name of dataset] returns a list of True and False of length $10^10$. I want only a single True if all terms of that dataset are positive and False otherwise.






list-manipulation expression-test




list-manipulation expression-test






share|improve this question









New contributor




a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 10 hours ago









mjw

1,05110




1,05110






New contributor




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asked 15 hours ago









a ba b

361




361




New contributor




a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






a b is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is slightly pensive♦
    15 hours ago






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    15 hours ago










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    14 hours ago












  • 3




    $begingroup$
    VectorQ[list, Positive]?
    $endgroup$
    – J. M. is slightly pensive♦
    15 hours ago






  • 1




    $begingroup$
    Use Apply as in And @@ Positive[list]
    $endgroup$
    – Bob Hanlon
    15 hours ago










  • $begingroup$
    How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
    $endgroup$
    – Roman
    14 hours ago







3




3




$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is slightly pensive♦
15 hours ago




$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is slightly pensive♦
15 hours ago




1




1




$begingroup$
Use Apply as in And @@ Positive[list]
$endgroup$
– Bob Hanlon
15 hours ago




$begingroup$
Use Apply as in And @@ Positive[list]
$endgroup$
– Bob Hanlon
15 hours ago












$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
$endgroup$
– Roman
14 hours ago




$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (use Positive), or do you need all terms to be zero or positive (use NonNegative)?
$endgroup$
– Roman
14 hours ago










4 Answers
4






active

oldest

votes


















10












$begingroup$

Alternate solution:



list = RandomReal[1, 10^6];
Min[list] >= 0





share|improve this answer









$endgroup$








  • 2




    $begingroup$
    ...i.e. NonNegative[Min[list]].
    $endgroup$
    – J. M. is slightly pensive♦
    13 hours ago






  • 2




    $begingroup$
    Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
    $endgroup$
    – Henrik Schumacher
    11 hours ago






  • 1




    $begingroup$
    This is about 100 times faster than any of the other solutions. Impressive!
    $endgroup$
    – Roman
    9 hours ago


















7












$begingroup$

Since you have a very large list, you should look at the timing



list = RandomReal[1, 10^6];

(And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

(* 0.050573, True *)

VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

(* 0.261642, True *)

(AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

(* 0.324062, True *)

And @@ (list /. x_?Negative -> False, 
 x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

(* 1.00664, True *)


EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



list2 = ReplacePart[list, 1000 -> -1];

(And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

(* 0.277642, False *)

VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

(* 0.000223, False *)

(AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

(* 0.000262, False *)

And @@ (list2 /. x_?Negative -> False, 
 x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

(* 1.43026, False *)





share|improve this answer











$endgroup$












  • $begingroup$
    Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
    $endgroup$
    – mjw
    14 hours ago






  • 1




    $begingroup$
    Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
    $endgroup$
    – mjw
    14 hours ago






  • 1




    $begingroup$
    @mjw, And[] does short-circuit evaluation.
    $endgroup$
    – J. M. is slightly pensive♦
    14 hours ago






  • 1




    $begingroup$
    @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
    $endgroup$
    – mjw
    13 hours ago






  • 1




    $begingroup$
    @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
    $endgroup$
    – J. M. is slightly pensive♦
    13 hours ago


















3












$begingroup$

Ah, maybe this is too simple, but works for exactly what you're doing:



data = Table[RandomReal[-1,1],i,1,1000];
AnyTrue[data,Negative] // Not
(*False*)

data2 = Table[RandomReal[], i, 1, 10^2];
AnyTrue[data2, Negative] // Not
(*True*)





share|improve this answer











$endgroup$












  • $begingroup$
    AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
    $endgroup$
    – Roman
    15 hours ago










  • $begingroup$
    @Roman - the poster is using AnyTrue not AllTrue
    $endgroup$
    – Bob Hanlon
    14 hours ago






  • 1




    $begingroup$
    Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
    $endgroup$
    – Roman
    14 hours ago











  • $begingroup$
    ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
    $endgroup$
    – morbo
    14 hours ago


















2












$begingroup$

list = 1, 2, 3, 4, -5, -6, -7;

list /. x_?Negative -> True, x_?Positive -> False





share|improve this answer









$endgroup$












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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    10












    $begingroup$

    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0





    share|improve this answer









    $endgroup$








    • 2




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago






    • 2




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      11 hours ago






    • 1




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      9 hours ago















    10












    $begingroup$

    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0





    share|improve this answer









    $endgroup$








    • 2




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago






    • 2




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      11 hours ago






    • 1




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      9 hours ago













    10












    10








    10





    $begingroup$

    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0





    share|improve this answer









    $endgroup$



    Alternate solution:



    list = RandomReal[1, 10^6];
    Min[list] >= 0






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 13 hours ago









    sakrasakra

    2,6781429




    2,6781429







    • 2




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago






    • 2




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      11 hours ago






    • 1




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      9 hours ago












    • 2




      $begingroup$
      ...i.e. NonNegative[Min[list]].
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago






    • 2




      $begingroup$
      Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
      $endgroup$
      – Henrik Schumacher
      11 hours ago






    • 1




      $begingroup$
      This is about 100 times faster than any of the other solutions. Impressive!
      $endgroup$
      – Roman
      9 hours ago







    2




    2




    $begingroup$
    ...i.e. NonNegative[Min[list]].
    $endgroup$
    – J. M. is slightly pensive♦
    13 hours ago




    $begingroup$
    ...i.e. NonNegative[Min[list]].
    $endgroup$
    – J. M. is slightly pensive♦
    13 hours ago




    2




    2




    $begingroup$
    Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
    $endgroup$
    – Henrik Schumacher
    11 hours ago




    $begingroup$
    Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
    $endgroup$
    – Henrik Schumacher
    11 hours ago




    1




    1




    $begingroup$
    This is about 100 times faster than any of the other solutions. Impressive!
    $endgroup$
    – Roman
    9 hours ago




    $begingroup$
    This is about 100 times faster than any of the other solutions. Impressive!
    $endgroup$
    – Roman
    9 hours ago











    7












    $begingroup$

    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)





    share|improve this answer











    $endgroup$












    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is slightly pensive♦
      14 hours ago






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      13 hours ago






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago















    7












    $begingroup$

    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)





    share|improve this answer











    $endgroup$












    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is slightly pensive♦
      14 hours ago






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      13 hours ago






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago













    7












    7








    7





    $begingroup$

    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)





    share|improve this answer











    $endgroup$



    Since you have a very large list, you should look at the timing



    list = RandomReal[1, 10^6];

    (And @@ Positive[list]) // AbsoluteTiming (* Hanlon *)

    (* 0.050573, True *)

    VectorQ[list, Positive] // AbsoluteTiming (* J.M. *)

    (* 0.261642, True *)

    (AnyTrue[list, Negative] // Not) // AbsoluteTiming (* Morbo *)

    (* 0.324062, True *)

    And @@ (list /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (* Alrubaie *)

    (* 1.00664, True *)


    EDIT: As suggested by mjw, encountering a nonpositive value early in the list significantly alters the results.



    list2 = ReplacePart[list, 1000 -> -1];

    (And @@ Positive[list2]) // AbsoluteTiming (*Hanlon*)

    (* 0.277642, False *)

    VectorQ[list2, Positive] // AbsoluteTiming (*J.M.*)

    (* 0.000223, False *)

    (AnyTrue[list2, Negative] // Not) // AbsoluteTiming (*Morbo*)

    (* 0.000262, False *)

    And @@ (list2 /. x_?Negative -> False, 
     x_?Positive -> True) // AbsoluteTiming (*Alrubaie*)

    (* 1.43026, False *)






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 14 hours ago

























    answered 14 hours ago









    Bob HanlonBob Hanlon

    61.1k33598




    61.1k33598











    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is slightly pensive♦
      14 hours ago






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      13 hours ago






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago
















    • $begingroup$
      Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
      $endgroup$
      – mjw
      14 hours ago






    • 1




      $begingroup$
      @mjw, And[] does short-circuit evaluation.
      $endgroup$
      – J. M. is slightly pensive♦
      14 hours ago






    • 1




      $begingroup$
      @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
      $endgroup$
      – mjw
      13 hours ago






    • 1




      $begingroup$
      @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
      $endgroup$
      – J. M. is slightly pensive♦
      13 hours ago















    $begingroup$
    Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
    $endgroup$
    – mjw
    14 hours ago




    $begingroup$
    Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
    $endgroup$
    – mjw
    14 hours ago




    1




    1




    $begingroup$
    Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
    $endgroup$
    – mjw
    14 hours ago




    $begingroup$
    Also, and I guess this depends on the probability of any entry being negative, it may make sense for the algorithm to stop as soon as it finds a negative (or non-positive) element in the list.
    $endgroup$
    – mjw
    14 hours ago




    1




    1




    $begingroup$
    @mjw, And[] does short-circuit evaluation.
    $endgroup$
    – J. M. is slightly pensive♦
    14 hours ago




    $begingroup$
    @mjw, And[] does short-circuit evaluation.
    $endgroup$
    – J. M. is slightly pensive♦
    14 hours ago




    1




    1




    $begingroup$
    @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
    $endgroup$
    – mjw
    13 hours ago




    $begingroup$
    @Bob, Thank you for your edit. Why, though, does it take longer for your method to work when there is a negative entry? I would have thought that in each case, it would take the same amount of time to go through the whole list.
    $endgroup$
    – mjw
    13 hours ago




    1




    1




    $begingroup$
    @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
    $endgroup$
    – J. M. is slightly pensive♦
    13 hours ago




    $begingroup$
    @Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation like And[True, True, False, True, True, ... True] will finish at once (and similar remarks apply for Or[]). Perhaps one can judiciously use Catch[]/Throw[]if an early-return test for long lists is desired.
    $endgroup$
    – J. M. is slightly pensive♦
    13 hours ago











    3












    $begingroup$

    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)





    share|improve this answer











    $endgroup$












    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      15 hours ago










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      14 hours ago






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      14 hours ago











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      14 hours ago















    3












    $begingroup$

    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)





    share|improve this answer











    $endgroup$












    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      15 hours ago










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      14 hours ago






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      14 hours ago











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      14 hours ago













    3












    3








    3





    $begingroup$

    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)





    share|improve this answer











    $endgroup$



    Ah, maybe this is too simple, but works for exactly what you're doing:



    data = Table[RandomReal[-1,1],i,1,1000];
    AnyTrue[data,Negative] // Not
    (*False*)

    data2 = Table[RandomReal[], i, 1, 10^2];
    AnyTrue[data2, Negative] // Not
    (*True*)






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 14 hours ago

























    answered 15 hours ago









    morbomorbo

    46418




    46418











    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      15 hours ago










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      14 hours ago






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      14 hours ago











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      14 hours ago
















    • $begingroup$
      AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
      $endgroup$
      – Roman
      15 hours ago










    • $begingroup$
      @Roman - the poster is using AnyTrue not AllTrue
      $endgroup$
      – Bob Hanlon
      14 hours ago






    • 1




      $begingroup$
      Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
      $endgroup$
      – Roman
      14 hours ago











    • $begingroup$
      ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
      $endgroup$
      – morbo
      14 hours ago















    $begingroup$
    AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
    $endgroup$
    – Roman
    15 hours ago




    $begingroup$
    AllTrue[data, Positive] to get the sign right. Or use Not on your solution.
    $endgroup$
    – Roman
    15 hours ago












    $begingroup$
    @Roman - the poster is using AnyTrue not AllTrue
    $endgroup$
    – Bob Hanlon
    14 hours ago




    $begingroup$
    @Roman - the poster is using AnyTrue not AllTrue
    $endgroup$
    – Bob Hanlon
    14 hours ago




    1




    1




    $begingroup$
    Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
    $endgroup$
    – Roman
    14 hours ago





    $begingroup$
    Yes @BobHanlon . In order to invert his solution to what the OP wants you have to either Not@AnyTrue[data,Negative] or (simpler) AllTrue[data,Positive] or AllTrue[data,NonNegative].
    $endgroup$
    – Roman
    14 hours ago













    $begingroup$
    ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
    $endgroup$
    – morbo
    14 hours ago




    $begingroup$
    ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
    $endgroup$
    – morbo
    14 hours ago











    2












    $begingroup$

    list = 1, 2, 3, 4, -5, -6, -7;

    list /. x_?Negative -> True, x_?Positive -> False





    share|improve this answer









    $endgroup$

















      2












      $begingroup$

      list = 1, 2, 3, 4, -5, -6, -7;

      list /. x_?Negative -> True, x_?Positive -> False





      share|improve this answer









      $endgroup$















        2












        2








        2





        $begingroup$

        list = 1, 2, 3, 4, -5, -6, -7;

        list /. x_?Negative -> True, x_?Positive -> False





        share|improve this answer









        $endgroup$



        list = 1, 2, 3, 4, -5, -6, -7;

        list /. x_?Negative -> True, x_?Positive -> False






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 15 hours ago









        AlrubaieAlrubaie

        31910




        31910




















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