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
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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
8
$begingroup$
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
edited Mar 27 at 19:01
mjw
1,26810
asked Mar 27 at 13:57
a ba b
712
$endgroup$
add a comment |
8
$begingroup$
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
edited Mar 27 at 19:01
mjw
1,26810
asked Mar 27 at 13:57
a ba b
712
$endgroup$
3
$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is away♦
Mar 27 at 14:26
1
$begingroup$
UseApplyas inAnd @@ Positive[list]
$endgroup$
– Bob Hanlon
Mar 27 at 14:52
$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (usePositive), or do you need all terms to be zero or positive (useNonNegative)?
$endgroup$
– Roman
Mar 27 at 15:01
add a comment |
8
8
8
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
edited Mar 27 at 19:01
mjw
1,26810
asked Mar 27 at 13:57
a ba b
712
$endgroup$
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
edited Mar 27 at 19:01
mjw
1,26810
asked Mar 27 at 13:57
a ba b
712
edited Mar 27 at 19:01
mjw
1,26810
asked Mar 27 at 13:57
a ba b
712
edited Mar 27 at 19:01
mjw
1,26810
edited Mar 27 at 19:01
mjw
1,26810
edited Mar 27 at 19:01
mjw
1,26810
1,26810
asked Mar 27 at 13:57
a ba b
712
asked Mar 27 at 13:57
a ba b
712
asked Mar 27 at 13:57
a ba b
712
712
3
$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is away♦
Mar 27 at 14:26
1
$begingroup$
UseApplyas inAnd @@ Positive[list]
$endgroup$
– Bob Hanlon
Mar 27 at 14:52
$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (usePositive), or do you need all terms to be zero or positive (useNonNegative)?
$endgroup$
– Roman
Mar 27 at 15:01
add a comment |
3
$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is away♦
Mar 27 at 14:26
1
$begingroup$
UseApplyas inAnd @@ Positive[list]
$endgroup$
– Bob Hanlon
Mar 27 at 14:52
$begingroup$
How do you want to deal with terms that are exactly zero? Do you need all terms to be positive (usePositive), or do you need all terms to be zero or positive (useNonNegative)?
$endgroup$
– Roman
Mar 27 at 15:01
3
3
$begingroup$
VectorQ[list, Positive]?$endgroup$
– J. M. is away♦
Mar 27 at 14:26
$begingroup$
VectorQ[list, Positive]?$endgroup$
– J. M. is away♦
Mar 27 at 14:26
1
1
$begingroup$
Use
Apply as in And @@ Positive[list]$endgroup$
– Bob Hanlon
Mar 27 at 14:52
$begingroup$
Use
Apply as in And @@ Positive[list]$endgroup$
– Bob Hanlon
Mar 27 at 14:52
$begingroup$
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
Mar 27 at 15:01
$begingroup$
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
Mar 27 at 15:01
add a comment |
4 Answers
4
active
oldest
votes
13
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered Mar 27 at 16:06
sakrasakra
2,7081429
$endgroup$
3
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is away♦
Mar 27 at 16:08
3
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
2
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
add a comment |
9
$begingroup$
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 *)
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
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
Mar 27 at 15:37
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is away♦
Mar 27 at 15:41
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
Mar 27 at 16:03
1
$begingroup$
@Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation likeAnd[True, True, False, True, True, ... True]will finish at once (and similar remarks apply forOr[]). Perhaps one can judiciously useCatch[]/Throw[]if an early-return test for long lists is desired.
$endgroup$
– J. M. is away♦
Mar 27 at 16:06
|
show 3 more comments
4
$begingroup$
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*)
answered Mar 27 at 14:29
morbomorbo
47428
$endgroup$
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
Mar 27 at 15:00
1
$begingroup$
Yes @BobHanlon . In order to invert his solution to what the OP wants you have to eitherNot@AnyTrue[data,Negative]or (simpler)AllTrue[data,Positive]orAllTrue[data,NonNegative].
$endgroup$
– Roman
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
add a comment |
3
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
13
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered Mar 27 at 16:06
sakrasakra
2,7081429
$endgroup$
3
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is away♦
Mar 27 at 16:08
3
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
2
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
add a comment |
13
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered Mar 27 at 16:06
sakrasakra
2,7081429
$endgroup$
3
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is away♦
Mar 27 at 16:08
3
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
2
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
add a comment |
13
13
13
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered Mar 27 at 16:06
sakrasakra
2,7081429
$endgroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered Mar 27 at 16:06
sakrasakra
2,7081429
answered Mar 27 at 16:06
sakrasakra
2,7081429
answered Mar 27 at 16:06
sakrasakra
2,7081429
answered Mar 27 at 16:06
sakrasakra
2,7081429
2,7081429
3
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is away♦
Mar 27 at 16:08
3
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
2
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
add a comment |
3
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is away♦
Mar 27 at 16:08
3
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
2
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
3
3
$begingroup$
...i.e.
NonNegative[Min[list]].$endgroup$
– J. M. is away♦
Mar 27 at 16:08
$begingroup$
...i.e.
NonNegative[Min[list]].$endgroup$
– J. M. is away♦
Mar 27 at 16:08
3
3
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
Mar 27 at 18:01
2
2
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
Mar 27 at 20:03
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
$begingroup$
Thank you very much.
$endgroup$
– a b
Mar 28 at 8:40
add a comment |
9
$begingroup$
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 *)
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
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
Mar 27 at 15:37
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is away♦
Mar 27 at 15:41
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
Mar 27 at 16:03
1
$begingroup$
@Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation likeAnd[True, True, False, True, True, ... True]will finish at once (and similar remarks apply forOr[]). Perhaps one can judiciously useCatch[]/Throw[]if an early-return test for long lists is desired.
$endgroup$
– J. M. is away♦
Mar 27 at 16:06
|
show 3 more comments
9
$begingroup$
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 *)
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
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
Mar 27 at 15:37
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is away♦
Mar 27 at 15:41
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
Mar 27 at 16:03
1
$begingroup$
@Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation likeAnd[True, True, False, True, True, ... True]will finish at once (and similar remarks apply forOr[]). Perhaps one can judiciously useCatch[]/Throw[]if an early-return test for long lists is desired.
$endgroup$
– J. M. is away♦
Mar 27 at 16:06
|
show 3 more comments
9
9
9
$begingroup$
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 *)
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
$endgroup$
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 *)
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
edited Mar 27 at 15:53
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
answered Mar 27 at 15:19
Bob HanlonBob Hanlon
61.4k33598
61.4k33598
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
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
Mar 27 at 15:37
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is away♦
Mar 27 at 15:41
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
Mar 27 at 16:03
1
$begingroup$
@Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation likeAnd[True, True, False, True, True, ... True]will finish at once (and similar remarks apply forOr[]). Perhaps one can judiciously useCatch[]/Throw[]if an early-return test for long lists is desired.
$endgroup$
– J. M. is away♦
Mar 27 at 16:06
|
show 3 more comments
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
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
Mar 27 at 15:37
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is away♦
Mar 27 at 15:41
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
Mar 27 at 16:03
1
$begingroup$
@Roman, I do not believe that any lazy evaluation is being done. My comment was more to point out that an evaluation likeAnd[True, True, False, True, True, ... True]will finish at once (and similar remarks apply forOr[]). Perhaps one can judiciously useCatch[]/Throw[]if an early-return test for long lists is desired.
$endgroup$
– J. M. is away♦
Mar 27 at 16:06
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
Mar 27 at 15:34
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
Mar 27 at 15:37
$begingroup$
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
Mar 27 at 15:37
1
1
$begingroup$
@mjw,
And[] does short-circuit evaluation.$endgroup$
– J. M. is away♦
Mar 27 at 15:41
$begingroup$
@mjw,
And[] does short-circuit evaluation.$endgroup$
– J. M. is away♦
Mar 27 at 15:41
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
Mar 27 at 16:03
$begingroup$
@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
Mar 27 at 16:03
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 away♦
Mar 27 at 16:06
$begingroup$
@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 away♦
Mar 27 at 16:06
|
show 3 more comments
4
$begingroup$
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*)
answered Mar 27 at 14:29
morbomorbo
47428
$endgroup$
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
Mar 27 at 15:00
1
$begingroup$
Yes @BobHanlon . In order to invert his solution to what the OP wants you have to eitherNot@AnyTrue[data,Negative]or (simpler)AllTrue[data,Positive]orAllTrue[data,NonNegative].
$endgroup$
– Roman
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
add a comment |
4
$begingroup$
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*)
answered Mar 27 at 14:29
morbomorbo
47428
$endgroup$
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
Mar 27 at 15:00
1
$begingroup$
Yes @BobHanlon . In order to invert his solution to what the OP wants you have to eitherNot@AnyTrue[data,Negative]or (simpler)AllTrue[data,Positive]orAllTrue[data,NonNegative].
$endgroup$
– Roman
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
add a comment |
4
4
4
$begingroup$
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*)
answered Mar 27 at 14:29
morbomorbo
47428
$endgroup$
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*)
answered Mar 27 at 14:29
morbomorbo
47428
edited Mar 27 at 15:03
answered Mar 27 at 14:29
morbomorbo
47428
answered Mar 27 at 14:29
morbomorbo
47428
answered Mar 27 at 14:29
morbomorbo
47428
47428
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
Mar 27 at 15:00
1
$begingroup$
Yes @BobHanlon . In order to invert his solution to what the OP wants you have to eitherNot@AnyTrue[data,Negative]or (simpler)AllTrue[data,Positive]orAllTrue[data,NonNegative].
$endgroup$
– Roman
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
add a comment |
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
Mar 27 at 15:00
1
$begingroup$
Yes @BobHanlon . In order to invert his solution to what the OP wants you have to eitherNot@AnyTrue[data,Negative]or (simpler)AllTrue[data,Positive]orAllTrue[data,NonNegative].
$endgroup$
– Roman
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
$begingroup$
AllTrue[data, Positive] to get the sign right. Or use Not on your solution.$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
AllTrue[data, Positive] to get the sign right. Or use Not on your solution.$endgroup$
– Roman
Mar 27 at 14:55
$begingroup$
@Roman - the poster is using
AnyTrue not AllTrue$endgroup$
– Bob Hanlon
Mar 27 at 15:00
$begingroup$
@Roman - the poster is using
AnyTrue not AllTrue$endgroup$
– Bob Hanlon
Mar 27 at 15:00
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
Mar 27 at 15:04
$begingroup$
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
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
Mar 27 at 15:04
add a comment |
3
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
$endgroup$
add a comment |
3
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
$endgroup$
add a comment |
3
3
3
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
$endgroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
answered Mar 27 at 14:31
AlrubaieAlrubaie
44011
44011
add a comment |
add a comment |
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3
$begingroup$
VectorQ[list, Positive]?$endgroup$
– J. M. is away♦
Mar 27 at 14:26
1
$begingroup$
Use
Applyas inAnd @@ Positive[list]$endgroup$
– Bob Hanlon
Mar 27 at 14:52
$begingroup$
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 (useNonNegative)?$endgroup$
– Roman
Mar 27 at 15:01