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How to check is there any negative term in a large list?
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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
edited 7 hours ago
mjw
1,04110
asked 12 hours ago
a ba b
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.
$endgroup$
add a comment |
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
edited 7 hours ago
mjw
1,04110
asked 12 hours ago
a ba b
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.
$endgroup$
3
$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is slightly pensive♦
11 hours ago
1
$begingroup$
UseApplyas inAnd @@ Positive[list]
$endgroup$
– Bob Hanlon
11 hours ago
$begingroup$
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
11 hours ago
add a comment |
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
edited 7 hours ago
mjw
1,04110
asked 12 hours ago
a ba b
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.
$endgroup$
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 7 hours ago
mjw
1,04110
asked 12 hours ago
a ba b
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.
edited 7 hours ago
mjw
1,04110
asked 12 hours ago
a ba b
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.
edited 7 hours ago
mjw
1,04110
edited 7 hours ago
mjw
1,04110
edited 7 hours ago
mjw
1,04110
1,04110
asked 12 hours ago
a ba b
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.
asked 12 hours ago
a ba b
361
asked 12 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♦
11 hours ago
1
$begingroup$
UseApplyas inAnd @@ Positive[list]
$endgroup$
– Bob Hanlon
11 hours ago
$begingroup$
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
11 hours ago
add a comment |
3
$begingroup$
VectorQ[list, Positive]?
$endgroup$
– J. M. is slightly pensive♦
11 hours ago
1
$begingroup$
UseApplyas inAnd @@ Positive[list]
$endgroup$
– Bob Hanlon
11 hours ago
$begingroup$
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
11 hours ago
3
3
$begingroup$
VectorQ[list, Positive]?$endgroup$
– J. M. is slightly pensive♦
11 hours ago
$begingroup$
VectorQ[list, Positive]?$endgroup$
– J. M. is slightly pensive♦
11 hours ago
1
1
$begingroup$
Use
Apply as in And @@ Positive[list]$endgroup$
– Bob Hanlon
11 hours ago
$begingroup$
Use
Apply as in And @@ Positive[list]$endgroup$
– Bob Hanlon
11 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
11 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
11 hours ago
add a comment |
4 Answers
4
active
oldest
votes
8
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered 9 hours ago
sakrasakra
2,6581428
$endgroup$
2
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is slightly pensive♦
9 hours ago
1
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
add a comment |
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 *)
answered 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
10 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
10 hours ago
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is slightly pensive♦
10 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
10 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 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 slightly pensive♦
9 hours ago
|
show 3 more comments
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*)
answered 11 hours ago
morbomorbo
46418
$endgroup$
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
11 hours ago
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
11 hours ago
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
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
add a comment |
2
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered 11 hours ago
AlrubaieAlrubaie
31910
$endgroup$
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
8
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered 9 hours ago
sakrasakra
2,6581428
$endgroup$
2
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is slightly pensive♦
9 hours ago
1
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
add a comment |
8
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered 9 hours ago
sakrasakra
2,6581428
$endgroup$
2
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is slightly pensive♦
9 hours ago
1
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
add a comment |
8
8
8
$begingroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered 9 hours ago
sakrasakra
2,6581428
$endgroup$
Alternate solution:
list = RandomReal[1, 10^6];
Min[list] >= 0
answered 9 hours ago
sakrasakra
2,6581428
answered 9 hours ago
sakrasakra
2,6581428
answered 9 hours ago
sakrasakra
2,6581428
answered 9 hours ago
sakrasakra
2,6581428
2,6581428
2
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is slightly pensive♦
9 hours ago
1
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
add a comment |
2
$begingroup$
...i.e.NonNegative[Min[list]].
$endgroup$
– J. M. is slightly pensive♦
9 hours ago
1
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
2
2
$begingroup$
...i.e.
NonNegative[Min[list]].$endgroup$
– J. M. is slightly pensive♦
9 hours ago
$begingroup$
...i.e.
NonNegative[Min[list]].$endgroup$
– J. M. is slightly pensive♦
9 hours ago
1
1
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
Very good solution! This avoids lists of Booleans and hence allows for vectorization. (Boolean arrays cannot be packed.)
$endgroup$
– Henrik Schumacher
8 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
$begingroup$
This is about 100 times faster than any of the other solutions. Impressive!
$endgroup$
– Roman
5 hours ago
add a comment |
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 *)
answered 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
10 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
10 hours ago
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is slightly pensive♦
10 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
10 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 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 slightly pensive♦
9 hours ago
|
show 3 more comments
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 *)
answered 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
$begingroup$
Looks like your method is five times faster than the next best! Can you give some insight into why this is? Thanks!
$endgroup$
– mjw
10 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
10 hours ago
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is slightly pensive♦
10 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
10 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 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 slightly pensive♦
9 hours ago
|
show 3 more comments
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 *)
answered 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
$endgroup$
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 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
edited 10 hours ago
answered 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
answered 10 hours ago
Bob HanlonBob Hanlon
61.1k33598
answered 10 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
10 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
10 hours ago
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is slightly pensive♦
10 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
10 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 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 slightly pensive♦
9 hours ago
|
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
10 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
10 hours ago
1
$begingroup$
@mjw,And[]does short-circuit evaluation.
$endgroup$
– J. M. is slightly pensive♦
10 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
10 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 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 slightly pensive♦
9 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
10 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
10 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
10 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
10 hours ago
1
1
$begingroup$
@mjw,
And[] does short-circuit evaluation.$endgroup$
– J. M. is slightly pensive♦
10 hours ago
$begingroup$
@mjw,
And[] does short-circuit evaluation.$endgroup$
– J. M. is slightly pensive♦
10 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
10 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
10 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♦
9 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♦
9 hours ago
|
show 3 more comments
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*)
answered 11 hours ago
morbomorbo
46418
$endgroup$
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
11 hours ago
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
11 hours ago
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
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
add a comment |
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*)
answered 11 hours ago
morbomorbo
46418
$endgroup$
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
11 hours ago
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
11 hours ago
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
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
add a comment |
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*)
answered 11 hours ago
morbomorbo
46418
$endgroup$
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 11 hours ago
morbomorbo
46418
edited 10 hours ago
answered 11 hours ago
morbomorbo
46418
answered 11 hours ago
morbomorbo
46418
answered 11 hours ago
morbomorbo
46418
46418
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
11 hours ago
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
11 hours ago
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
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
add a comment |
$begingroup$
AllTrue[data, Positive]to get the sign right. Or useNoton your solution.
$endgroup$
– Roman
11 hours ago
$begingroup$
@Roman - the poster is usingAnyTruenotAllTrue
$endgroup$
– Bob Hanlon
11 hours ago
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
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
$begingroup$
AllTrue[data, Positive] to get the sign right. Or use Not on your solution.$endgroup$
– Roman
11 hours ago
$begingroup$
AllTrue[data, Positive] to get the sign right. Or use Not on your solution.$endgroup$
– Roman
11 hours ago
$begingroup$
@Roman - the poster is using
AnyTrue not AllTrue$endgroup$
– Bob Hanlon
11 hours ago
$begingroup$
@Roman - the poster is using
AnyTrue not AllTrue$endgroup$
– Bob Hanlon
11 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
10 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
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
$begingroup$
ah, signs are reversed, missed that part. I updated the code to reflect questioners exact question.
$endgroup$
– morbo
10 hours ago
add a comment |
2
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered 11 hours ago
AlrubaieAlrubaie
31910
$endgroup$
add a comment |
2
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered 11 hours ago
AlrubaieAlrubaie
31910
$endgroup$
add a comment |
2
2
2
$begingroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered 11 hours ago
AlrubaieAlrubaie
31910
$endgroup$
list = 1, 2, 3, 4, -5, -6, -7;
list /. x_?Negative -> True, x_?Positive -> False
answered 11 hours ago
AlrubaieAlrubaie
31910
answered 11 hours ago
AlrubaieAlrubaie
31910
answered 11 hours ago
AlrubaieAlrubaie
31910
answered 11 hours ago
AlrubaieAlrubaie
31910
31910
add a comment |
add a comment |
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3
$begingroup$
VectorQ[list, Positive]?$endgroup$
– J. M. is slightly pensive♦
11 hours ago
1
$begingroup$
Use
Applyas inAnd @@ Positive[list]$endgroup$
– Bob Hanlon
11 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 (useNonNegative)?$endgroup$
– Roman
11 hours ago