Removing lists with zero at any position

up vote
8
down vote

favorite

I want to delete the complete list within a list containing 0 at any position.

I tried using:

DeleteCases[Tuples[Range[-2, 2], 2], 0 ..]

-2, -2, -2, -1, -2, 0, -2, 1, -2, 2, -1, -2, -1, -1, -1, 0,
-1, 1, -1, 2, 0, -2, 0, -1, 0, 1, 0, 2, 1, -2, 1, -1, 1, 0, 1, 1,
1, 2, 2, -2, 2, -1, 2, 0, 2, 1, 2, 2

But only the lists starting with 0 are removed.

How can I remove the entire list that contains 0 at any place?

edited 15 hours ago

kglr

159k8184384

asked 19 hours ago

dykes

836

add a commentÂ |Â

up vote
8
down vote

favorite

I want to delete the complete list within a list containing 0 at any position.

I tried using:

DeleteCases[Tuples[Range[-2, 2], 2], 0 ..]

-2, -2, -2, -1, -2, 0, -2, 1, -2, 2, -1, -2, -1, -1, -1, 0,
-1, 1, -1, 2, 0, -2, 0, -1, 0, 1, 0, 2, 1, -2, 1, -1, 1, 0, 1, 1,
1, 2, 2, -2, 2, -1, 2, 0, 2, 1, 2, 2

But only the lists starting with 0 are removed.

How can I remove the entire list that contains 0 at any place?

edited 15 hours ago

kglr

159k8184384

asked 19 hours ago

dykes

836

add a commentÂ |Â

up vote
8
down vote

favorite

I want to delete the complete list within a list containing 0 at any position.

I tried using:

DeleteCases[Tuples[Range[-2, 2], 2], 0 ..]

-2, -2, -2, -1, -2, 0, -2, 1, -2, 2, -1, -2, -1, -1, -1, 0,
-1, 1, -1, 2, 0, -2, 0, -1, 0, 1, 0, 2, 1, -2, 1, -1, 1, 0, 1, 1,
1, 2, 2, -2, 2, -1, 2, 0, 2, 1, 2, 2

But only the lists starting with 0 are removed.

How can I remove the entire list that contains 0 at any place?

edited 15 hours ago

kglr

159k8184384

asked 19 hours ago

dykes

836

I want to delete the complete list within a list containing 0 at any position.

I tried using:

DeleteCases[Tuples[Range[-2, 2], 2], 0 ..]

-2, -2, -2, -1, -2, 0, -2, 1, -2, 2, -1, -2, -1, -1, -1, 0,
-1, 1, -1, 2, 0, -2, 0, -1, 0, 1, 0, 2, 1, -2, 1, -1, 1, 0, 1, 1,
1, 2, 2, -2, 2, -1, 2, 0, 2, 1, 2, 2

But only the lists starting with 0 are removed.

How can I remove the entire list that contains 0 at any place?

list-manipulation array

edited 15 hours ago

kglr

159k8184384

asked 19 hours ago

dykes

836

edited 15 hours ago

kglr

159k8184384

asked 19 hours ago

dykes

836

edited 15 hours ago

kglr

159k8184384

edited 15 hours ago

kglr

159k8184384

edited 15 hours ago

kglr

159k8184384

asked 19 hours ago

dykes

836

asked 19 hours ago

dykes

836

asked 19 hours ago

dykes

836

add a commentÂ |Â

3 Answers
3

active

oldest

votes

up vote
8
down vote

 DeleteCases[Tuples[Range[-2, 2], 2], ___, 0, ___] (* or *)
 DeleteCases[data, 0, 0|1,0|0,1]

-2, -2, -2, -1, -2, 1, -2, 2, -1, -2, -1, -1, -1,
1, -1, 2, 1, -2, 1, -1, 1, 1, 1, 2, 2, -2, 2, -1, 2,
1, 2, 2

Alternatively, if you can, remove zeros from the input lists before using Tuples:

Tuples[Join[-Reverse@#, #]&@Range[2],2]

same result

Even if data = Tuples[Range[-2, 2], 2] is already created, processing data to create a zero-free input list and using Tuples on it is (to my surprise) faster than using Pick to remove entries with zeros from pairs of data:

data = Tuples[Range[-100, 100], 2];

ra = Tuples[Join[-Reverse @ #, #]& @ Range[100],2]; // RepeatedTiming // First

0.0000524

rb = With[r = DeleteCases[DeleteDuplicates @ Flatten[data], 0],
 Tuples[r, 2]];// RepeatedTiming // First

0.000178

rc = Pick[data, Unitize[data].1, 1, 2];//RepeatedTiming//First (*from @Henrik's answer*)

0.0013

rd = DeleteCases[data, 0, 0|1,0|0,1];// RepeatedTiming // First

0.017

re = DeleteCases[data, ___, 0, ___];// RepeatedTiming // First

0.026

rf = Select[data, FreeQ[0] ];// RepeatedTiming // First (*from David's answer *)

0.060

ra == rb == rc == rd == re == rf

True

edited 9 hours ago

answered 19 hours ago

kglr

159k8184384

Well.... if you're free to delete 0 before using Tuples simply use Tuples[-2, -1, 1, 2, 2].
â€“Â David G. Stork
18 hours ago

@David, yes; i used the more cumbersome way to deal with cases like Range[-100, 100]
â€“Â kglr
18 hours ago

1

Oh, method b and c... nice out-of-the-box-thinking!
â€“Â Henrik Schumacher
16 hours ago

DeleteCases[DeleteDuplicates @ Flatten[data], 0] is not packed, while Join[-Reverse @ #, #]& @ Range[100] is, explaining the difference in timing
â€“Â Carl Woll
15 hours ago

@CarlWoll, good point. But in this case, DeleteDuplicates @ Flatten[data] (which is packed) already takes 0.00011 seconds (RepeatedTiming) compared to 3.06*10^-6 for Join[-Reverse @ #, #]& @ Range[100], simply because the former processes a list with dimensions 40401,2 and the latter with just 200.
â€“Â kglr
14 hours ago

add a commentÂ |Â

up vote
7
down vote

Here a more cumbersome (not a code golfer) but faster approach:

data = Tuples[Range[-100, 100], 2];
a = DeleteCases[data, ___, 0, ___]; // RepeatedTiming // First
b = Pick[data, Unitize[data].1, 1, 2]; // RepeatedTiming // First
a == b

0.028

0.00098

True

This skips pattern matching in favor of faster vectorized functions.

edited 19 hours ago

answered 19 hours ago

Henrik Schumacher

37.6k249107

add a commentÂ |Â

up vote
7
down vote

Or...

Select[Tuples[Range[-2, 2], 2], ! MemberQ[#, 0] &]

or as @ThatGravityGuy points out...

Select[Tuples[Range[-2, 2], 2], FreeQ[#, 0] &]

edited 18 hours ago

answered 19 hours ago

David G. Stork

21.4k11646

1

Instead of !MemberQ couldn't you just use FreeQ?
â€“Â That Gravity Guy
18 hours ago

Oooh... of course. Yes. Thanks.
â€“Â David G. Stork
18 hours ago

You can use the operator form FreeQ[0] instead of FreeQ[#, 0] &
â€“Â kglr
17 hours ago

.. or operator forms of both Select and FreeQ: Select[FreeQ@0]@data
â€“Â kglr
15 hours ago

add a commentÂ |Â

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
8
down vote

 DeleteCases[Tuples[Range[-2, 2], 2], ___, 0, ___] (* or *)
 DeleteCases[data, 0, 0|1,0|0,1]

-2, -2, -2, -1, -2, 1, -2, 2, -1, -2, -1, -1, -1,
1, -1, 2, 1, -2, 1, -1, 1, 1, 1, 2, 2, -2, 2, -1, 2,
1, 2, 2

Alternatively, if you can, remove zeros from the input lists before using Tuples:

Tuples[Join[-Reverse@#, #]&@Range[2],2]

same result

data = Tuples[Range[-100, 100], 2];

ra = Tuples[Join[-Reverse @ #, #]& @ Range[100],2]; // RepeatedTiming // First

0.0000524

rb = With[r = DeleteCases[DeleteDuplicates @ Flatten[data], 0],
 Tuples[r, 2]];// RepeatedTiming // First

0.000178

rc = Pick[data, Unitize[data].1, 1, 2];//RepeatedTiming//First (*from @Henrik's answer*)

0.0013

rd = DeleteCases[data, 0, 0|1,0|0,1];// RepeatedTiming // First

0.017

re = DeleteCases[data, ___, 0, ___];// RepeatedTiming // First

0.026

rf = Select[data, FreeQ[0] ];// RepeatedTiming // First (*from David's answer *)

0.060

ra == rb == rc == rd == re == rf

True

edited 9 hours ago

answered 19 hours ago

kglr

159k8184384

Well.... if you're free to delete 0 before using Tuples simply use Tuples[-2, -1, 1, 2, 2].
â€“Â David G. Stork
18 hours ago

@David, yes; i used the more cumbersome way to deal with cases like Range[-100, 100]
â€“Â kglr
18 hours ago

1

Oh, method b and c... nice out-of-the-box-thinking!
â€“Â Henrik Schumacher
16 hours ago

DeleteCases[DeleteDuplicates @ Flatten[data], 0] is not packed, while Join[-Reverse @ #, #]& @ Range[100] is, explaining the difference in timing
â€“Â Carl Woll
15 hours ago

@CarlWoll, good point. But in this case, DeleteDuplicates @ Flatten[data] (which is packed) already takes 0.00011 seconds (RepeatedTiming) compared to 3.06*10^-6 for Join[-Reverse @ #, #]& @ Range[100], simply because the former processes a list with dimensions 40401,2 and the latter with just 200.
â€“Â kglr
14 hours ago

add a commentÂ |Â

up vote
8
down vote

 DeleteCases[Tuples[Range[-2, 2], 2], ___, 0, ___] (* or *)
 DeleteCases[data, 0, 0|1,0|0,1]

-2, -2, -2, -1, -2, 1, -2, 2, -1, -2, -1, -1, -1,
1, -1, 2, 1, -2, 1, -1, 1, 1, 1, 2, 2, -2, 2, -1, 2,
1, 2, 2

Alternatively, if you can, remove zeros from the input lists before using Tuples:

Tuples[Join[-Reverse@#, #]&@Range[2],2]

same result

data = Tuples[Range[-100, 100], 2];

ra = Tuples[Join[-Reverse @ #, #]& @ Range[100],2]; // RepeatedTiming // First

0.0000524

rb = With[r = DeleteCases[DeleteDuplicates @ Flatten[data], 0],
 Tuples[r, 2]];// RepeatedTiming // First

0.000178

rc = Pick[data, Unitize[data].1, 1, 2];//RepeatedTiming//First (*from @Henrik's answer*)

0.0013

rd = DeleteCases[data, 0, 0|1,0|0,1];// RepeatedTiming // First

0.017

re = DeleteCases[data, ___, 0, ___];// RepeatedTiming // First

0.026

rf = Select[data, FreeQ[0] ];// RepeatedTiming // First (*from David's answer *)

0.060

ra == rb == rc == rd == re == rf

True

edited 9 hours ago

answered 19 hours ago

kglr

159k8184384

Well.... if you're free to delete 0 before using Tuples simply use Tuples[-2, -1, 1, 2, 2].
â€“Â David G. Stork
18 hours ago

@David, yes; i used the more cumbersome way to deal with cases like Range[-100, 100]
â€“Â kglr
18 hours ago

1

Oh, method b and c... nice out-of-the-box-thinking!
â€“Â Henrik Schumacher
16 hours ago

DeleteCases[DeleteDuplicates @ Flatten[data], 0] is not packed, while Join[-Reverse @ #, #]& @ Range[100] is, explaining the difference in timing
â€“Â Carl Woll
15 hours ago

@CarlWoll, good point. But in this case, DeleteDuplicates @ Flatten[data] (which is packed) already takes 0.00011 seconds (RepeatedTiming) compared to 3.06*10^-6 for Join[-Reverse @ #, #]& @ Range[100], simply because the former processes a list with dimensions 40401,2 and the latter with just 200.
â€“Â kglr
14 hours ago

add a commentÂ |Â

up vote
8
down vote

 DeleteCases[Tuples[Range[-2, 2], 2], ___, 0, ___] (* or *)
 DeleteCases[data, 0, 0|1,0|0,1]

-2, -2, -2, -1, -2, 1, -2, 2, -1, -2, -1, -1, -1,
1, -1, 2, 1, -2, 1, -1, 1, 1, 1, 2, 2, -2, 2, -1, 2,
1, 2, 2

Alternatively, if you can, remove zeros from the input lists before using Tuples:

Tuples[Join[-Reverse@#, #]&@Range[2],2]

same result

data = Tuples[Range[-100, 100], 2];

ra = Tuples[Join[-Reverse @ #, #]& @ Range[100],2]; // RepeatedTiming // First

0.0000524

rb = With[r = DeleteCases[DeleteDuplicates @ Flatten[data], 0],
 Tuples[r, 2]];// RepeatedTiming // First

0.000178

rc = Pick[data, Unitize[data].1, 1, 2];//RepeatedTiming//First (*from @Henrik's answer*)

0.0013

rd = DeleteCases[data, 0, 0|1,0|0,1];// RepeatedTiming // First

0.017

re = DeleteCases[data, ___, 0, ___];// RepeatedTiming // First

0.026

rf = Select[data, FreeQ[0] ];// RepeatedTiming // First (*from David's answer *)

0.060

ra == rb == rc == rd == re == rf

True

edited 9 hours ago

answered 19 hours ago

kglr

159k8184384

 DeleteCases[Tuples[Range[-2, 2], 2], ___, 0, ___] (* or *)
 DeleteCases[data, 0, 0|1,0|0,1]

-2, -2, -2, -1, -2, 1, -2, 2, -1, -2, -1, -1, -1,
1, -1, 2, 1, -2, 1, -1, 1, 1, 1, 2, 2, -2, 2, -1, 2,
1, 2, 2

Alternatively, if you can, remove zeros from the input lists before using Tuples:

Tuples[Join[-Reverse@#, #]&@Range[2],2]

same result

data = Tuples[Range[-100, 100], 2];

ra = Tuples[Join[-Reverse @ #, #]& @ Range[100],2]; // RepeatedTiming // First

0.0000524

rb = With[r = DeleteCases[DeleteDuplicates @ Flatten[data], 0],
 Tuples[r, 2]];// RepeatedTiming // First

0.000178

rc = Pick[data, Unitize[data].1, 1, 2];//RepeatedTiming//First (*from @Henrik's answer*)

0.0013

rd = DeleteCases[data, 0, 0|1,0|0,1];// RepeatedTiming // First

0.017

re = DeleteCases[data, ___, 0, ___];// RepeatedTiming // First

0.026

rf = Select[data, FreeQ[0] ];// RepeatedTiming // First (*from David's answer *)

0.060

ra == rb == rc == rd == re == rf

True

edited 9 hours ago

answered 19 hours ago

kglr

159k8184384

edited 9 hours ago

answered 19 hours ago

kglr

159k8184384

answered 19 hours ago

kglr

159k8184384

answered 19 hours ago

kglr

159k8184384

Well.... if you're free to delete 0 before using Tuples simply use Tuples[-2, -1, 1, 2, 2].
â€“Â David G. Stork
18 hours ago

@David, yes; i used the more cumbersome way to deal with cases like Range[-100, 100]
â€“Â kglr
18 hours ago

1

Oh, method b and c... nice out-of-the-box-thinking!
â€“Â Henrik Schumacher
16 hours ago

DeleteCases[DeleteDuplicates @ Flatten[data], 0] is not packed, while Join[-Reverse @ #, #]& @ Range[100] is, explaining the difference in timing
â€“Â Carl Woll
15 hours ago

@CarlWoll, good point. But in this case, DeleteDuplicates @ Flatten[data] (which is packed) already takes 0.00011 seconds (RepeatedTiming) compared to 3.06*10^-6 for Join[-Reverse @ #, #]& @ Range[100], simply because the former processes a list with dimensions 40401,2 and the latter with just 200.
â€“Â kglr
14 hours ago

add a commentÂ |Â

Well.... if you're free to delete 0 before using Tuples simply use Tuples[-2, -1, 1, 2, 2].
â€“Â David G. Stork
18 hours ago

@David, yes; i used the more cumbersome way to deal with cases like Range[-100, 100]
â€“Â kglr
18 hours ago

1

Oh, method b and c... nice out-of-the-box-thinking!
â€“Â Henrik Schumacher
16 hours ago

DeleteCases[DeleteDuplicates @ Flatten[data], 0] is not packed, while Join[-Reverse @ #, #]& @ Range[100] is, explaining the difference in timing
â€“Â Carl Woll
15 hours ago

@CarlWoll, good point. But in this case, DeleteDuplicates @ Flatten[data] (which is packed) already takes 0.00011 seconds (RepeatedTiming) compared to 3.06*10^-6 for Join[-Reverse @ #, #]& @ Range[100], simply because the former processes a list with dimensions 40401,2 and the latter with just 200.
â€“Â kglr
14 hours ago

Well.... if you're free to delete 0 before using Tuples simply use Tuples[-2, -1, 1, 2, 2].
â€“Â David G. Stork
18 hours ago

@David, yes; i used the more cumbersome way to deal with cases like Range[-100, 100]
â€“Â kglr
18 hours ago

Oh, method b and c... nice out-of-the-box-thinking!
â€“Â Henrik Schumacher
16 hours ago

DeleteCases[DeleteDuplicates @ Flatten[data], 0] is not packed, while Join[-Reverse @ #, #]& @ Range[100] is, explaining the difference in timing
â€“Â Carl Woll
15 hours ago

@CarlWoll, good point. But in this case, DeleteDuplicates @ Flatten[data] (which is packed) already takes 0.00011 seconds (RepeatedTiming) compared to 3.06*10^-6 for Join[-Reverse @ #, #]& @ Range[100], simply because the former processes a list with dimensions 40401,2 and the latter with just 200.
â€“Â kglr
14 hours ago

add a commentÂ |Â

up vote
7
down vote

Here a more cumbersome (not a code golfer) but faster approach:

data = Tuples[Range[-100, 100], 2];
a = DeleteCases[data, ___, 0, ___]; // RepeatedTiming // First
b = Pick[data, Unitize[data].1, 1, 2]; // RepeatedTiming // First
a == b

0.028

0.00098

True

This skips pattern matching in favor of faster vectorized functions.

edited 19 hours ago

answered 19 hours ago

Henrik Schumacher

37.6k249107

add a commentÂ |Â

up vote
7
down vote

Here a more cumbersome (not a code golfer) but faster approach:

data = Tuples[Range[-100, 100], 2];
a = DeleteCases[data, ___, 0, ___]; // RepeatedTiming // First
b = Pick[data, Unitize[data].1, 1, 2]; // RepeatedTiming // First
a == b

0.028

0.00098

True

This skips pattern matching in favor of faster vectorized functions.

edited 19 hours ago

answered 19 hours ago

Henrik Schumacher

37.6k249107

add a commentÂ |Â

up vote
7
down vote

Here a more cumbersome (not a code golfer) but faster approach:

data = Tuples[Range[-100, 100], 2];
a = DeleteCases[data, ___, 0, ___]; // RepeatedTiming // First
b = Pick[data, Unitize[data].1, 1, 2]; // RepeatedTiming // First
a == b

0.028

0.00098

True

This skips pattern matching in favor of faster vectorized functions.

edited 19 hours ago

answered 19 hours ago

Henrik Schumacher

37.6k249107

Here a more cumbersome (not a code golfer) but faster approach:

data = Tuples[Range[-100, 100], 2];
a = DeleteCases[data, ___, 0, ___]; // RepeatedTiming // First
b = Pick[data, Unitize[data].1, 1, 2]; // RepeatedTiming // First
a == b

0.028

0.00098

True

This skips pattern matching in favor of faster vectorized functions.

edited 19 hours ago

answered 19 hours ago

Henrik Schumacher

37.6k249107

edited 19 hours ago

answered 19 hours ago

Henrik Schumacher

37.6k249107

answered 19 hours ago

Henrik Schumacher

37.6k249107

answered 19 hours ago

Henrik Schumacher

37.6k249107

add a commentÂ |Â

up vote
7
down vote

Or...

Select[Tuples[Range[-2, 2], 2], ! MemberQ[#, 0] &]

or as @ThatGravityGuy points out...

Select[Tuples[Range[-2, 2], 2], FreeQ[#, 0] &]

edited 18 hours ago

answered 19 hours ago

David G. Stork

21.4k11646

1

Instead of !MemberQ couldn't you just use FreeQ?
â€“Â That Gravity Guy
18 hours ago

Oooh... of course. Yes. Thanks.
â€“Â David G. Stork
18 hours ago

You can use the operator form FreeQ[0] instead of FreeQ[#, 0] &
â€“Â kglr
17 hours ago

.. or operator forms of both Select and FreeQ: Select[FreeQ@0]@data
â€“Â kglr
15 hours ago

add a commentÂ |Â

up vote
7
down vote

Or...

Select[Tuples[Range[-2, 2], 2], ! MemberQ[#, 0] &]

or as @ThatGravityGuy points out...

Select[Tuples[Range[-2, 2], 2], FreeQ[#, 0] &]

edited 18 hours ago

answered 19 hours ago

David G. Stork

21.4k11646

1

Instead of !MemberQ couldn't you just use FreeQ?
â€“Â That Gravity Guy
18 hours ago

Oooh... of course. Yes. Thanks.
â€“Â David G. Stork
18 hours ago

You can use the operator form FreeQ[0] instead of FreeQ[#, 0] &
â€“Â kglr
17 hours ago

.. or operator forms of both Select and FreeQ: Select[FreeQ@0]@data
â€“Â kglr
15 hours ago

add a commentÂ |Â

up vote
7
down vote

Or...

Select[Tuples[Range[-2, 2], 2], ! MemberQ[#, 0] &]

or as @ThatGravityGuy points out...

Select[Tuples[Range[-2, 2], 2], FreeQ[#, 0] &]

edited 18 hours ago

answered 19 hours ago

David G. Stork

21.4k11646

Or...

Select[Tuples[Range[-2, 2], 2], ! MemberQ[#, 0] &]

or as @ThatGravityGuy points out...

Select[Tuples[Range[-2, 2], 2], FreeQ[#, 0] &]

edited 18 hours ago

answered 19 hours ago

David G. Stork

21.4k11646

edited 18 hours ago

answered 19 hours ago

David G. Stork

21.4k11646

answered 19 hours ago

David G. Stork

21.4k11646

answered 19 hours ago

David G. Stork

21.4k11646

1

Instead of !MemberQ couldn't you just use FreeQ?
â€“Â That Gravity Guy
18 hours ago

Oooh... of course. Yes. Thanks.
â€“Â David G. Stork
18 hours ago

You can use the operator form FreeQ[0] instead of FreeQ[#, 0] &
â€“Â kglr
17 hours ago

.. or operator forms of both Select and FreeQ: Select[FreeQ@0]@data
â€“Â kglr
15 hours ago

add a commentÂ |Â

1

Instead of !MemberQ couldn't you just use FreeQ?
â€“Â That Gravity Guy
18 hours ago

Oooh... of course. Yes. Thanks.
â€“Â David G. Stork
18 hours ago

You can use the operator form FreeQ[0] instead of FreeQ[#, 0] &
â€“Â kglr
17 hours ago

.. or operator forms of both Select and FreeQ: Select[FreeQ@0]@data
â€“Â kglr
15 hours ago

Instead of !MemberQ couldn't you just use FreeQ?
â€“Â That Gravity Guy
18 hours ago

Oooh... of course. Yes. Thanks.
â€“Â David G. Stork
18 hours ago

You can use the operator form FreeQ[0] instead of FreeQ[#, 0] &
â€“Â kglr
17 hours ago

.. or operator forms of both Select and FreeQ: Select[FreeQ@0]@data
â€“Â kglr
15 hours ago

add a commentÂ |Â

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