Mdthbs

What is the difference between nullifying your vote and not going to vote at all?

My player wants to cast multiple charges of magic missile from a wand

Is there an evolutionary advantage to having two heads?

Expenditure in Poland - Forex doesn't have Zloty

In what episode of TOS did a character on the bridge make a comment about raising the number 1 to some power?

Socratic Paradox

How was Apollo supposed to rendezvous in the case of a lunar abort?

Draw a checker pattern with a black X in the center

The qvolume of an integer

How to properly maintain eye contact with people that have distinctive facial features?

Why do Russians call their women expensive ("дорогая")?

What is the 中 in ダウンロード中？

Can a rogue effectively triple their speed by combining Dash and Ready?

Is this light switch installation safe and legal?

Fastest way to perform complex search on pandas dataframe

find the Integer value after a string from a file

Where can I find the list of all tendons in the human body?

Can a helicopter mask itself from Radar?

Is it possible to kill all life on Earth?

Why does the UK have more political parties than the US?

Biblical Basis for 400 years of silence between old and new testament

Uncommanded roll at high speed

Infinitely many hats

If a massive object like Jupiter flew past the Earth how close would it need to come to pull people off of the surface?

Select row of data if next row contains zero

Data Table Manipulation in MathematicaData Table Manipulation in Mathematica: Step 2Using levels in MapIndexedCompute the evolution of percentages from a data fileGenerate a new listSort columns in a TableFormmaking two columns from 1D-horizontal dataLoading and formatting data for plotting standard deviation bars on top of dataSelect rows of data that satisfies both conditions simultaneouslySelecting matrices based on existence of unique column and row of all equal terms

6

$begingroup$

I am a new user of Mathematica, and just faced a problem.
I have a data set of three columns, and want to select only those rows that go before the rows, which don't contain 0 in the third column.
Eventually I want to create a new data set containing only these selected rows.

Can it be performed in Mathematica?
Thank you in advance!

(edited)

list-manipulation

share|improve this question

edited 7 hours ago

user64494

3,90211323

asked 10 hours ago

AnnaAnna

313

New contributor

Anna 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 | 

6

$begingroup$

I am a new user of Mathematica, and just faced a problem.
I have a data set of three columns, and want to select only those rows that go before the rows, which don't contain 0 in the third column.
Eventually I want to create a new data set containing only these selected rows.

Can it be performed in Mathematica?
Thank you in advance!

(edited)

list-manipulation

share|improve this question

edited 7 hours ago

user64494

3,90211323

asked 10 hours ago

AnnaAnna

313

New contributor

Anna 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 | 

$begingroup$

I am a new user of Mathematica, and just faced a problem.
I have a data set of three columns, and want to select only those rows that go before the rows, which don't contain 0 in the third column.
Eventually I want to create a new data set containing only these selected rows.

Can it be performed in Mathematica?
Thank you in advance!

(edited)

list-manipulation

share|improve this question

edited 7 hours ago

user64494

3,90211323

asked 10 hours ago

AnnaAnna

313

New contributor

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

$endgroup$

share|improve this question

edited 7 hours ago

user64494

3,90211323

asked 10 hours ago

AnnaAnna

313

New contributor

Anna 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

edited 7 hours ago

user64494

3,90211323

asked 10 hours ago

AnnaAnna

313

New contributor

Anna 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

user64494

3,90211323

edited 7 hours ago

user64494

3,90211323

asked 10 hours ago

AnnaAnna

313

New contributor

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

asked 10 hours ago

AnnaAnna

313

3 Answers
3

active

oldest

votes

10

$begingroup$

You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.

data = {{1, 2}, {8, 4}, {5, 0}, {3, 2}, {8, 9}, {7, 0}, {2, 3}};



SequenceCases[data, {a_, {___, 0}} :> a]

(* {{8, 4}, {8, 9}} *)

Here {a_, {___, 0}} is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.

share|improve this answer

edited 7 hours ago

answered 10 hours ago

Jason B.Jason B.

49.6k391199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great solution! There's no need to name b, you can use the pattern {a_, {_, 0}} or even {a_, {___, 0}} (to be more flexible, see @kglr's comment below other solution).
  $endgroup$
  – Roman
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the suggestions!
  $endgroup$
  – Jason B.
  7 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

You can also use ReplaceList:

data = {{1, 1, 2}, {2, 8, 4}, {3, 5, 0}, {4, 3, 2}, {5, 8, 9}, {6, 7, 0}, {7, 2, 3}};

ReplaceList[data, {___, a_, {__, 0}, ___} :> a]

{{2, 8, 4}, {5, 8, 9}}

Or combinations of

Extract and Position:

Extract[data, Position[data[[2 ;;, -1]], 0]]

{{2, 8, 4}, {5, 8, 9}}

PositionIndex and Part:

data[[PositionIndex[data[[2 ;;, -1]]][0]]]

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 10 hours ago

kglrkglr

195k10216438

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
  $endgroup$
  – Anna
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anna, made a small change ( replaced {_,0} with {__,0}) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
  $endgroup$
  – kglr
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], {, a_, {, 0}} :> a] returns {} for me.
  $endgroup$
  – Christopher Lamb
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChristopherLamb, thank you; it is fixed now.
  $endgroup$
  – kglr
  6 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

data //Pick[Most@#, #[[2;;,3]],0]&

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 6 hours ago

user1066user1066

6,19822033

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

Anna is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f199260%2fselect-row-of-data-if-next-row-contains-zero%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

10

$begingroup$

You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.

data = {{1, 2}, {8, 4}, {5, 0}, {3, 2}, {8, 9}, {7, 0}, {2, 3}};



SequenceCases[data, {a_, {___, 0}} :> a]

(* {{8, 4}, {8, 9}} *)

Here {a_, {___, 0}} is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.

share|improve this answer

edited 7 hours ago

answered 10 hours ago

Jason B.Jason B.

49.6k391199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great solution! There's no need to name b, you can use the pattern {a_, {_, 0}} or even {a_, {___, 0}} (to be more flexible, see @kglr's comment below other solution).
  $endgroup$
  – Roman
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the suggestions!
  $endgroup$
  – Jason B.
  7 hours ago
  

  
  
  
  

add a comment | 

10

$begingroup$

You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.

data = {{1, 2}, {8, 4}, {5, 0}, {3, 2}, {8, 9}, {7, 0}, {2, 3}};



SequenceCases[data, {a_, {___, 0}} :> a]

(* {{8, 4}, {8, 9}} *)

Here {a_, {___, 0}} is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.

share|improve this answer

edited 7 hours ago

answered 10 hours ago

Jason B.Jason B.

49.6k391199

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great solution! There's no need to name b, you can use the pattern {a_, {_, 0}} or even {a_, {___, 0}} (to be more flexible, see @kglr's comment below other solution).
  $endgroup$
  – Roman
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for the suggestions!
  $endgroup$
  – Jason B.
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use SequenceCases to define a pattern of two elements, the second of which has a 0 in the second column.

data = {{1, 2}, {8, 4}, {5, 0}, {3, 2}, {8, 9}, {7, 0}, {2, 3}};



SequenceCases[data, {a_, {___, 0}} :> a]

(* {{8, 4}, {8, 9}} *)

Here {a_, {___, 0}} is the pattern for two rows, and the :> a says that we want to extract the first part of the pattern.

share|improve this answer

edited 7 hours ago

answered 10 hours ago

Jason B.Jason B.

49.6k391199

$endgroup$

data = {{1, 2}, {8, 4}, {5, 0}, {3, 2}, {8, 9}, {7, 0}, {2, 3}};



SequenceCases[data, {a_, {___, 0}} :> a]

(* {{8, 4}, {8, 9}} *)

share|improve this answer

edited 7 hours ago

answered 10 hours ago

Jason B.Jason B.

49.6k391199

answered 10 hours ago

Jason B.Jason B.

49.6k391199

answered 10 hours ago

Jason B.Jason B.

49.6k391199

7

$begingroup$

You can also use ReplaceList:

data = {{1, 1, 2}, {2, 8, 4}, {3, 5, 0}, {4, 3, 2}, {5, 8, 9}, {6, 7, 0}, {7, 2, 3}};

ReplaceList[data, {___, a_, {__, 0}, ___} :> a]

{{2, 8, 4}, {5, 8, 9}}

Or combinations of

Extract and Position:

Extract[data, Position[data[[2 ;;, -1]], 0]]

{{2, 8, 4}, {5, 8, 9}}

PositionIndex and Part:

data[[PositionIndex[data[[2 ;;, -1]]][0]]]

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 10 hours ago

kglrkglr

195k10216438

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
  $endgroup$
  – Anna
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anna, made a small change ( replaced {_,0} with {__,0}) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
  $endgroup$
  – kglr
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], {, a_, {, 0}} :> a] returns {} for me.
  $endgroup$
  – Christopher Lamb
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChristopherLamb, thank you; it is fixed now.
  $endgroup$
  – kglr
  6 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

You can also use ReplaceList:

data = {{1, 1, 2}, {2, 8, 4}, {3, 5, 0}, {4, 3, 2}, {5, 8, 9}, {6, 7, 0}, {7, 2, 3}};

ReplaceList[data, {___, a_, {__, 0}, ___} :> a]

{{2, 8, 4}, {5, 8, 9}}

Or combinations of

Extract and Position:

Extract[data, Position[data[[2 ;;, -1]], 0]]

{{2, 8, 4}, {5, 8, 9}}

PositionIndex and Part:

data[[PositionIndex[data[[2 ;;, -1]]][0]]]

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 10 hours ago

kglrkglr

195k10216438

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you a lot!! It works perfectly for my data set. I wonder now how the command Extract would work for 3 columns (zero value is in the 3rd column)?
  $endgroup$
  – Anna
  9 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Anna, made a small change ( replaced {_,0} with {__,0}) so that all methods should work for any number of columns (as long as the criterion is "the last column is of the next row is 0")
  $endgroup$
  – kglr
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kglr For what it's worth: Cases[Split[data, #[[2]] != 0 &], {, a_, {, 0}} :> a] returns {} for me.
  $endgroup$
  – Christopher Lamb
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChristopherLamb, thank you; it is fixed now.
  $endgroup$
  – kglr
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

You can also use ReplaceList:

data = {{1, 1, 2}, {2, 8, 4}, {3, 5, 0}, {4, 3, 2}, {5, 8, 9}, {6, 7, 0}, {7, 2, 3}};

ReplaceList[data, {___, a_, {__, 0}, ___} :> a]

{{2, 8, 4}, {5, 8, 9}}

Or combinations of

Extract and Position:

Extract[data, Position[data[[2 ;;, -1]], 0]]

{{2, 8, 4}, {5, 8, 9}}

PositionIndex and Part:

data[[PositionIndex[data[[2 ;;, -1]]][0]]]

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 10 hours ago

kglrkglr

195k10216438

$endgroup$

data = {{1, 1, 2}, {2, 8, 4}, {3, 5, 0}, {4, 3, 2}, {5, 8, 9}, {6, 7, 0}, {7, 2, 3}};

ReplaceList[data, {___, a_, {__, 0}, ___} :> a]

Extract[data, Position[data[[2 ;;, -1]], 0]]

data[[PositionIndex[data[[2 ;;, -1]]][0]]]

share|improve this answer

edited 1 hour ago

answered 10 hours ago

kglrkglr

195k10216438

answered 10 hours ago

kglrkglr

195k10216438

answered 10 hours ago

kglrkglr

195k10216438

0

$begingroup$

data //Pick[Most@#, #[[2;;,3]],0]&

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 6 hours ago

user1066user1066

6,19822033

$endgroup$

add a comment | 

0

$begingroup$

data //Pick[Most@#, #[[2;;,3]],0]&

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 6 hours ago

user1066user1066

6,19822033

$endgroup$

add a comment | 

$begingroup$

data //Pick[Most@#, #[[2;;,3]],0]&

{{2, 8, 4}, {5, 8, 9}}

share|improve this answer

edited 1 hour ago

answered 6 hours ago

user1066user1066

6,19822033

$endgroup$

data //Pick[Most@#, #[[2;;,3]],0]&

share|improve this answer

edited 1 hour ago

answered 6 hours ago

user1066user1066

6,19822033

answered 6 hours ago

user1066user1066

6,19822033

answered 6 hours ago

user1066user1066

6,19822033