Eliminate empty elements from a list with a specific patternList operation on specific elementsSelect elements from list with given headSelecting elements from a list with nullsReplace empty list elements with patternReplacing Non-Constant Elements from List with Patternselecting elements from a list with two numbersRemoving elements of a specific length from a listEliminate types of elements from the listListPlot3D with empty matrix elementsDelete a large pattern from one list
Is ipsum/ipsa/ipse a third person pronoun, or can it serve other functions?
Shall I use personal or official e-mail account when registering to external websites for work purpose?
Crop image to path created in TikZ?
How to manage monthly salary
Patience, young "Padovan"
Can a planet have a different gravitational pull depending on its location in orbit around its sun?
Unbreakable Formation vs. Cry of the Carnarium
Add an angle to a sphere
Ideas for 3rd eye abilities
Lied on resume at previous job
Is it wise to focus on putting odd beats on left when playing double bass drums?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Need help identifying/translating a plaque in Tangier, Morocco
extract characters between two commas?
What are the advantages and disadvantages of running one shots compared to campaigns?
LWC and complex parameters
How can I plot a Farey diagram?
Is this food a bread or a loaf?
When blogging recipes, how can I support both readers who want the narrative/journey and ones who want the printer-friendly recipe?
Are objects structures and/or vice versa?
What is the offset in a seaplane's hull?
Landing in very high winds
Can the Produce Flame cantrip be used to grapple, or as an unarmed strike, in the right circumstances?
How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?
Eliminate empty elements from a list with a specific pattern
List operation on specific elementsSelect elements from list with given headSelecting elements from a list with nullsReplace empty list elements with patternReplacing Non-Constant Elements from List with Patternselecting elements from a list with two numbersRemoving elements of a specific length from a listEliminate types of elements from the listListPlot3D with empty matrix elementsDelete a large pattern from one list
$begingroup$
I am really new in this patterns part of Mathematica. Basically what I need to do is eliminate null elements from a list but that has a specific name before the empty element. For example, my list is:
list="a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2"
and I need to obtain
list="a12-b11-1", "d33-c22-2"
The list was created using
list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <>
ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]],
i, 1, 4, j, 1, 4], 1]
and for some values it writtes because there is not a value equal to $0.5$. Until now I have been able to do it term by term as
list//."a11-b11-"-> Sequence[]
but the real list contains a lot of elements and could be almost impossible to do it that way to solve the problem. I think my main problem is that I am not sure how to specify the pattern search (something like " *-name " in gnu/linux). Is there a wise way to do this?. Thanks in advance.
list-manipulation filtering
$endgroup$
add a comment |
$begingroup$
I am really new in this patterns part of Mathematica. Basically what I need to do is eliminate null elements from a list but that has a specific name before the empty element. For example, my list is:
list="a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2"
and I need to obtain
list="a12-b11-1", "d33-c22-2"
The list was created using
list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <>
ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]],
i, 1, 4, j, 1, 4], 1]
and for some values it writtes because there is not a value equal to $0.5$. Until now I have been able to do it term by term as
list//."a11-b11-"-> Sequence[]
but the real list contains a lot of elements and could be almost impossible to do it that way to solve the problem. I think my main problem is that I am not sure how to specify the pattern search (something like " *-name " in gnu/linux). Is there a wise way to do this?. Thanks in advance.
list-manipulation filtering
$endgroup$
$begingroup$
Have a look atDeleteCases
andStringMatchQ
orStringContainsQ
.
$endgroup$
– b.gatessucks
14 hours ago
$begingroup$
I think it would be more efficient to first filter out the unwanted cases in theint
function, and then construct strings only from the remaining ones.
$endgroup$
– Roman
13 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
13 hours ago
$begingroup$
@Roman You are right, but I am new ih this cases stuf in Mathematica and I did no know how to do it when I created the list.
$endgroup$
– mors
13 hours ago
add a comment |
$begingroup$
I am really new in this patterns part of Mathematica. Basically what I need to do is eliminate null elements from a list but that has a specific name before the empty element. For example, my list is:
list="a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2"
and I need to obtain
list="a12-b11-1", "d33-c22-2"
The list was created using
list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <>
ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]],
i, 1, 4, j, 1, 4], 1]
and for some values it writtes because there is not a value equal to $0.5$. Until now I have been able to do it term by term as
list//."a11-b11-"-> Sequence[]
but the real list contains a lot of elements and could be almost impossible to do it that way to solve the problem. I think my main problem is that I am not sure how to specify the pattern search (something like " *-name " in gnu/linux). Is there a wise way to do this?. Thanks in advance.
list-manipulation filtering
$endgroup$
I am really new in this patterns part of Mathematica. Basically what I need to do is eliminate null elements from a list but that has a specific name before the empty element. For example, my list is:
list="a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2"
and I need to obtain
list="a12-b11-1", "d33-c22-2"
The list was created using
list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <>
ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]],
i, 1, 4, j, 1, 4], 1]
and for some values it writtes because there is not a value equal to $0.5$. Until now I have been able to do it term by term as
list//."a11-b11-"-> Sequence[]
but the real list contains a lot of elements and could be almost impossible to do it that way to solve the problem. I think my main problem is that I am not sure how to specify the pattern search (something like " *-name " in gnu/linux). Is there a wise way to do this?. Thanks in advance.
list-manipulation filtering
list-manipulation filtering
edited 8 hours ago
Roman
4,66511129
4,66511129
asked 14 hours ago
morsmors
496
496
$begingroup$
Have a look atDeleteCases
andStringMatchQ
orStringContainsQ
.
$endgroup$
– b.gatessucks
14 hours ago
$begingroup$
I think it would be more efficient to first filter out the unwanted cases in theint
function, and then construct strings only from the remaining ones.
$endgroup$
– Roman
13 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
13 hours ago
$begingroup$
@Roman You are right, but I am new ih this cases stuf in Mathematica and I did no know how to do it when I created the list.
$endgroup$
– mors
13 hours ago
add a comment |
$begingroup$
Have a look atDeleteCases
andStringMatchQ
orStringContainsQ
.
$endgroup$
– b.gatessucks
14 hours ago
$begingroup$
I think it would be more efficient to first filter out the unwanted cases in theint
function, and then construct strings only from the remaining ones.
$endgroup$
– Roman
13 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
13 hours ago
$begingroup$
@Roman You are right, but I am new ih this cases stuf in Mathematica and I did no know how to do it when I created the list.
$endgroup$
– mors
13 hours ago
$begingroup$
Have a look at
DeleteCases
and StringMatchQ
or StringContainsQ
.$endgroup$
– b.gatessucks
14 hours ago
$begingroup$
Have a look at
DeleteCases
and StringMatchQ
or StringContainsQ
.$endgroup$
– b.gatessucks
14 hours ago
$begingroup$
I think it would be more efficient to first filter out the unwanted cases in the
int
function, and then construct strings only from the remaining ones.$endgroup$
– Roman
13 hours ago
$begingroup$
I think it would be more efficient to first filter out the unwanted cases in the
int
function, and then construct strings only from the remaining ones.$endgroup$
– Roman
13 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
13 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
13 hours ago
$begingroup$
@Roman You are right, but I am new ih this cases stuf in Mathematica and I did no know how to do it when I created the list.
$endgroup$
– mors
13 hours ago
$begingroup$
@Roman You are right, but I am new ih this cases stuf in Mathematica and I did no know how to do it when I created the list.
$endgroup$
– mors
13 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If the list elements are strings, as it appears after your comment, you can use Select
with a string pattern:
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringMatchQ[__ ~~ ""]]
"a12-b11-1", "d33-c22-2"
You could also Select
before making the strings:
L = DeleteCases[
Flatten[
Table[
namea[[i]], nameb[[j]], Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1],
i, 4, j, 4],
1],
_, _, ];
and then make these into strings:
StringRiffle[ToString /@ #, "-"] & /@ L
I can't check this because you didn't supply functioning code.
$endgroup$
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
add a comment |
$begingroup$
In 10.1, two functions were added to handle a pair of very common cases: StringStartsQ
and StringEndsQ
which return True
if the string matches a pattern at the beginning or end, respectively. So, while Roman's answer gives you the full general form, most of the pattern can be eliminated by using
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringEndsQ[""]]
instead.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
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
);
);
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194815%2feliminate-empty-elements-from-a-list-with-a-specific-pattern%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the list elements are strings, as it appears after your comment, you can use Select
with a string pattern:
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringMatchQ[__ ~~ ""]]
"a12-b11-1", "d33-c22-2"
You could also Select
before making the strings:
L = DeleteCases[
Flatten[
Table[
namea[[i]], nameb[[j]], Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1],
i, 4, j, 4],
1],
_, _, ];
and then make these into strings:
StringRiffle[ToString /@ #, "-"] & /@ L
I can't check this because you didn't supply functioning code.
$endgroup$
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
add a comment |
$begingroup$
If the list elements are strings, as it appears after your comment, you can use Select
with a string pattern:
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringMatchQ[__ ~~ ""]]
"a12-b11-1", "d33-c22-2"
You could also Select
before making the strings:
L = DeleteCases[
Flatten[
Table[
namea[[i]], nameb[[j]], Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1],
i, 4, j, 4],
1],
_, _, ];
and then make these into strings:
StringRiffle[ToString /@ #, "-"] & /@ L
I can't check this because you didn't supply functioning code.
$endgroup$
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
add a comment |
$begingroup$
If the list elements are strings, as it appears after your comment, you can use Select
with a string pattern:
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringMatchQ[__ ~~ ""]]
"a12-b11-1", "d33-c22-2"
You could also Select
before making the strings:
L = DeleteCases[
Flatten[
Table[
namea[[i]], nameb[[j]], Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1],
i, 4, j, 4],
1],
_, _, ];
and then make these into strings:
StringRiffle[ToString /@ #, "-"] & /@ L
I can't check this because you didn't supply functioning code.
$endgroup$
If the list elements are strings, as it appears after your comment, you can use Select
with a string pattern:
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringMatchQ[__ ~~ ""]]
"a12-b11-1", "d33-c22-2"
You could also Select
before making the strings:
L = DeleteCases[
Flatten[
Table[
namea[[i]], nameb[[j]], Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1],
i, 4, j, 4],
1],
_, _, ];
and then make these into strings:
StringRiffle[ToString /@ #, "-"] & /@ L
I can't check this because you didn't supply functioning code.
edited 9 hours ago
answered 14 hours ago
RomanRoman
4,66511129
4,66511129
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
add a comment |
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
$begingroup$
Thank you, I forgot to specify the way the list is created. I create the list as 'list = Flatten[Table[ ToString[namea[[i]] <> "-" <> nameb[[j]] <> "-" <> ToString[ Select[int[i, j, 0.5], Abs[#] == 0.5 &, 1] ]], i, 1, 4, j, 1, 4], 1] '
$endgroup$
– mors
14 hours ago
add a comment |
$begingroup$
In 10.1, two functions were added to handle a pair of very common cases: StringStartsQ
and StringEndsQ
which return True
if the string matches a pattern at the beginning or end, respectively. So, while Roman's answer gives you the full general form, most of the pattern can be eliminated by using
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringEndsQ[""]]
instead.
$endgroup$
add a comment |
$begingroup$
In 10.1, two functions were added to handle a pair of very common cases: StringStartsQ
and StringEndsQ
which return True
if the string matches a pattern at the beginning or end, respectively. So, while Roman's answer gives you the full general form, most of the pattern can be eliminated by using
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringEndsQ[""]]
instead.
$endgroup$
add a comment |
$begingroup$
In 10.1, two functions were added to handle a pair of very common cases: StringStartsQ
and StringEndsQ
which return True
if the string matches a pattern at the beginning or end, respectively. So, while Roman's answer gives you the full general form, most of the pattern can be eliminated by using
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringEndsQ[""]]
instead.
$endgroup$
In 10.1, two functions were added to handle a pair of very common cases: StringStartsQ
and StringEndsQ
which return True
if the string matches a pattern at the beginning or end, respectively. So, while Roman's answer gives you the full general form, most of the pattern can be eliminated by using
list = "a11-b11-", "a12-b11-1", "c11-d22-", "d33-c22-2";
Select[list, Not@*StringEndsQ[""]]
instead.
answered 7 hours ago
rcollyerrcollyer
28.6k674166
28.6k674166
add a comment |
add a comment |
Thanks for contributing an answer to Mathematica Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f194815%2feliminate-empty-elements-from-a-list-with-a-specific-pattern%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Have a look at
DeleteCases
andStringMatchQ
orStringContainsQ
.$endgroup$
– b.gatessucks
14 hours ago
$begingroup$
I think it would be more efficient to first filter out the unwanted cases in the
int
function, and then construct strings only from the remaining ones.$endgroup$
– Roman
13 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
13 hours ago
$begingroup$
@Roman You are right, but I am new ih this cases stuf in Mathematica and I did no know how to do it when I created the list.
$endgroup$
– mors
13 hours ago