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
How to answer pointed "are you quitting" questioning when I don't want them to suspect
How can I fix this gap between bookcases I made?
Is "plugging out" electronic devices an American expression?
What is the command to reset a PC without deleting any files
Mapping arrows in commutative diagrams
Is domain driven design an anti-SQL pattern?
What's the difference between repeating elections every few years and repeating a referendum after a few years?
How to move the player while also allowing forces to affect it
New order #4: World
How to make payment on the internet without leaving a money trail?
Patience, young "Padovan"
Latin words with no plurals in English
What to wear for invited talk in Canada
aging parents with no investments
Can a planet have a different gravitational pull depending on its location in orbit around its sun?
How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?
Why doesn't a const reference extend the life of a temporary object passed via a function?
What does "enim et" mean?
Denied boarding due to overcrowding, Sparpreis ticket. What are my rights?
I see my dog run
Doomsday-clock for my fantasy planet
Re-submission of rejected manuscript without informing co-authors
What is the meaning of "of trouble" in the following sentence?
Why airport relocation isn't done gradually?
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
8 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
8 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
8 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
8 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 3 hours ago
Roman
4,65511128
4,65511128
asked 8 hours ago
morsmors
496
496
$begingroup$
Have a look atDeleteCases
andStringMatchQ
orStringContainsQ
.
$endgroup$
– b.gatessucks
8 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
8 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
8 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
8 hours ago
add a comment |
$begingroup$
Have a look atDeleteCases
andStringMatchQ
orStringContainsQ
.
$endgroup$
– b.gatessucks
8 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
8 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
8 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
8 hours ago
$begingroup$
Have a look at
DeleteCases
and StringMatchQ
or StringContainsQ
.$endgroup$
– b.gatessucks
8 hours ago
$begingroup$
Have a look at
DeleteCases
and StringMatchQ
or StringContainsQ
.$endgroup$
– b.gatessucks
8 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
8 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
8 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
8 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
8 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
8 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
8 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
8 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
8 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
8 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 4 hours ago
answered 8 hours ago
RomanRoman
4,65511128
4,65511128
$begingroup$
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
8 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
8 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
8 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
8 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 2 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
8 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
8 hours ago
$begingroup$
@b.gatessucks Thank you, I will look those option in Mathematica.
$endgroup$
– mors
8 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
8 hours ago