Ggthjy

What is the wife of a henpecked husband called?

How to satisfy a player character's curiosity about another player character?

Criticizing long fiction. How is it different from short?

Crystal compensation for temp and voltage

What is the purpose of easy combat scenarios that don't need resource expenditure?

How can I mix up weapons for large groups of similar monsters/characters?

Is there a way to help users from having to clicking emails twice before logging into a new sandbox

How do Japanese speakers determine the implied topic when none has been mentioned?

Meth dealer reference in Family Guy

What happens if a wizard reaches level 20 but has no 3rd-level spells that they can use with the Signature Spells feature?

Where was Karl Mordo in Infinity War?

How Should I Define/Declare String Constants

Should I choose Itemized or Standard deduction?

When does coming up with an idea constitute sufficient contribution for authorship?

Is my plan for fixing my water heater leak bad?

Meaning of すきっとした

What is Crew Dragon approaching in this picture?

What's a good word to describe a public place that looks like it wouldn't be rough?

Yeshiva University RIETS Semicha Yorei and Yadin

How to add multiple differently colored borders around a node?

Table enclosed in curly brackets

Is it a fallacy if someone claims they need an explanation for every word of your argument to the point where they don't understand common terms?

What can I substitute for soda pop in a sweet pork recipe?

How to use a mathematical expression as xticklable

How can I get the count of how many times a string appears in my list?

How to extract a consecutive list of substring from a string?Reading binary strings from file as stringsHighlighting or coloring certain words / substrings appearing in a larger stringHow do I pad numbers to the left so that all are the same length?Counting the rows of a list where the order of numbers in the row doesn't matterDetermine the number of times a function will be calledUsing Position to get position of string of a list of stringsRegex named groups — how to refer back to them in the replacement stringMost common prefixes in an English corpus (Hamlet)Simple way to apply a rule (from list of rules) based on the presence of the a rule key within a string

1

$begingroup$

I have a list of strings, like so:

{"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}

I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?

string-manipulation counting

share|improve this question

edited 32 mins ago

m_goldberg

87k872197

asked 3 hours ago

zongxianzongxian

1074

$endgroup$

add a comment | 

1

$begingroup$

I have a list of strings, like so:

{"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}

I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?

string-manipulation counting

share|improve this question

edited 32 mins ago

m_goldberg

87k872197

asked 3 hours ago

zongxianzongxian

1074

$endgroup$

add a comment | 

$begingroup$

I have a list of strings, like so:

{"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}

I know WordCounts will return the numbers of every single word, but I want to the word group numbers, i.e., "aa bb" 3 times, and "cc dd" 2 times. How can I do this?

string-manipulation counting

share|improve this question

edited 32 mins ago

m_goldberg

87k872197

asked 3 hours ago

zongxianzongxian

1074

$endgroup$

{"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss", "kk mm"}

share|improve this question

edited 32 mins ago

m_goldberg

87k872197

asked 3 hours ago

zongxianzongxian

1074

share|improve this question

edited 32 mins ago

m_goldberg

87k872197

asked 3 hours ago

zongxianzongxian

1074

edited 32 mins ago

m_goldberg

87k872197

edited 32 mins ago

m_goldberg

87k872197

asked 3 hours ago

zongxianzongxian

1074

asked 3 hours ago

zongxianzongxian

1074

1 Answer
1

active

oldest

votes

3

$begingroup$

You can try Tally

lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};

Tally[lst]

Edit by m_goldberg

As J.M. says in his comment below, Counts will give the same information as an association.

Counts[data]

<|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>

This is equivalent to

Rule @@@ Tally[data] // Association

share|improve this answer

edited 20 mins ago

m_goldberg

87k872197

answered 3 hours ago

NasserNasser

58k489206

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Counts[] should also work.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

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
});


}
});

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%2f192562%2fhow-can-i-get-the-count-of-how-many-times-a-string-appears-in-my-list%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

You can try Tally

lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};

Tally[lst]

Edit by m_goldberg

As J.M. says in his comment below, Counts will give the same information as an association.

Counts[data]

<|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>

This is equivalent to

Rule @@@ Tally[data] // Association

share|improve this answer

edited 20 mins ago

m_goldberg

87k872197

answered 3 hours ago

NasserNasser

58k489206

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Counts[] should also work.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You can try Tally

lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};

Tally[lst]

Edit by m_goldberg

As J.M. says in his comment below, Counts will give the same information as an association.

Counts[data]

<|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>

This is equivalent to

Rule @@@ Tally[data] // Association

share|improve this answer

edited 20 mins ago

m_goldberg

87k872197

answered 3 hours ago

NasserNasser

58k489206

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Counts[] should also work.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

You can try Tally

lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};

Tally[lst]

Edit by m_goldberg

As J.M. says in his comment below, Counts will give the same information as an association.

Counts[data]

<|"aa bb" -> 3, "cc dd" -> 2, "ww ss" -> 1, "ss ss" -> 1, "kk mm" -> 1|>

This is equivalent to

Rule @@@ Tally[data] // Association

share|improve this answer

edited 20 mins ago

m_goldberg

87k872197

answered 3 hours ago

NasserNasser

58k489206

$endgroup$

lst = {"aa bb", "cc dd", "aa bb", "aa bb", "cc dd", "ww ss", "ss ss",   "kk mm"};

Tally[lst]

Counts[data]

Rule @@@ Tally[data] // Association

share|improve this answer

edited 20 mins ago

m_goldberg

87k872197

answered 3 hours ago

NasserNasser

58k489206

edited 20 mins ago

m_goldberg

87k872197

edited 20 mins ago

m_goldberg

87k872197

answered 3 hours ago

NasserNasser

58k489206

answered 3 hours ago

NasserNasser

58k489206