How to just get key only on GNU Smalltalk?
I am currently using sortedCollection that stores dictionary of character (key) and number of occurrence for that character (value). When iterating through sortedCollection, how do I access the key value only?
e.g.
[que last notNil] whileTrue: [
stdout << 'current character is ' *key* << ' and occurs << *val* << ' times.' << nl.
]
Where que is the sortedCollection that sorts dictionary by value.
My goal is following: let's say que has:
[$a:20, $e:100] where first letter is the key of dictionary and second number is the value of dictionary. My output should look something like this:
current character is a and occurs 20 times.
current character is e and occurs 100 times.
I am not sure how to get a, or the key in dictionary since keys are arbitrary.
smalltalk gnu-smalltalk
asked Nov 12 at 14:07
selfPointer
264
add a comment |
I am currently using sortedCollection that stores dictionary of character (key) and number of occurrence for that character (value). When iterating through sortedCollection, how do I access the key value only?
e.g.
[que last notNil] whileTrue: [
stdout << 'current character is ' *key* << ' and occurs << *val* << ' times.' << nl.
]
Where que is the sortedCollection that sorts dictionary by value.
My goal is following: let's say que has:
[$a:20, $e:100] where first letter is the key of dictionary and second number is the value of dictionary. My output should look something like this:
current character is a and occurs 20 times.
current character is e and occurs 100 times.
I am not sure how to get a, or the key in dictionary since keys are arbitrary.
smalltalk gnu-smalltalk
asked Nov 12 at 14:07
selfPointer
264
You don't show how do you build the collection que, but it seems that to retrieve keys and values, you should use a SortedDictionary or a sorted collection of Associations
– Carlos E. Ferro
Nov 12 at 14:56
I'm sorry,queis defined as following:que = sortedCollection new .que := add: nodeLeaf.wherenodeLeaf := Dictionary new.and iterate overkeyAndValuesDo, which has the character and its number of occurrence I understand that dictionary is not sorted yet, but I think I saw how to sort dictionary by value on different post, so I'll try that first.
– selfPointer
Nov 12 at 15:31
1
@selfPointer Can you update the question with the actual code you have, please? For example, there is no mention of keysAndValuesDo: and how you use it in your question.
– JayK
Nov 12 at 17:12
So you're sayingqueis a sorted collection of dictionaries?
– lurker
Nov 21 at 19:06
add a comment |
I am currently using sortedCollection that stores dictionary of character (key) and number of occurrence for that character (value). When iterating through sortedCollection, how do I access the key value only?
e.g.
[que last notNil] whileTrue: [
stdout << 'current character is ' *key* << ' and occurs << *val* << ' times.' << nl.
]
Where que is the sortedCollection that sorts dictionary by value.
My goal is following: let's say que has:
[$a:20, $e:100] where first letter is the key of dictionary and second number is the value of dictionary. My output should look something like this:
current character is a and occurs 20 times.
current character is e and occurs 100 times.
I am not sure how to get a, or the key in dictionary since keys are arbitrary.
smalltalk gnu-smalltalk
asked Nov 12 at 14:07
selfPointer
264
I am currently using sortedCollection that stores dictionary of character (key) and number of occurrence for that character (value). When iterating through sortedCollection, how do I access the key value only?
e.g.
[que last notNil] whileTrue: [
stdout << 'current character is ' *key* << ' and occurs << *val* << ' times.' << nl.
]
Where que is the sortedCollection that sorts dictionary by value.
My goal is following: let's say que has:
[$a:20, $e:100] where first letter is the key of dictionary and second number is the value of dictionary. My output should look something like this:
current character is a and occurs 20 times.
current character is e and occurs 100 times.
I am not sure how to get a, or the key in dictionary since keys are arbitrary.
smalltalk gnu-smalltalk
smalltalk gnu-smalltalk
asked Nov 12 at 14:07
selfPointer
264
asked Nov 12 at 14:07
selfPointer
264
asked Nov 12 at 14:07
selfPointer
264
asked Nov 12 at 14:07
selfPointer
264
asked Nov 12 at 14:07
selfPointer
264
264
You don't show how do you build the collection que, but it seems that to retrieve keys and values, you should use a SortedDictionary or a sorted collection of Associations
– Carlos E. Ferro
Nov 12 at 14:56
I'm sorry,queis defined as following:que = sortedCollection new .que := add: nodeLeaf.wherenodeLeaf := Dictionary new.and iterate overkeyAndValuesDo, which has the character and its number of occurrence I understand that dictionary is not sorted yet, but I think I saw how to sort dictionary by value on different post, so I'll try that first.
– selfPointer
Nov 12 at 15:31
1
@selfPointer Can you update the question with the actual code you have, please? For example, there is no mention of keysAndValuesDo: and how you use it in your question.
– JayK
Nov 12 at 17:12
So you're sayingqueis a sorted collection of dictionaries?
– lurker
Nov 21 at 19:06
add a comment |
You don't show how do you build the collection que, but it seems that to retrieve keys and values, you should use a SortedDictionary or a sorted collection of Associations
– Carlos E. Ferro
Nov 12 at 14:56
I'm sorry,queis defined as following:que = sortedCollection new .que := add: nodeLeaf.wherenodeLeaf := Dictionary new.and iterate overkeyAndValuesDo, which has the character and its number of occurrence I understand that dictionary is not sorted yet, but I think I saw how to sort dictionary by value on different post, so I'll try that first.
– selfPointer
Nov 12 at 15:31
1
@selfPointer Can you update the question with the actual code you have, please? For example, there is no mention of keysAndValuesDo: and how you use it in your question.
– JayK
Nov 12 at 17:12
So you're sayingqueis a sorted collection of dictionaries?
– lurker
Nov 21 at 19:06
You don't show how do you build the collection que, but it seems that to retrieve keys and values, you should use a SortedDictionary or a sorted collection of Associations
– Carlos E. Ferro
Nov 12 at 14:56
You don't show how do you build the collection que, but it seems that to retrieve keys and values, you should use a SortedDictionary or a sorted collection of Associations
– Carlos E. Ferro
Nov 12 at 14:56
I'm sorry,
que is defined as following: que = sortedCollection new . que := add: nodeLeaf. where nodeLeaf := Dictionary new. and iterate over keyAndValuesDo, which has the character and its number of occurrence I understand that dictionary is not sorted yet, but I think I saw how to sort dictionary by value on different post, so I'll try that first.– selfPointer
Nov 12 at 15:31
I'm sorry,
que is defined as following: que = sortedCollection new . que := add: nodeLeaf. where nodeLeaf := Dictionary new. and iterate over keyAndValuesDo, which has the character and its number of occurrence I understand that dictionary is not sorted yet, but I think I saw how to sort dictionary by value on different post, so I'll try that first.– selfPointer
Nov 12 at 15:31
1
1
@selfPointer Can you update the question with the actual code you have, please? For example, there is no mention of keysAndValuesDo: and how you use it in your question.
– JayK
Nov 12 at 17:12
@selfPointer Can you update the question with the actual code you have, please? For example, there is no mention of keysAndValuesDo: and how you use it in your question.
– JayK
Nov 12 at 17:12
So you're saying
que is a sorted collection of dictionaries?– lurker
Nov 21 at 19:06
So you're saying
que is a sorted collection of dictionaries?– lurker
Nov 21 at 19:06
add a comment |
1 Answer
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Assuming that que is a sorted collection and each element is an Association with the character as the key and the count as the value, as in your [$a:20, $e:100] example, you could do it like this:
que do: [:each | stdout << 'current character is ' << each key
<< ' and occurs ' << each value << ' times.' << nl]
If que is a Dictionary, use que keysAndValuesDo: [:char :count | stdout << "..." char << "..." count << "..." nl].
Depending on your overall application, you could also use a Bag, which is an unordered collection of elements that can appear multiple times (as opposed to a Set, which contains each element only once).
characters := 'hello world' asBag.
characters asSet do: [:each |
stdout << 'current character is ' << each
<< ' and occurs ' << (characters occurrencesOf: each)
<< ' times.' << nl]
You could also have a look at the #sortedByCount method and see whether it suits your case. I cannot tell from the reference how exactly its returned collection is structured, so I will not provide you with guessed example code.
answered Nov 12 at 17:30
JayK
1,3411919
add a comment |
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Assuming that que is a sorted collection and each element is an Association with the character as the key and the count as the value, as in your [$a:20, $e:100] example, you could do it like this:
que do: [:each | stdout << 'current character is ' << each key
<< ' and occurs ' << each value << ' times.' << nl]
If que is a Dictionary, use que keysAndValuesDo: [:char :count | stdout << "..." char << "..." count << "..." nl].
Depending on your overall application, you could also use a Bag, which is an unordered collection of elements that can appear multiple times (as opposed to a Set, which contains each element only once).
characters := 'hello world' asBag.
characters asSet do: [:each |
stdout << 'current character is ' << each
<< ' and occurs ' << (characters occurrencesOf: each)
<< ' times.' << nl]
You could also have a look at the #sortedByCount method and see whether it suits your case. I cannot tell from the reference how exactly its returned collection is structured, so I will not provide you with guessed example code.
answered Nov 12 at 17:30
JayK
1,3411919
add a comment |
Assuming that que is a sorted collection and each element is an Association with the character as the key and the count as the value, as in your [$a:20, $e:100] example, you could do it like this:
que do: [:each | stdout << 'current character is ' << each key
<< ' and occurs ' << each value << ' times.' << nl]
If que is a Dictionary, use que keysAndValuesDo: [:char :count | stdout << "..." char << "..." count << "..." nl].
Depending on your overall application, you could also use a Bag, which is an unordered collection of elements that can appear multiple times (as opposed to a Set, which contains each element only once).
characters := 'hello world' asBag.
characters asSet do: [:each |
stdout << 'current character is ' << each
<< ' and occurs ' << (characters occurrencesOf: each)
<< ' times.' << nl]
You could also have a look at the #sortedByCount method and see whether it suits your case. I cannot tell from the reference how exactly its returned collection is structured, so I will not provide you with guessed example code.
answered Nov 12 at 17:30
JayK
1,3411919
add a comment |
Assuming that que is a sorted collection and each element is an Association with the character as the key and the count as the value, as in your [$a:20, $e:100] example, you could do it like this:
que do: [:each | stdout << 'current character is ' << each key
<< ' and occurs ' << each value << ' times.' << nl]
If que is a Dictionary, use que keysAndValuesDo: [:char :count | stdout << "..." char << "..." count << "..." nl].
Depending on your overall application, you could also use a Bag, which is an unordered collection of elements that can appear multiple times (as opposed to a Set, which contains each element only once).
characters := 'hello world' asBag.
characters asSet do: [:each |
stdout << 'current character is ' << each
<< ' and occurs ' << (characters occurrencesOf: each)
<< ' times.' << nl]
You could also have a look at the #sortedByCount method and see whether it suits your case. I cannot tell from the reference how exactly its returned collection is structured, so I will not provide you with guessed example code.
answered Nov 12 at 17:30
JayK
1,3411919
Assuming that que is a sorted collection and each element is an Association with the character as the key and the count as the value, as in your [$a:20, $e:100] example, you could do it like this:
que do: [:each | stdout << 'current character is ' << each key
<< ' and occurs ' << each value << ' times.' << nl]
If que is a Dictionary, use que keysAndValuesDo: [:char :count | stdout << "..." char << "..." count << "..." nl].
Depending on your overall application, you could also use a Bag, which is an unordered collection of elements that can appear multiple times (as opposed to a Set, which contains each element only once).
characters := 'hello world' asBag.
characters asSet do: [:each |
stdout << 'current character is ' << each
<< ' and occurs ' << (characters occurrencesOf: each)
<< ' times.' << nl]
You could also have a look at the #sortedByCount method and see whether it suits your case. I cannot tell from the reference how exactly its returned collection is structured, so I will not provide you with guessed example code.
answered Nov 12 at 17:30
JayK
1,3411919
answered Nov 12 at 17:30
JayK
1,3411919
answered Nov 12 at 17:30
JayK
1,3411919
answered Nov 12 at 17:30
JayK
1,3411919
1,3411919
add a comment |
add a comment |
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You don't show how do you build the collection que, but it seems that to retrieve keys and values, you should use a SortedDictionary or a sorted collection of Associations
– Carlos E. Ferro
Nov 12 at 14:56
I'm sorry,
queis defined as following:que = sortedCollection new .que := add: nodeLeaf.wherenodeLeaf := Dictionary new.and iterate overkeyAndValuesDo, which has the character and its number of occurrence I understand that dictionary is not sorted yet, but I think I saw how to sort dictionary by value on different post, so I'll try that first.– selfPointer
Nov 12 at 15:31
1
@selfPointer Can you update the question with the actual code you have, please? For example, there is no mention of keysAndValuesDo: and how you use it in your question.
– JayK
Nov 12 at 17:12
So you're saying
queis a sorted collection of dictionaries?– lurker
Nov 21 at 19:06