sort+counting

girlzone · 05-30-2003, 11:42 AM

Hi, I have a program that should read a line of text and output the number of words in the line and the number of occurrences of each letter. the output of the letters must be in alphabetical order counting both upper and lowercase versions .(ex input: Hello Hi. And the output: there are 2 words
1 e
2 h
1 i
2 l
1 o
I've done the first part but I couldn't solve the second part. first, I changed all the uppercase to lowercase.then,I found this sort but I have the spaces and other punctuations showing with it..

Code:

 do
 {
   finish = 0;
    for ( i = 0; i < x - 1; i++)      // sorting the string..
    {
      if (s[i] > s[i + 1])
      {
        hold     = s[i];
        s[i]     = s[i + 1];
        s[i + 1] = hold;

        finish = 1;
      }
    }
 }while (finish != 0) ;

* so I decided to delete everything that is not a letter but it didn't work…

Code:

 for (i=0; i<size; i++)                // string contains only letters..
	  if (isalpha(s[i]))
		  temp[i]=s[i] ;

* and I thought of another way that also didn't work..

Code:

 for (i=0; i<size; i++)            // string contains only letters..
	  if (!isalpha(s[i]))
		  s[i]=s[i+1] ;

* finally I decided to do the counting any way, but I have the occurrences of each letter repeated as many times as it occurred, as well as ,counting punctuations.

Code:

 for ( i=0;i<x;i++)                     //counting the occurrences of each letter..
	{	for(j=0;j<x;j++)
		{
			if (s[i]==s[j])
				a++ ;
		}
    cout << s[i] << " occured : " <<a << endl;
	a=0;
	}
	cout << endl; 
	return 0;
}

what can i do ??!!

PJYelton · 05-30-2003, 11:59 AM

Its real easy to create a function that kills punctuation. Just do something like this:

Code:

string killpunc(string s)
{
   string retString="";
   for (int x=0; x<s.length(); x++)
   {
        if (isalpha(s[x]))
            retString+=s[x]+"";
   }
   return retString;
}

Wledge · 05-30-2003, 12:00 PM

Another approach

Code:

#include <map>
#include <cctype>
...
std::map<char,int> abc;
const int size = 100;
char s[size]
....
for(int i = 0;i < size;++i)
{
    if(std::isalpha(s[i]))
        abc[std::tolower(s[i])]++;
}

for(std::map<char,int>::const_iterator iter = abc.begin();iter != abc.end();++iter)
    cout << iter->first << " " << iter->second << endl;

PJYelton · 05-30-2003, 12:01 PM

lol, I think that is a little advanced for her

elad · 05-30-2003, 12:10 PM

you can try to use the asci representation of the characters to identify an index and then increment the value stored at that index. I think it goes something like this:

char string[800];//string to read
int array[26] = {0};//array to hold char count all elements set to zero initially

//loop to read each char in string
//in loop do something like this:
if(isalpha[string[i])//count just letters
{
tolower(string[i]);//convert to lower case
array[string[i] - 'a']++;//increment index obtained by substracting the asci value of 'a' from string[i].
}

in the end the number of a's will be at index 0, b's at index 1, c's at index 2, etc.

girlzone · 05-30-2003, 12:17 PM

yes

i need the program to be simple.we didn't take all the functions!!i know i might sound stupid but what does this mean?

retString+=s[x]+"";
how can you make a string=string+char?

and when i tried it i had an error C2297: '+=' : illegal, right operand has type 'char [1]'

Zach L. · 05-30-2003, 12:23 PM

Both PJYelton and Wledge's examples use STL. The 'string' class defined in <string> has numerous operators defined for various types. For examples, the + operator will concatenate two strings. Those operators are not similarly defined for cahr and char*s.

Wledge's is a little more complex, and probably a bit more elegant. The map is an associative container, meaning it has key/value pairs. In this case, the keys are characters, and ints (which count the number of occurences) are the values. An entry is not created until it is accessed, and the map will initialize ints (as values) to 0, which is why this code maintains an accurate count.

Hope thats a decent explanation.

Cheers

PJYelton · 05-30-2003, 12:36 PM

Oops, I am so used to using strings that I forget not everybody uses them

I take it your input is in the form of a char array? I know there is a function that concatenates two char arrays, but without my book I'm not sure exactly what it is. You can just use that in place of my += line and use char arrays instead of strings.

Zach L. · 05-30-2003, 12:43 PM

From <cstring>
std::strcat(char*, char*);

girlzone · 05-30-2003, 12:48 PM

ok thanks but isn't there another way??

PJYelton · 05-30-2003, 12:57 PM

To do the whole problem or just kill the punctuation?

Zach L. · 05-30-2003, 12:59 PM

There are plenty of solutions... What type are you looking for?

girlzone · 05-30-2003, 01:15 PM

well...if you know a way to count the number of occurrences of each letter in that string in alphabetical order without the need to kill the punctuations it is ok... but that was the way i was thinking to solve the program...

PJYelton · 05-30-2003, 01:32 PM

Well, personally I wouldn't care about the punctuation. If given the problem I probably would have right off the bat done what elad did.

Create an int array of size 26, all initialized to zero. This is where you will hold the number of each letter. Spot 0 will be the number of A's, spot 1 will be the number of B's, etc.

Now go through each letter in the input, if its a non-alpha then skip it and go to the next letter. Whatever letter you are on, increment your int array by one at the corresponding letter. So if you hit the letter Z, then increment array[25] by one. Now, an easy way to do this without having 26 if statements is to simply subtract the letter 'a' from from the current letter. a-a = 0, b-a=1, c-a=2, etc, which happens to be exactly the index in your array you want to increment. So all you need to do is this:

Code:

array[s[i]-'a']++;

So at the end, all you have to do is loop through your array, printing out the amounts of each letter if the value is greater than zero.

Hope that helps, about as far as I go without writing the entire program for ya

hpy_gilmore8 · 10-23-2003, 04:52 PM

I have quick question along the same lines as the original question. What if a person had to count how many times individual letters occurred in a string of text, but did not want to declare it as a char and wanted to use string instead?

For instance instead of this:

Code:

char string[800];//string to read
int array[26] = {0};//array to hold char count all elements set to zero initially

//loop to read each char in string
//in loop do something like this:
if(isalpha[string[i])//count just letters
{
tolower(string[i]);//convert to lower case
array[string[i] - 'a']++;//increment index obtained by substracting the asci value of 'a' from string[i].
}

It would be more along the lines of this:

Code:

string line;//string to read
int array[26] = {0};//array to hold char count all elements set to zero initially

cout << "Enter a line of text:\n";
getline(cin,line);
//loop to read each char in string
//in loop do something like this:
if(isalpha[line[i])//count just letters
{
tolower(line[i]);//convert to lower case
array[line[i] - 'a']++;//increment index obtained by substracting the asci value of 'a' from string[i].
}

I am assuming this doesn't work right because of the desire to declare the variable as string instead of char. Is there a way to declare it as string and then perhaps do a static_cast<char> to the variable somewhere else along the line of code so that the same concept could apply? If not, is there a better way that someone could suggest to do it if they wanted (or perhaps were required) to declare the variable as a string instead of char?

Thanks very much in advance!

05-30-2003, 11:42 AM	#1
girlzone Registered User Join Date: Apr 2003 Posts: 25	sort+counting Hi, I have a program that should read a line of text and output the number of words in the line and the number of occurrences of each letter. the output of the letters must be in alphabetical order counting both upper and lowercase versions .(ex input: Hello Hi. And the output: there are 2 words 1 e 2 h 1 i 2 l 1 o I've done the first part but I couldn't solve the second part. first, I changed all the uppercase to lowercase.then,I found this sort but I have the spaces and other punctuations showing with it.. Code: do { finish = 0; for ( i = 0; i < x - 1; i++) // sorting the string.. { if (s[i] > s[i + 1]) { hold = s[i]; s[i] = s[i + 1]; s[i + 1] = hold; finish = 1; } } }while (finish != 0) ; * so I decided to delete everything that is not a letter but it didn't work… Code: for (i=0; i<size; i++) // string contains only letters.. if (isalpha(s[i])) temp[i]=s[i] ; * and I thought of another way that also didn't work.. Code: for (i=0; i<size; i++) // string contains only letters.. if (!isalpha(s[i])) s[i]=s[i+1] ; * finally I decided to do the counting any way, but I have the occurrences of each letter repeated as many times as it occurred, as well as ,counting punctuations. Code: for ( i=0;i<x;i++) //counting the occurrences of each letter.. { for(j=0;j<x;j++) { if (s[i]==s[j]) a++ ; } cout << s[i] << " occured : " <<a << endl; a=0; } cout << endl; return 0; } what can i do ??!!
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05-30-2003, 11:59 AM	#2
PJYelton Cheesy Poofs! Join Date: Sep 2002 Location: Boulder Posts: 1,728	Its real easy to create a function that kills punctuation. Just do something like this: Code: string killpunc(string s) { string retString=""; for (int x=0; x<s.length(); x++) { if (isalpha(s[x])) retString+=s[x]+""; } return retString; }
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05-30-2003, 12:01 PM	#4
PJYelton Cheesy Poofs! Join Date: Sep 2002 Location: Boulder Posts: 1,728	lol, I think that is a little advanced for her
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05-30-2003, 12:10 PM	#5
elad Registered User Join Date: Mar 2002 Posts: 1,595	you can try to use the asci representation of the characters to identify an index and then increment the value stored at that index. I think it goes something like this: char string[800];//string to read int array[26] = {0};//array to hold char count all elements set to zero initially //loop to read each char in string //in loop do something like this: if(isalpha[string[i])//count just letters { tolower(string[i]);//convert to lower case array[string[i] - 'a']++;//increment index obtained by substracting the asci value of 'a' from string[i]. } in the end the number of a's will be at index 0, b's at index 1, c's at index 2, etc. Last edited by elad; 05-30-2003 at 12:13 PM.
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05-30-2003, 12:17 PM	#6
girlzone Registered User Join Date: Apr 2003 Posts: 25	didn't get it yes i need the program to be simple.we didn't take all the functions!!i know i might sound stupid but what does this mean? retString+=s[x]+""; how can you make a string=string+char? and when i tried it i had an error C2297: '+=' : illegal, right operand has type 'char [1]' __________________ have a nice day
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05-30-2003, 12:23 PM	#7
Zach L. Toaster Join Date: Aug 2001 Posts: 2,686	Both PJYelton and Wledge's examples use STL. The 'string' class defined in <string> has numerous operators defined for various types. For examples, the + operator will concatenate two strings. Those operators are not similarly defined for cahr and chars. Wledge's is a little more complex, and probably a bit more elegant. The map is an associative container, meaning it has key/value pairs. In this case, the keys are characters, and ints (which count the number of occurences) are the values. An entry is not created until it is accessed, and the map will initialize ints (as values) to 0, which is why this code maintains an accurate count. Hope thats a decent explanation. Cheers __________________ The word rap* as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.
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05-30-2003, 12:36 PM	#8
PJYelton Cheesy Poofs! Join Date: Sep 2002 Location: Boulder Posts: 1,728	Oops, I am so used to using strings that I forget not everybody uses them I take it your input is in the form of a char array? I know there is a function that concatenates two char arrays, but without my book I'm not sure exactly what it is. You can just use that in place of my += line and use char arrays instead of strings.
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05-30-2003, 12:43 PM	#9
Zach L. Toaster Join Date: Aug 2001 Posts: 2,686	From <cstring> std::strcat(char, char); __________________ The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.
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05-30-2003, 12:48 PM	#10
girlzone Registered User Join Date: Apr 2003 Posts: 25	ok thanks but isn't there another way?? __________________ have a nice day
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05-30-2003, 12:57 PM	#11
PJYelton Cheesy Poofs! Join Date: Sep 2002 Location: Boulder Posts: 1,728	To do the whole problem or just kill the punctuation?
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05-30-2003, 12:59 PM	#12
Zach L. Toaster Join Date: Aug 2001 Posts: 2,686	There are plenty of solutions... What type are you looking for? __________________ The word rap as it applies to music is the result of a peculiar phonological rule which has stripped the word of its initial voiceless velar stop.
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05-30-2003, 01:15 PM	#13
girlzone Registered User Join Date: Apr 2003 Posts: 25	anything... well...if you know a way to count the number of occurrences of each letter in that string in alphabetical order without the need to kill the punctuations it is ok... but that was the way i was thinking to solve the program... __________________ have a nice day
girlzone is offline

05-30-2003, 01:32 PM	#14
PJYelton Cheesy Poofs! Join Date: Sep 2002 Location: Boulder Posts: 1,728	Well, personally I wouldn't care about the punctuation. If given the problem I probably would have right off the bat done what elad did. Create an int array of size 26, all initialized to zero. This is where you will hold the number of each letter. Spot 0 will be the number of A's, spot 1 will be the number of B's, etc. Now go through each letter in the input, if its a non-alpha then skip it and go to the next letter. Whatever letter you are on, increment your int array by one at the corresponding letter. So if you hit the letter Z, then increment array[25] by one. Now, an easy way to do this without having 26 if statements is to simply subtract the letter 'a' from from the current letter. a-a = 0, b-a=1, c-a=2, etc, which happens to be exactly the index in your array you want to increment. So all you need to do is this: Code: array[s[i]-'a']++; So at the end, all you have to do is loop through your array, printing out the amounts of each letter if the value is greater than zero. Hope that helps, about as far as I go without writing the entire program for ya
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10-23-2003, 04:52 PM	#15
hpy_gilmore8 Registered User Join Date: Mar 2003 Posts: 72	I have quick question along the same lines as the original question. What if a person had to count how many times individual letters occurred in a string of text, but did not want to declare it as a char and wanted to use string instead? For instance instead of this: Code: char string[800];//string to read int array[26] = {0};//array to hold char count all elements set to zero initially //loop to read each char in string //in loop do something like this: if(isalpha[string[i])//count just letters { tolower(string[i]);//convert to lower case array[string[i] - 'a']++;//increment index obtained by substracting the asci value of 'a' from string[i]. } It would be more along the lines of this: Code: string line;//string to read int array[26] = {0};//array to hold char count all elements set to zero initially cout << "Enter a line of text:\n"; getline(cin,line); //loop to read each char in string //in loop do something like this: if(isalpha[line[i])//count just letters { tolower(line[i]);//convert to lower case array[line[i] - 'a']++;//increment index obtained by substracting the asci value of 'a' from string[i]. } I am assuming this doesn't work right because of the desire to declare the variable as string instead of char. Is there a way to declare it as string and then perhaps do a static_cast<char> to the variable somewhere else along the line of code so that the same concept could apply? If not, is there a better way that someone could suggest to do it if they wanted (or perhaps were required) to declare the variable as a string instead of char? Thanks very much in advance!
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