Programming Forums > Application Development > C

aznluvsmc · Aug 19th, 2005, 9:46 PM

Hi,

I'm working on this problem on how to count the average number of letters per word for a given input. The input is ended with a EOF in stead of the <Enter> key.

The following is my current code.

(Toggle Plain Text)

#include <stdio.h>
#include <ctype.h>

int main(void)
        {
        unsigned int letters = 0, words = 0;
        int ch;

        printf("Enter something: ");
        while((ch = getchar()) != EOF)
                {
                words += !!isspace(ch);
                letters += (!!isupper(ch) || !!islower(ch));
                }
        printf("\nNumber of words: %u", words);
        printf("\nThe average letters per word is %.2f\n", (float)letters / words);

        return 0;
        }

The problem is that my logic of determing a word has having a space following it is not correct as multiple spaces will cause the word count to be off. Do you guys have any algorithms for doing a word count?

Sane · Aug 19th, 2005, 10:08 PM

Finding a letter will turn a gate to True.

Finding a space will turn the gate to False.

Right before the gate is turned False, if it used to be True, then the wordcount += 1.

:p

Since I don't know C, this is it implemented in Python (because I'm bored):

(Toggle Plain Text)

x = """
Try  to  count   the   number of words in   this!


Yo..."""



x = x.replace('\n',' ') + ' ' #Python registers newline as a word

words = 0
gate = False

for a in x:
    if a == ' ':
        gate = True
    else:
        if gate:
            words += 1
        gate = False

print words

>>> 10

Then the average number of letters could be calculated by changing:
gate = True

to:
gate += 1

, and then registering gate to an array before it resets to 0.

aznluvsmc · Aug 19th, 2005, 11:29 PM

I'm not really sure what you mean there since I don't know Python.

Sane · Aug 20th, 2005, 2:02 AM

Just do what I said: :p

The gate variable starts at 0.

Finding a letter will turn the gate to 1.

Finding a space will turn the gate to 0.

Right before the gate is turned 0 when a space is found, if the gate == 1: then the wordcount += 1.

Ooble · Aug 20th, 2005, 6:14 AM

Why do you use !!isspace(ch)? I take it this is to make sure it only returns 1 or 0, but wouldn't this be better done by using the ternary operator, rather than resorting to some hack?

(Toggle Plain Text)

words += isspace(ch) ? 1 : 0;

Or, as you're not incrementing the variable at all when isspace(ch) returns 0, why bother with that, even?

(Toggle Plain Text)

if (isspace(ch))
{
    words++;
}

DaWei · Aug 20th, 2005, 6:27 AM

You need to tokenize on more than spaces. I don't think you'd want to assign the '.' to the letter count of "spaces" in the foregoing sentence. How many words is, "I went to the fair.I came back."? How many words is, "He has a Ph.D." Counting unique words is even less trivial. How many different words is, "He is a frog among frogs."?

stevengs · Aug 20th, 2005, 6:30 AM

Quote:

Originally Posted by DaWei

"He is a frog among frogs."?

is his name Jeremiah ?

DaWei · Aug 20th, 2005, 6:41 AM

Dang you, Steven, that'll be running through my head all cottonpickin' day now!

aznluvsmc · Aug 20th, 2005, 9:05 AM

Quote:

Originally Posted by Ooble

Why do you use !!isspace(ch)?

This was a programming trick I learned in school. I just wanted to employ it so that I don't have to use an if statement. There isn't any real good reason to use a double negation. I just wanted to be fancy.

Still learning what I can and can't do with C. :o

stevengs · Aug 20th, 2005, 10:24 AM

Quote:

Originally Posted by DaWei

Dang you, Steven, that'll be running through my head all cottonpickin' day now!

Well, there are certainly worse tunes to have boppin' around up there. (like: "There Coming to Take Me Away" - Napoleon XIV.... colleagues start getting worried when you sing that from within your cubicle

)

Aug 19th, 2005, 9:46 PM	#1
aznluvsmc Hobbyist Programmer Join Date: Aug 2005 Posts: 137 Rep Power: 4	How to determine if it's a word Hi, I'm working on this problem on how to count the average number of letters per word for a given input. The input is ended with a EOF in stead of the <Enter> key. The following is my current code. (Toggle Plain Text) #include <stdio.h> #include <ctype.h> int main(void) { unsigned int letters = 0, words = 0; int ch; printf("Enter something: "); while((ch = getchar()) != EOF) { words += !!isspace(ch); letters += (!!isupper(ch) \|\| !!islower(ch)); } printf("\nNumber of words: %u", words); printf("\nThe average letters per word is %.2f\n", (float)letters / words); return 0; } #include <stdio.h> #include <ctype.h> int main(void) { unsigned int letters = 0, words = 0; int ch; printf("Enter something: "); while((ch = getchar()) != EOF) { words += !!isspace(ch); letters += (!!isupper(ch) \|\| !!islower(ch)); } printf("\nNumber of words: %u", words); printf("\nThe average letters per word is %.2f\n", (float)letters / words); return 0; } The problem is that my logic of determing a word has having a space following it is not correct as multiple spaces will cause the word count to be off. Do you guys have any algorithms for doing a word count?

Aug 19th, 2005, 10:08 PM	#2
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,886 Rep Power: 5	Finding a letter will turn a gate to True. Finding a space will turn the gate to False. Right before the gate is turned False, if it used to be True, then the wordcount += 1. :p Since I don't know C, this is it implemented in Python (because I'm bored): (Toggle Plain Text) x = """ Try to count the number of words in this! Yo...""" x = x.replace('\n',' ') + ' ' #Python registers newline as a word words = 0 gate = False for a in x: if a == ' ': gate = True else: if gate: words += 1 gate = False print words x = """ Try to count the number of words in this! Yo...""" x = x.replace('\n',' ') + ' ' #Python registers newline as a word words = 0 gate = False for a in x: if a == ' ': gate = True else: if gate: words += 1 gate = False print words >>> 10 Then the average number of letters could be calculated by changing: gate = True to: gate += 1 , and then registering gate to an array before it resets to 0. __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

Aug 20th, 2005, 2:02 AM	#4
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,886 Rep Power: 5	Just do what I said: :p The gate variable starts at 0. Finding a letter will turn the gate to 1. Finding a space will turn the gate to 0. Right before the gate is turned 0 when a space is found, if the gate == 1: then the wordcount += 1. __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

Aug 20th, 2005, 6:14 AM	#5
Ooble I eat cake for breakfast. Join Date: Jul 2004 Location: In my box. Posts: 4,434 Rep Power: 9	Why do you use !!isspace(ch)? I take it this is to make sure it only returns 1 or 0, but wouldn't this be better done by using the ternary operator, rather than resorting to some hack? (Toggle Plain Text) words += isspace(ch) ? 1 : 0; words += isspace(ch) ? 1 : 0; Or, as you're not incrementing the variable at all when isspace(ch) returns 0, why bother with that, even? (Toggle Plain Text) if (isspace(ch)) { words++; } if (isspace(ch)) { words++; } __________________ Me :: You :: Them

Aug 20th, 2005, 6:27 AM	#6
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	You need to tokenize on more than spaces. I don't think you'd want to assign the '.' to the letter count of "spaces" in the foregoing sentence. How many words is, "I went to the fair.I came back."? How many words is, "He has a Ph.D." Counting unique words is even less trivial. How many different words is, "He is a frog among frogs."? __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Aug 19th, 2005, 11:29 PM	#3
aznluvsmc Hobbyist Programmer Join Date: Aug 2005 Posts: 137 Rep Power: 4	I'm not really sure what you mean there since I don't know Python.

Aug 20th, 2005, 6:41 AM	#8
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Dang you, Steven, that'll be running through my head all cottonpickin' day now! __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

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Programming Forums > Application Development > C

How to determine if it's a word