Histogram question

[zaturn]

To student:



The this the new assignment.



Ex7 VERTICAL HISTOGRAM

Write a program that reads text and prints a vertical histogram (bar graph) representing the occurrences of each letter of the alphabet (A-Z) in the text. Characters other than the letters (A-Z) are to be ignored. You can assume that no lower case letters appear in the text.



Text to be tested:



"THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG. THIS IS AN EXAMPLE OF HOW TO TEST YOUR HISTOGRAM

PROGRAM. YOU CAN USE THIS EXAMPLE."



Sample Run

*

*

* *

* * *

* * * **

* * * * ***

* * ** * * ****

* * ** * ** ****

* * *** ***** **** **

* * ***** ***** **** ***

**************************

ABCDEFGHIJKLMNOPQRSTUVWXYZ


End of problem.



I don't know .How do i start?

[gpraghuram]

First have a 2 dimensional array which stores the occurances of every alphabet.
Increment the value depending on the character u read.
Finally you need to print it

Raghu

[newb16]

@gpraghuram
There is no need in two-dimensional array, one-dimensional array is enough. No need in 2d array during print phase as well.

[JosAH]

@OP: I changed your thread title. Next time use a proper title please; text speak is not much appreciated in this forum.

kind regards,

Jos (moderator)

[weaknessforcats]

When you do your display, to get the image correctly, make a pass of the array and find the largest number.

Then for the actual print, you make a pass of the array and print a space for each element whose value is less than the largest number. Only the element equal to the largest number is printed as an *.

Repeat the above after reducing the largest value by 1.

When yo get ot zero, print the letters corresponding to your array elements.

[whodgson]

Try puting the sentence into an array like:

char dog[]={"THE QUICK BROWN FOX etc"};
//Now use a for loop to traverse the array like:
int j=char(65);
while(j<91)    
  for(int i=0;i<27;i++)
     {
       if(array[i]==char( j ))cout<<"*"<<"\n";//prints a new line after each splat
       if(array[i]==26)cout<<"\t";//prints a tab after each pass
       j++;
    } 

Now print the uppercase alphbet on one line with a tab between each.
Might work.

[whodgson]

I think the following would produce an histogram based on the y axis

int j=65;//the char A
while(j<91)
    {
     for(int i=0;i<44;i++)//the number of letters and spaces in the sentence
     {if(i<1)cout<<"\n";//staarts a new line before every traverse of fox[]
     if(fox[i]==j)cout<<char (j)<<" ";//prints the letter not *
     else cout<<""; }
      j++;
 
    }
 

[zaturn]

Thx Everybody This my histrogram

 
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
 
int main ( void )
{start:
int num[26] = {};
int height=0,;
int y1,x1;
int a=0,b=0,c=0,d=0,e=0,f=0,g=0,h=0,i=0,j=0,k=0,l=0,m=0,n=0,o=0,p=0,q=0,
r=0,s=0,t=0,u=0,v=0,w=0,x=0,y=0,z=0;
string text;
cout<<"Input 0 to exit or Type your text : ";
getline(cin,text);
if(text=="0")goto finish;
cout<<'\n';
for(int count=0;count<text.length();++count)
{
if(text.substr(count,1)=="A"||text.substr(count,1)=="a"){++a;num[0]=a;}
if(text.substr(count,1)=="B"||text.substr(count,1)=="b"){++b;num[1]=b;}
if(text.substr(count,1)=="C"||text.substr(count,1)=="c"){++c;num[2]=c;}
if(text.substr(count,1)=="D"||text.substr(count,1)=="d"){++d;num[3]=d;}
if(text.substr(count,1)=="E"||text.substr(count,1)=="e"){++e;num[4]=e;}
if(text.substr(count,1)=="F"||text.substr(count,1)=="f"){++f;num[5]=f;}
if(text.substr(count,1)=="G"||text.substr(count,1)=="g"){++g;num[6]=g;}
if(text.substr(count,1)=="H"||text.substr(count,1)=="h"){++h;num[7]=h;}
if(text.substr(count,1)=="I"||text.substr(count,1)=="i"){++i;num[8]=i;}
if(text.substr(count,1)=="J"||text.substr(count,1)=="j"){++j;num[9]=j;}
if(text.substr(count,1)=="K"||text.substr(count,1)=="k"){++k;num[10]=k;}
if(text.substr(count,1)=="L"||text.substr(count,1)=="l"){++l;num[11]=l;}
if(text.substr(count,1)=="M"||text.substr(count,1)=="m"){++m;num[12]=m;}
if(text.substr(count,1)=="N"||text.substr(count,1)=="n"){++n;num[13]=n;}
if(text.substr(count,1)=="O"||text.substr(count,1)=="o"){++o;num[14]=o;}
if(text.substr(count,1)=="P"||text.substr(count,1)=="p"){++p;num[15]=p;}
if(text.substr(count,1)=="Q"||text.substr(count,1)=="q"){++q;num[16]=q;}
if(text.substr(count,1)=="R"||text.substr(count,1)=="r"){++r;num[17]=r;}
if(text.substr(count,1)=="S"||text.substr(count,1)=="s"){++s;num[18]=s;}
if(text.substr(count,1)=="T"||text.substr(count,1)=="t"){++t;num[19]=t;}
if(text.substr(count,1)=="U"||text.substr(count,1)=="u"){++u;num[20]=u;}
if(text.substr(count,1)=="V"||text.substr(count,1)=="v"){++v;num[21]=v;}
if(text.substr(count,1)=="W"||text.substr(count,1)=="w"){++w;num[22]=w;}
if(text.substr(count,1)=="X"||text.substr(count,1)=="x"){++x;num[23]=x;}
if(text.substr(count,1)=="Y"||text.substr(count,1)=="y"){++y;num[24]=y;}
if(text.substr(count,1)=="Z"||text.substr(count,1)=="z"){++z;num[25]=z;}
}
for (x1=0; x1<26; x1++) height = max(height, num[x1]);
for (y1=height; y1>=1; y1--)
{
cout << setw(3) << y1 << "|";
for (x1=0; x1<26; x1++)
if (num[x1]>=y1) cout << " *"; else cout << " ";
cout << endl;
}
cout << "---+" << string(26*2, '-') << endl;
cout << " |";
for (x1=0; x1<26; x1++)
{
if (x1<26)
cout << " " << char('A'+x1);
 
}
cout << endl << endl;goto start;
finish:
 
system("pause");
}
 

[newb16]

Use text[count] and compare it with char literal ( single quotes, 'A' instead of "A" )
Do no unroll inner loop 'A' .. 'Z' - i.e. use loop from 'A' to 'Z' and make variables a .. z an array ( oh wait.. there is already num[], why do you need another set of variables? )
use toUpper().

[whodgson]

This is the output from thread#7
Forms a histogram of letters in a sentence

THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG

A
B
C
D
E E E
F
G
H H
I
J
K
L
M
N
O O O O
P
Q
R R
S
T T
U U
V
W
X
Y
Z
*/
Satisfactory?

[JosAH]

@whodgson
Nope; the OP wants that histogram rotated 90 degrees.

kind regards,

Jos

[whodgson]

ok JosAH
how about this:

 int jcount=0;
  //get the quantity of each letter and record in dog array.
    j=65;//'A"
    while(j<91) //'Z'
    { 
      jcount=0;
      for(int i=0;i<80;i++)//number of letters & spaces in sentence
         { if(fox[i]==j)jcount++;
           dog[j]=jcount;
         }
      j++;
     }
 
    for(int j=65;j<91;j++)//print the count of each letter   
       { cout<<dog[j]<<" ";}
        cout<<"\n";
    int count = 0;
    while(count<8)//print the rows of asterisks
       {for(int j=65;j<91;j++) 
         if(dog[j]>count)cout<<"* ";
         else cout<<"  ";
         if(j>=90)cout<<"\n";
         count++;
         if(count==8)break;
       }
 

which produces this:
Forms a vertical histogram of the quantity of each
letter in a sentence

THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
WHICH IS A GOOD WAY TO FRIGHTEN DOGS





A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
3 1 2 3 4 2 4 5 4 1 1 1 1 2 8 1 1 3 3 4 2 1 3 1 2 1
* * * * * * * * * * * * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * *
* * * * * * * * * * *
* * * * * *
* *
*
*
*


.............never again!

. .

[JosAH]

@whodgson
But now it is upside down; you have to flip it 180 degrees ;-)

kind regards,

Jos

[whodgson]

ok Jos
180° flip cummin up;

 int count = 8;
    while(count>-1)//print the histogram asterisks
       {for(int j=65;j<91;j++) 
         if(dog[j]> count)cout<<"* ";
         else cout<<"  ";
         if(j>=90)cout<<"\n";
         count--;
         if(count==-1)break;
       }

which produces this:
THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG
WHICH IS A GOOD WAY TO FRIGHTEN DOGS
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
3 1 2 3 4 2 4 5 4 1 1 1 1 2 8 1 1 3 3 4 2 1 3 1 2 1

*
*
*
* *
* * * * * *
* * * * * * * * * * *
* * * * * * * * * * * * * * * *
* * * * * * * * * * * * * * * * * * * * * * * * * *
ok?

[JosAH]

@whodgson
Almost; according to the OP's example the 'A B ... Z' line should be at the bottom, but that change is futile; well done and happy New Year!

kind regards,

Jos

[whodgson]

@JosAH
Many thanks your patience (& exactitude!) also compliments of new season.
w.