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justvince
Level II

Convert decimal to Binary

I need to convert Columns of data (decimal) to binary (8 bit) , then concantenate say the 4 LSB to the next column of data with the 4 MSB.  Is there a way to at least convert decimal data to binary in JMP? 

1 ACCEPTED SOLUTION

Accepted Solutions
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Re: Convert decimal to Binary

As previously stated:

 

Hi - can you adapt the following to do what you need?

 

 

// Function to convert an integer into any positive base up to hexadecimal;
base = Function( {n, nbase}, {default local},        ndash = "";        digits= "0123456789ABCDEF";      While( n > 0,               remainder= Mod( n, nbase );               newdigit= Substr( digits, (1 + remainder), 1 );               ndash= newdigit || ndash;               n= (n - remainder)/ nbase;        ); ndash; );   // Print out a few numbers to show it working in binary;   For( i = 1, i <= 32, i++, Show( Char( i ) || " = " || base( i, 2 ) ) );   // Or create some new columns in a data table and fill them;   dt = New Table( "Number Bases", New Column( "N", numeric, values( Index( 1, 32 ) ) ) );   numberbase = {2, 3, 8, 10, 12, 16}; // Some familiar number bases;   For( i = 1, i <= N Items( numberbase ), i++,      k= numberbase[i];      dt<< New Column( "Base "|| Char( k ), character );      For( j = 1, j <= N Row( dt ), j++,      Column( dt, ("Base "|| Char( k )) )[j] = base( j, k ) );      );   // End of program;

 

 

And:

 

No problem - an extra "if" clause will treat n=0 as a special case:

 

 

base = function({n, base}, {default local}, ndash = ""; digits = "0123456789ABCDEF";
if(n > 0, while(n > 0, remainder = mod(n, base); newdigit = substr(digits, (1 + remainder), 1);
ndash = newdigit || ndash; n = (n - remainder) / base), ndash = "0"); ndash);

 

 

(I'm assuming that n is a non-negative integer, and that base is a positive integer.)

Duane Hayes

View solution in original post

4 REPLIES 4
Highlighted

Re: Convert decimal to Binary

Hi - can you adapt the following to do what you need?

// Function to convert an integer into any positive base up to hexadecimal;

base = Function( {n, nbase}, {default local},

       ndash = "";

       digits= "0123456789ABCDEF";

     While( n > 0,

              remainder= Mod( n, nbase );

              newdigit= Substr( digits, (1 + remainder), 1 );

              ndash= newdigit || ndash;

              n= (n - remainder)/ nbase;

       );

ndash;

);

// Print out a few numbers to show it working in binary;

For( i = 1, i <= 32, i++, Show( Char( i ) || " = " || base( i, 2 ) ) );

// Or create some new columns in a data table and fill them;

dt = New Table( "Number Bases", New Column( "N", numeric, values( Index( 1, 32 ) ) ) );

numberbase = {2, 3, 8, 10, 12, 16}; // Some familiar number bases;

For( i = 1, i <= N Items( numberbase ), i++,

     k= numberbase[i];

     dt<< New Column( "Base "|| Char( k ), character );

     For( j = 1, j <= N Row( dt ), j++,

     Column( dt, ("Base "|| Char( k )) )[j] = base( j, k ) );

     );

// End of program;

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justvince
Level II

Re: Convert decimal to Binary

It is a start.  But it appears the function base will not convert a zero to binary 0,

Highlighted

Re: Convert decimal to Binary

No problem - an extra "if" clause will treat n=0 as a special case:

base = function({n, base}, {default local}, ndash = ""; digits = "0123456789ABCDEF";

if(n > 0, while(n > 0, remainder = mod(n, base); newdigit = substr(digits, (1 + remainder), 1);

ndash = newdigit || ndash; n = (n - remainder) / base), ndash = "0"); ndash);

(I'm assuming that n is a non-negative integer, and that base is a positive integer.)

Highlighted

Re: Convert decimal to Binary

As previously stated:

 

Hi - can you adapt the following to do what you need?

 

 

// Function to convert an integer into any positive base up to hexadecimal;
base = Function( {n, nbase}, {default local},        ndash = "";        digits= "0123456789ABCDEF";      While( n > 0,               remainder= Mod( n, nbase );               newdigit= Substr( digits, (1 + remainder), 1 );               ndash= newdigit || ndash;               n= (n - remainder)/ nbase;        ); ndash; );   // Print out a few numbers to show it working in binary;   For( i = 1, i <= 32, i++, Show( Char( i ) || " = " || base( i, 2 ) ) );   // Or create some new columns in a data table and fill them;   dt = New Table( "Number Bases", New Column( "N", numeric, values( Index( 1, 32 ) ) ) );   numberbase = {2, 3, 8, 10, 12, 16}; // Some familiar number bases;   For( i = 1, i <= N Items( numberbase ), i++,      k= numberbase[i];      dt<< New Column( "Base "|| Char( k ), character );      For( j = 1, j <= N Row( dt ), j++,      Column( dt, ("Base "|| Char( k )) )[j] = base( j, k ) );      );   // End of program;

 

 

And:

 

No problem - an extra "if" clause will treat n=0 as a special case:

 

 

base = function({n, base}, {default local}, ndash = ""; digits = "0123456789ABCDEF";
if(n > 0, while(n > 0, remainder = mod(n, base); newdigit = substr(digits, (1 + remainder), 1);
ndash = newdigit || ndash; n = (n - remainder) / base), ndash = "0"); ndash);

 

 

(I'm assuming that n is a non-negative integer, and that base is a positive integer.)

Duane Hayes

View solution in original post

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