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Print count of unique values


 
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# 1  
Print count of unique values

Hello experts,

I am converting a number into its binary output as :

Code:
 read n
echo "obase=2;$n" | bc

I wish to count the maximum continuous occurrences of the digit 1.

Code:
 
 Example : 
  
 1. The binary equivalent of 5 = 101. Hence the output must be 1.
 2. The binary equivalent of 6 = 110. Hence the output must be 2.
 3. The binary equivalent of 9 = 1001. Hence the output must be 1.
 4. The binary equivalent of 13 = 1101. Hence the output must be 2

Could you please help.

Thanks,
Haider
# 2  
how about this ?
Code:
$ echo "111110000111000" | tr '0' '\n' | sort -nr | awk 'NR==1{print length}'
5

$ echo "111001100111000" | tr '0' '\n' | sort -nr | awk 'NR==1{print length}'
3

$ echo "1110011001111000" | tr '0' '\n' | sort -nr | awk 'NR==1{print length}'
4

# 3  
Hello H squared,

Could you please try following.
Code:
cat script.ksh
echo "Enter a number here.."
read num
echo "obase=2;$num" | bc | awk '{print $0;while(match($0,/1+/)){;q=q>substr($0,RSTART,RLENGTH)?q:substr($0,RSTART,RLENGTH);$0=substr($0,RLENGTH+1)}} END{print length(q)}'

So let's say we will run it as follows.
Code:
./script.ksh
Enter a number here..
12312446577991213
101011101111100001101011011100100110001101111000101101
5

./script.ksh
Enter a number here..
12111113312
1011010001111000001110110001100000
4

EDIT: Adding a non-one liner form of solution on same now.
Code:
echo "Enter a number here.."
read num
echo "obase=2;$num" | bc | awk '{print $0;
                                 while(match($0,/1+/)){
                                                        q=q>substr($0,RSTART,RLENGTH)?q:substr($0,RSTART,RLENGTH);
                                                        $0=substr($0,RLENGTH+1)
                                                      }
                                }
                                 END                  {
                                                        print length(q)
                                                      }
                               '

Thanks,
R. Singh

Last edited by RavinderSingh13; 08-26-2016 at 04:06 AM.. Reason: Added a non-one liner form of solution now.
This User Gave Thanks to RavinderSingh13 For This Post:
# 4  
In a bash script, you could split the record on zeros then count the length of each 'field' it creates, perhaps like this:-
Code:
#!/bin/bash

read -p "Enter a number here.. " num

one_strings="$(echo \"obase=2\;$num\" | bc | tr '0' ' ' )"

max_found_length=0
for a_string in $one_strings        # Unquoted to ensure it splits up for the read
do
   a_string_length="${#a_string}"
   if [ $a_string_length -gt $max_found_length ]
   then
      max_found_length=$a_string_length
   fi
done

echo "Max length of consecutive 1's found was $max_found_length"

Apart from the conversion to binary, this is all internal to the shell, so should be quick.

For ksh you will need to change the read:-
Code:
read  num?"Enter a number here.. "



I hope that this helps,
Robin

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