[SOLVED] Arithmetic operation between 2 Arrays to generate a new Array

astalavista2000 · 10-17-2013, 04:39 PM

Hi, I have defined to different arrays, ARRAY1 and ARRAY2. Each of them has 20 elements (integer numbers).

Code:

#!/bin/bash

ARRAY1=(0 3 7 3 9 10 6 8 8 8 7 3 8 4 4 7 1 6 7 6)

ARRAY2=(1 4 4 2 1 5 4 6 1 3 5 5 3 6 1 1 6 1 2 2)

for ((i; i < 20; i++))
do

   if [ ${ARRAY1[i]} -ge ${ARRAY2[i-1]} ]
    then
     ARRAY3+=(${ARRAY2[i]})
    else
     ARRAY3+=$(( (${ARRAY2[i-1]}) - (${ARRAY1[i]}) + (${ARRAY2[i]}) ))

        
#Double brackets in each side to do with arithmetic operation
    fi
done

echo ${ARRAY3[@]}

I have to compare each element in the position "i" in ARRAY1 to the element in position "i-1" in ARRAY2. If the element in ARRAY1 is either equal or greater than the element in ARRAY2, this element in ARRAY2 is added to a new ARRAY3. Otherwise I add to ARRAY 3 (element i-1 in ARRAY2 - element i in Array1 + element i in ARRAY3)

As an example.
ARRAY1[1]=3
ARRAY2[1-1]=1
ARRAY2[1]=4

Then element in ARRAY3 = ARRAY2[1] = 4

If if only test one set of elements,

Code:

  if [ ${ARRAY1[12]} -ge ${ARRAY2[11]} ]
    then
        ARRAY3+=(${ARRAY2[12]})
    else
        ARRAY3+=$(( (${ARRAY2[11]}) - (${ARRAY1[12]}) + (${ARRAY2[12]}) ))
   fi

the code function as expected, because it gives 3 as solution.
But when put the code inside of the for loop it doesn't work.
I think that the problem lays on how I put [i-1] inside the if clause.
Another doubt that I have is how is going to behave the code in the first iteration, because there is no [i-1] in first element of ARRAY2.

I hope you can shade some light in this issue.
Thanks!!

suicidaleggroll · 10-17-2013, 04:54 PM

Why are you calling 3 an array? The way you're using it in your code it's just a number.

Anyway, you're not initializing "i" to anything, so of course it's not working. You'd probably want to start it at 1, since there is no "-1" element in array2.

astalavista2000 · 10-17-2013, 07:48 PM

Quote:

Originally Posted by suicidaleggroll

Why are you calling 3 an array? The way you're using it in your code it's just a number.

Anyway, you're not initializing "i" to anything, so of course it's not working. You'd probably want to start it at 1, since there is no "-1" element in array2.

3 is the result of the example iteration. In that case it acts as a simple variable, but it is just and example.
What I need is to generate an array, that's way I put the code inside a for loop.

suicidaleggroll, can you explain to me how do I correctly initialize "i"?
Thanks!

suicidaleggroll · 10-17-2013, 08:10 PM

Quote:

Originally Posted by astalavista2000

3 is the result of the example iteration. In that case it acts as a simple variable, but it is just and example.
What I need is to generate an array, that's way I put the code inside a for loop.

Yet you're still treating "ARRAY3" as a scalar. Where's the array indexing that lets the program know in which index it needs to place the result of your calculation?

Quote:

Originally Posted by astalavista2000

suicidaleggroll, can you explain to me how do I correctly initialize "i"?

By using the correct syntax in your for loop:
http://www.freeos.com/guides/lsst/ch03sec06.html

The way you're using the for loop in your code is literally no different than the way you do a for loop in any other language. There are three things you need to supply to any for loop:
1) Where to start
2) Where to stop
3) How to increment

You have #2 and #3 down, but not #1. In other words, you've told it to stop when i>=20, and to increment by 1 (i++), but you never told it where to start.

astalavista2000 · 10-17-2013, 09:07 PM

Quote:

Originally Posted by suicidaleggroll

Yet you're still treating "ARRAY3" as a scalar. Where's the array indexing that lets the program know in which index it needs to place the result of your calculation?

By using the correct syntax in your for loop:
http://www.freeos.com/guides/lsst/ch03sec06.html

The way you're using the for loop in your code is literally no different than the way you do a for loop in any other language. There are three things you need to supply to any for loop:
1) Where to start
2) Where to stop
3) How to increment

You have #2 and #3 down, but not #1. In other words, you've told it to stop when i>=20, and to increment by 1 (i++), but you never told it where to start.

OK. Now I understand what you are saying. I thought that there is no need to specify the index because it puts the result of the calculation in the "i" position determined by the for loop. And each time it iterates, there is a new available position.
Can you explain to me how do I treat ARRAY3 as an array, and how do I correct this line?
I really appreciate the time that you're taking to help me!

This is how I corrected the for loop. The code is almost ready.

Code:

for ((i=0; i<=20; i++))
 do

   if [ ${ARRAY1[i]} -ge ${ARRAY2[i-1]} ]
    then
        ARRAY3+=(${ARRAY2[i]})
    else
        ARRAY3+=$(( (${ARRAY2[i-1]}) - (${ARRAY1[i]}) + (${ARRAY2[i]}) ))
   fi
done

PD. If I put i=1 instead of i=0 in the for loop I will avoid the situation that I described in which there is no i-1, won't I?

suicidaleggroll · 10-17-2013, 10:56 PM

You simply need to assign the value to the correct index, and remove the "+=" since you're not incrementing anything, you're assigning:

Code:

ARRAY3[$i]=${ARRAY2[$i]}

etc.

astalavista2000 · 10-18-2013, 01:53 AM

Quote:

Originally Posted by suicidaleggroll

You simply need to assign the value to the correct index, and remove the "+=" since you're not incrementing anything, you're assigning:

Code:

ARRAY3[$i]=${ARRAY2[$i]}

etc.

I'm almost there.

Code:

 for ((i=1; i<=20; i++))
                do



[LINE 22]               if [ ${ARRAY1[i]} -ge ${ARRAY2[i-1]} ]
                        then
                        Vector[$i]=${ARRAY2[i]}
                        else
[LINE 26]               Vector[$i]=$(( (${ARRAY2[i-1]}) - (${ARRAY1[i]}) + (${ARRAY2[i]}) ))
                        fi


                done

#I add the first position, where i-1 doesn't exist
ARRAY3[0]=${ARRAY2[0]}




echo ${ARRAY3[*]}

This is works as expected but with some errors. I got the following result

Code:

linux:/home/admin/simulacion# ./comparaVec.sh 
./comparaVec.sh: line 22: [: -ge: unary operator expected
./comparaVec.sh: line 26: (2) - () + () : syntax error: operand expected (error token is ") + () ")
1 4 4 3 1 5 4 6 1 3 5 7 3 6 3 1 6 1 2 2

As you can see the ARRAY3 is showing the correct 20 numbers that are generated after the operations between ARRAY1 and ARRAY2.
I just need to correct the errors. Can you help me with that?
Thanks!

grail · 10-18-2013, 02:49 AM

What is the value in either array 1 or 2 when i = 20??

astalavista2000 · 10-18-2013, 05:29 AM

Quote:

Originally Posted by grail

What is the value in either array 1 or 2 when i = 20??

Grail, all the values of ARRAY1 and ARRAY2 are at the begining of the code.
Here is what you've asked:

ARRAY1[20]=6
ARRAY2[20]=2
ARRAY2[20-1]=2

if [ARRAY1[20] >= ARRAY2[20-1]]
Then
ARRAY3[20]=ARRAY2[20] that is "2"

You can see that the result of this operation is the same that the last element of ARRAY3 (I wrote the output in my previous post.
Thanks

astalavista2000 · 10-18-2013, 07:42 AM

Quote:

Originally Posted by astalavista2000

Grail, all the values of ARRAY1 and ARRAY2 are at the begining of the code.
Here is what you've asked:

ARRAY1[20]=6
ARRAY2[20]=2
ARRAY2[20-1]=2

if [ARRAY1[20] >= ARRAY2[20-1]]
Then
ARRAY3[20]=ARRAY2[20] that is "2"

You can see that the result of this operation is the same that the last element of ARRAY3 (I wrote the output in my previous post.
Thanks

SOLVED!

I debug the code and re-execute the script.
The Error was this: "./comparaVec.sh: line 22: [: -ge: unary operator expected"
And the last iteration in the debug mode showed this: + '[' -ge 2 ']'
-ge is a binary operator and I need two arguments. But, when i=20 ARRAY1 is empty! So I simple replace the i<=20 with i<20.
Thanks to both of you, suicidaleggroll for your help!

ntubski · 10-18-2013, 08:44 AM

By the way, += does work to append to an array.

3.4 Shell Parameters:

Quote:

When ‘+=’ is applied to an array variable using compound assignment (see Arrays), the variable’s value is not unset (as it is when using ‘=’), and new values are appended to the array beginning at one greater than the array’s maximum index (for indexed arrays), or added as additional key-value pairs in an associative array.

astalavista2000 · 10-19-2013, 02:41 AM

Quote:

Originally Posted by ntubski

By the way, += does work to append to an array.

3.4 Shell Parameters:

So ARRAY3[i]= and ARRAY3+= are the same, aren't they?

grail · 10-19-2013, 07:49 AM

Quote:

Originally Posted by astalavista2000

So ARRAY3[i]= and ARRAY3+= are the same, aren't they?

No they are not the same unless 'i' is exactly in order with single index spacing.

So if you compare adding the first element to an array:

Code:

ARRAY3[3]="foo" # this places the string in the fourth index position

ARRAY4+=( "foo" ) # this places the string at the first index position

suicidaleggroll · 10-19-2013, 09:44 AM

Quote:

Originally Posted by ntubski

By the way, += does work to append to an array.

I did not realize that, thanks.

astalavista2000 · 10-19-2013, 09:44 PM

Quote:

Originally Posted by grail

No they are not the same unless 'i' is exactly in order with single index spacing.

So if you compare adding the first element to an array:

Code:

ARRAY3[3]="foo" # this places the string in the fourth index position

ARRAY4+=( "foo" ) # this places the string at the first index position

Great, I see the difference now. Thanks for being so clear grail.
I'm marking this thread as solved.
Thanks to all of you for being so helpful.