How to identify the position of a sequence of elements in an array in bash - arrays

I'm learning bash and I'm trying identify a sequence elements (patch or subarray) position into a array.
For example:
array=(9 5 8 3 2 7 5 9 0 1 1 5 4 3 8 9 6 2 6 5 7 9 8);
patch=(0 1 1 5)
I would like to obtain a output equals to 8 (start position of my patch in relation to array) or 11 (final position).

bash doesn't really have any built-in facility to do this; you need to walk the array yourself:
for ((i=0; i<${#array[#]}; i++)); do
for ((j=0; j<${#patch[#]}; j++)); do
# Make sure the corresponding elements
# match, or give up. RHS is quoted to ensure
# actual string equality, rather than just pattern matching
[[ ${array[i+j] == "${patch[j]}" ]] || break
done
if [[ $j == ${#patch[#] ]]; then
# All the comparisons succeeded!
start=$i
finish=$((i+j-1))
break
fi
done

Related

'if' in array w/ Shell Script

I have an array with different values, and if the value is 3 (integer) this value will be exchanged for 999 (integer). But I have a syntax problem in the 'if' block of the array.
The correct return would be: 1 999 5 7 9 999
vetor=(1 3 5 7 9 3)
for i in ${vetor[*]}
do
if [[ ${vetor[i]} = 3 ]]; then
${vetor[i]} = 999
fi
echo $i
done
This produces the correct output in Bash:
vetor=(1 3 5 7 9 3);
for i in ${!vetor[*]};
do
if [[ ${vetor[i]} -eq 3 ]]; then
vetor[i]=999;
fi;
echo ${vetor[i]};
done
I added ! in the for loop expression to get the indices of vetor instead of the values, and I removed the ${} around the assignment in the if condition (this was giving "if 3 is not a typo" warnings). Also changed the echo to get the value at vetor[i] instead of printing the index.

delete elements in array bash

im trying to delete two elements at same time in my array in bash script
my code is
elegidos = (1 2 3 4 5 6 7)
i=0
j=${#elegidos[#]}
delete=($i $j)
while [ $i -le $j ]; do
#elegidosmenosdos=${elegidos[#]/$i:$j}
echo ${elegidos[#]/$delete}
delete=($i $j)
let "i++"
let "j--"
done
the output that i have is
1 2 3 4 5 6 7
1 2 3 4 5 6 7
2 3 4 5 6 7
1 3 4 5 6 7
and i need 21 different combinations with five elements using seven numbers
output example
1 2 3 4 5
2 3 4 5 6
7 6 5 4 3
.
.
.
.
.(21 MORE)
Well, it took a minute to suss out that you were simply wanting to do an end-to-middle delete of two-elements at a time working from the ends of the array, deleting the end nodes at the same time, increment/decrement your counters and repeat until you reached the middle. Why you want to do it this way is a bit of a mystery. You can, of course, simply unset elegidos and unset all values at once. However, if you want to work in from both ends -- that fine too if you have a purpose for it.
You have several problems in your script. In bash, all arrays are zero indexed. Your array has 7-elements, so the valid indexes are 0-6. Therefore, you j was wrong to begin with. You needed to subtract 1 from the number of elements to get the index for the end-element, e.g.
i=0
j=$((${#elegidos[#]} - 1))
bash provides a C-style loop that can greatly simplify your task. While you are free to use a while a C-style loop can handle the index increment and decrement seamlessly, e.g.
for ((; i <= j; i++, j--)); do
note the i <= j. If you have an odd-number of elements in the array, on your last iteration, you will simply be deleting one-value instead of two. To handle that condition you need a simple test within the loop to check whether [ "$i" = "$j" ] (or using the arithmetic comparison (( i == j ))).
Putting that altogether, you could refine your element removal to empty the array two-elements at a time to something similar to the following:
#!/bin/bash
elegidos=(1 2 3 4 5 6 7)
i=0
j=$((${#elegidos[#]} - 1))
delete=($i $j)
for ((; i <= j; i++, j--)); do
declare -p elegidos
unset elegidos[$i]
[ "$i" != "$j" ] && unset elegidos[$j]
delete=($i $j)
done
Example Use/Output
$ bash array_del.sh
declare -a elegidos='([0]="1" [1]="2" [2]="3" [3]="4" [4]="5" [5]="6" [6]="7")'
declare -a elegidos='([1]="2" [2]="3" [3]="4" [4]="5" [5]="6")'
declare -a elegidos='([2]="3" [3]="4" [4]="5")'
declare -a elegidos='([3]="4")'
You can see above, that on the first 3-iterations, both end-elements are removed. However, notice on the last removal (there originally being and odd number of elements, only the single-last value is removed on the final iteration.
Look things over and let me know if I captured what you were attempting and whether you have any further questions. If your intent was something else, drop a comment and I'm happy to help further.

How to add a value into the middle of an array?

Hello I have the following array:
array=(1 2 3 4 5 7 8 9 12 13)
I execute this for loop:
for w in ${!array[#]}
do
comp=$(echo "${array[w+1]} - ${array[w]} " | bc)
if [ $comp = 1 ]; then
/*???*/
else
/*???*/
fi
done
What I would like to do is to insert a value when the difference between two consecutive elements is not = 1
How can I do it?
Many thanks.
Just create a loop from the minimum to the maximum values and fill the gaps:
array=(1 2 3 4 5 7 8 9 12 13)
min=${array[0]}
max=${array[-1]}
new_array=()
for ((i=min; i<=max; i++)); do
if [[ " ${array[#]} " =~ " $i " ]]; then
new_array+=($i)
else
new_array+=(0)
fi
done
echo "${new_array[#]}"
This creates a new array $new_array with the values:
1 2 3 4 5 0 7 8 9 0 0 12 13
This uses the trick in Check if an array contains a value.
You can select parts of the original array with ${arr[#]:index:count}.
Select the start, insert a new element, add the end.
To insert an element after index i=5 (the fifth element)
$ array=(1 2 3 4 5 7 8 9 12 13)
$ i=5
$ arr=("${array[#]:0:i}") ### take the start of the array.
$ arr+=( 0 ) ### add a new value ( may use $((i+1)) )
$ arr+=("${array[#]:i}") ### copy the tail of the array.
$ array=("${arr[#]}") ### transfer the corrected array.
$ printf '<%s>' "${array[#]}"; echo
<1><2><3><4><5><6><7><8><9><12><13>
To process all the elements, just do a loop:
#!/bin/bash
array=(1 2 3 4 5 7 8 9 12 13)
for (( i=1;i<${#array[#]};i++)); do
if (( array[i] != i+1 ));
then arr=("${array[#]:0:i}") ### take the start of the array.
arr+=( "0" ) ### add a new value
arr+=("${array[#]:i}") ### copy the tail of the array.
# echo "head $i ${array[#]:0:i}" ### see the array.
# echo "tail $i ${array[#]:i}"
array=("${arr[#]}") ### transfer the corrected array.
fi
done
printf '<%s>' "${array[#]}"; echo
$ chmod u+x ./script.sh
$ ./script.sh
<1><2><3><4><5><0><7><8><9><10><0><0><13>
There does not seem to be a way to insert directly into an array. You can append elements to another array instead though:
result=()
for w in ${!array[#]}; do
result+=("${array[w]}")
comp=$(echo "${array[w+1]} - ${array[w]} " | bc)
if [ $comp = 1 ]; then
/* probably leave empty */
else
/* handle missing digits */
fi
done
As a final step, you can assign result back to the original array.

bad substitution shell- trying to use variable as name of array

#!/bin/bash
array=( 2 4 5 8 15 )
a_2=( 2 4 8 10 )
a_4=( 2 4 8 10 )
a_5=( 10 12 )
a_8=( 8 12 )
a_15=( 2 4 )
numberOfTests=5
while [ $i -lt ${#array[#]} ]; do
j=0
currentArray =${array[$i]}
*while [ $j -lt ${#a_$currentArray [#]} ]; do #### this line i get ->>>> bad substitution*
./test1.sh "${array[$i]}" -c "${a_"$currentArray "[$j]}" &
let j=j+1
done
let i=i+1
done
so Im trying this code, loop over an array(called array), The array should point out
the array number we are now looping(a_X). And every time to point out the current place and value.
can anybody help me how im using the $currentArray to work properly so I can know the length of the array and the value?
I get in the line I marked an error.
Thank you guys!
The simplest solution is to store the full names of the arrays, not just the numerical suffix, in array. Then you can use indirect parameter expansion while iterating directly over the values, not the indices, of the arrays.
# Omitting numberOfTests has it does not seem to be used
array=(a_2 a_4 a_5 a_8 a_15)
a_2=( 2 4 8 10 )
a_4=( 2 4 8 10 )
a_5=( 10 12 )
a_8=( 8 12 )
a_15=( 2 4 )
for arr in "${array[#]}"; do
currentArray=$arr[#]
for value in "${!currentArray}"; do
./test1.h "${arr#a_}" -c "$value" &
done
done

Sorting an array in a BASH Script by columns and rows while keeping them intact

I have a test file that looks like this:
1 1 1 1 1
9 3 4 5 5
6 7 8 9 7
3 6 8 9 1
3 4 2 1 4
6 4 4 7 7
Each row is supposed to represent a students grades. So the user puts in either an 'r' or a 'c' into the command line to choose to sort by rows or columns, followed by the file name. Sorting by rows would represent getting a students average and sorting my columns would represent a particular assignments average.
I am not doing anything with the choice variable yet because I need to get the array sorted first so I can take the averages and then get the median for each column and row.
So im not sure how I can choose to sort by those specific options. Here is what I have so far:
#!/bin/bash
choice="$1"
filename="$2"
c=0
if [ -e "$filename" ]
then
while read line
do
myArray[$c]=$line
c=$(expr $c + 1)
done < "$filename"
else
echo "File does not exist"
fi
printf -- '%s\n' "${myArray[#]}"
FS=$'\n' sorted=($(sort -n -k 1,1<<<"${myArray[*]}"))
echo " "
printf '%s\n' "${sorted[#]}"
This is only sorting the first column though and im not sure why its even doing that. Any push in the right direction would be appreciated. Examples would help a ton, thanks!
UPDATE:
With the changes that were suggested I have this so far:
#!/bin/sh
IFS=$'\n';
choice="$1"
filename="$2"
if [ -e "$filename" ]
then
while read line
do
myArray[$c]=$line
c=$(expr $c + 1)
done < "$filename"
else
echo "File does not exist."
fi
printf -- '%s\n' "${myArray[#]}"
width=${myArray[0]// /}
width=${#width}
height=${#myArray[#]}
bar=()
for w in $(seq 0 1 $((${width}-1)))
do
tmp=($(sort -n <<<"${myArray[*]}"))
for h in $(seq 0 1 $((${height}-1)))
do
myArray[h]=${myArray[h]#* }
bar[h]="${bar[h]} ${tmp[h]%% *}"
bar[h]="${bar[h]# }"
done
done
printf -- '%s\n' "${bar[*]}"
But now I am getting some really strange output of way more numbers than i started with and in a seemingly random order.
actually it is sorting $line(s) which are strings. you need to initialize the column to sort correctly, so that it is an array
UPDATE:
the following code is really straight forward. no performance aspects are regarded. so for large datasets this will take a while to sort column wise. your datasets have to contain lines of numbers seperated by single spaces to make this work.
#!/bin/bash
IFS=$'\n';
# here you can place your read line function
ar[0]="5 3 2 8"
ar[1]="1 1 1 1"
ar[2]="3 2 4 5"
printf -- '%s\n' "${ar[*]}" # print the column wise unsorted ar
echo
# sorting
width=${ar[0]// /}
width=${#width}
height=${#ar[#]}
bar=()
for w in $(seq 0 1 $((${width}-1))); do # for each column
#sort -n <<<"${ar[*]}" # debug, see first column removal
tmp=($(sort -n <<<"${ar[*]}")) # this just sorts lexigraphically by "first column"
# rows are strings, see initial definition of ar
#echo
for h in $(seq 0 1 $((${height}-1))); do # update first column
ar[h]=${ar[h]#* } # strip first column
bar[h]="${bar[h]} ${tmp[h]%% *}" # add sorted column to new array
bar[h]="${bar[h]# }" # trim leading space
done
#printf -- '%s\n' "${bar[*]}" # debug, see growing bar
#echo "---"
done
printf -- '%s\n' "${bar[*]}" # print the column wise sorted ar
prints out the unsorted and sorted array
5 3 2 8
1 1 1 1
3 2 4 5
1 1 1 1
3 2 2 5
5 3 4 8

Resources