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

Question

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.

Answer 1

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