Bash associative array sorting by value - bash

Question

I get the following output:
Pushkin - 100500
Gogol - 23
Dostoyevsky - 9999
Which is the result of the following script:
for k in "${!authors[#]}"
do
echo $k ' - ' ${authors["$k"]}
done
All I want is to get the output like this:
Pushkin - 100500
Dostoyevsky - 9999
Gogol - 23
which means that the keys in associative array should be sorted by value. Is there an easy method to do so?

Answer 1

You can easily sort your output, in descending numerical order of the 3rd field:
for k in "${!authors[#]}"
do
echo $k ' - ' ${authors["$k"]}
done |
sort -rn -k3
See sort(1) for more about the sort command. This just sorts output lines; I don't know of any way to sort an array directly in bash.
I also can't see how the above can give you names ("Pushkin" et al.) as array keys. In bash, array keys are always integers.

Answer 2

Alternatively you can sort the indexes and use the sorted list of indexes to loop through the array:
authors_indexes=( ${!authors[#]} )
IFS=$'\n' authors_sorted=( $(echo -e "${authors_indexes[#]/%/\n}" | sed -r -e 's/^ *//' -e '/^$/d' | sort) )
for k in "${authors_sorted[#]}"; do
echo $k ' - ' ${authors["$k"]}
done

Answer 3

Extending the answer from #AndrewSchulman, using -rn as a global sort option reverses all columns. In this example, authors with the same associative array value will be output by reverse order of name.
For example
declare -A authors
authors=( [Pushkin]=10050 [Gogol]=23 [Dostoyevsky]=9999 [Tolstoy]=23 )
for k in "${!authors[#]}"
do
echo $k ' - ' ${authors["$k"]}
done | sort -rn -k3
will output
Pushkin - 10050
Dostoyevsky - 9999
Tolstoy - 23
Gogol - 23
Options for sorting specific columns can be provided after the column specifier.
i.e. sort -k3rn
Note that keys can be specified as spans. Here -k3 happens to be fine because it is the final span, but to use only column 3 explicitly (in case further columns were added), it should be specified as -k3,3,
Similarly to sort by column three in descending order, and then column one in ascending order (which is probably what is desired in this example):
declare -A authors
authors=( [Pushkin]=10050 [Gogol]=23 [Dostoyevsky]=9999 [Tolstoy]=23 )
for k in "${!authors[#]}"
do
echo $k ' - ' ${authors["$k"]}
done | sort -k3,3rn -k1,1
will output
Pushkin - 10050
Dostoyevsky - 9999
Gogol - 23
Tolstoy - 23

Answer 4

The best way to sort a bash associative array by VALUE is to NOT sort it.
Instead, get the list of VALUE:::KEYS, sort that list into a new KEY LIST, and iterate through the list.
declare -A ADDR
ADDR[192.168.1.3]="host3"
ADDR[192.168.1.1]="host1"
ADDR[192.168.1.2]="host2"
KEYS=$(
for KEY in ${!ADDR[#]}; do
echo "${ADDR[$KEY]}:::$KEY"
done | sort | awk -F::: '{print $2}'
)
for KEY in $KEYS; do
VAL=${ADDR[$KEY]}
echo "KEY=[$KEY] VAL=[$VAL]"
done
output:
KEY=[192.168.1.1] VAL=[host1]
KEY=[192.168.1.2] VAL=[host2]
KEY=[192.168.1.3] VAL=[host3]

Answer 5

Do something with unsorted keys:
for key in ${!Map[#]}; do
echo $key
done
Do something with sorted keys:
for key in $(for x in ${!Map[#]}; do echo $x; done | sort); do
echo $key
done
Stored sorted keys as array:
Keys=($(for x in ${!Map[#]}; do echo $x; done | sort))

Answer 6

If you can assume the value is always a number (no spaces), but want to allow for the possibility of spaces in the key:
for k in "${!authors[#]}"; do
echo "${authors["$k"]} ${k}"
done | sort -rn | while read number author; do
echo "${author} - ${number}"
done
Example:
$ declare -A authors
$ authors=(['Shakespeare']=1 ['Kant']=2 ['Von Neumann']=3 ['Von Auersperg']=4)
$ for k in "${!authors[#]}"; do echo "${authors["$k"]} ${k}"; done | sort -rn | while read number author; do echo "${author} - ${number}"; done
Von Auersperg - 4
Von Neumann - 3
Kant - 2
Shakespeare - 1
$
The chosen answer seems to work if there are no spaces in the keys, but fails if there are:
$ declare -A authors
$ authors=(['Shakespeare']=1 ['Kant']=2 ['Von Neumann']=3 ['Von Auersperg']=4)
$ for k in "${!authors[#]}"; do echo $k ' - ' ${authors["$k"]}; done | sort -rn -k 3
Kant - 2
Shakespeare - 1
Von Neumann - 3
Von Auersperg - 4
$