使用命令行工具按排序序列计算重复项

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if order is not important

# echo "100 100 100 99 99 26 25 24 24" | awk '{for(i=1;i<=NF;i++)a[$i]++}END{for(o in a) printf "%s %s ",o,a[o]}'
26 1 100 3 99 2 24 2 25 1

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最佳答案

how about;

$ echo "100 100 100 99 99 26 25 24 24" \
| tr " " "\n" \
| sort \
| uniq -c \
| sort -k2nr \
| awk '{printf("%s\t%s\n",$2,$1)}END{print}'

The result is :

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uniq -c works for GNU uniq 8.23 at least, and does exactly what you want (assuming sorted input).

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Numerically sort the numbers in reverse, then count the duplicates, then swap the left and the right words. Align into columns.

printf '%d\n' 100 99 26 25 100 24 100 24 99 \
| sort -nr | uniq -c | awk '{printf "%-8s%s\n", $2, $1}'

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In Bash, we can use an associative array to count instances of each input value. Assuming we have the command $cmd1, e.g.

#!/bin/bash


cmd1='printf %d\n 100 99 26 25 100 24 100 24 99'

Then we can count values in the array variable a using the ++ mathematical operator on the relevant array entries:

while read i
do
((++a["$i"]))
done < <($cmd1)

We can print the resulting values:

for i in "${!a[@]}"
do
echo "$i ${a[$i]}"
done

If the order of output is important, we might need an external sort of the keys:

for i in $(printf '%s\n' "${!a[@]}" | sort -nr)
do
echo "$i ${a[$i]}"
done

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In case you have input stored in my_file you can do:

sort -nr my_file | uniq -c | awk ' { t = $1; $1 = $2; $2 = t; print; } '

Otherwise just pipe the input to be processed to the same cmd.

Explanation: