Re: Most frequent history words

[Sebastian Gniazdowski <sgniazdowski@gmail.com>]

The code works. Thanks. Had only to change (on) to (On). Interesting
trick with the sorting on "$v=$k". Here is a complete version for
someone to quickly reuse:


typeset -A uniq
for k in ${historywords[@]}; do
    uniq[$k]=$(( ${uniq[$k]:-0} + 1 ))
done

vk=()
for k v in ${(kv)uniq}; do
    vk+="$v=$k"
done
print -rl -- ${${${(On)vk}#<->=}[1,10]}


Interesingly, changing ${historywords[@]} to ${history[@]} and
"${history[@]}" doesn't change script's output, it still outputs most
frequent words, not history entries.

Best regards,
Sebastian Gniazdowski


On 25 April 2016 at 21:49, Bart Schaefer <schaefer@brasslantern.com> wrote:
> On Apr 25, 12:36pm, Sebastian Gniazdowski wrote:
> } Subject: Most frequent history words
> }
> } Hello,
> } can the following be made less coreutils dependent, i.e. more pure-Zsh code?
> }
> } print -rl "${historywords[@]}" | sort | uniq -c | sort -k1,1nr -k2,2  | head
>
> It can, but it's probably not very efficient.
>
> The pipe to sort can be replaced with
>
>     print -rl -- ${(o)historywords[@]}
>
> The "uniq -c" would have to be replaced by a loop building a hash whose
> keys are words and whose values are the count thereof (making the initial
> sort irrelevant).
>
>     typeset -A uniq
>     for k in ${historywords[@]}
>     do uniq[$k]=$(( ${uniq[$k]:-0} + 1 ))
>     done
>
> Some quoting on $k such as ${(b)k} is probably required there, this is
> the shakiest part of the process.
>
> Then the final "sort -k..." would have to be done by iterating over the
> hash, with "head" just taking an array slice.
>
>     vk=()
>     for k v in ${(kv)uniq}
>     do vk+="$v=$k"
>     done
>     print -rl -- ${${${(on)vk}#<->=}[1,10]}
>
> Plus unwrapping there whatever quoting on $k you did in the first loop.