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FinnAPL idiom library: Difference between revisions
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|colspan=2 style="background-color: #F5F5F5"|<source lang=apl inline>(X,'*')[(⎕IO+⍴X)⌊⍋(⍳⍴X),(Y×⍴G)⍴G]</source> | |colspan=2 style="background-color: #F5F5F5"|<source lang=apl inline>(X,'*')[(⎕IO+⍴X)⌊⍋(⍳⍴X),(Y×⍴G)⍴G]</source> | ||
|- style="background-color: #FFFFFF" | |- style="background-color: #FFFFFF" | ||
|rowspan=2| 36. || Median<ref>Note: it doesn't average the middle two elements as per median's definition.</ref> ||style="text-align: right;"|<source lang=apl inline>X←D1</source> | |rowspan=2| 36. || Median<ref>Note: it doesn't average the middle two elements as per median's definition. A more correct idiomatic expression is <source lang=apl inline>0.5×+/X[(⍋X)[|⌈¯0.5 0.5×1+⍴X]]</source></ref> ||style="text-align: right;"|<source lang=apl inline>X←D1</source> | ||
|- style="background-color: #FFFFFF" | |- style="background-color: #FFFFFF" | ||
|colspan=2 style="background-color: #FFFFFF"|<source lang=apl inline>X[(⍋X)[⌈.5×⍴X]]</source> | |colspan=2 style="background-color: #FFFFFF"|<source lang=apl inline>X[(⍋X)[⌈.5×⍴X]]</source> | ||
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|colspan=2 style="background-color: #F5F5F5"|<source lang=apl inline>(¯1⌽1↓(X≠¯1⌽X),1)/X</source> | |colspan=2 style="background-color: #F5F5F5"|<source lang=apl inline>(¯1⌽1↓(X≠¯1⌽X),1)/X</source> | ||
|- style="background-color: #FFFFFF" | |- style="background-color: #FFFFFF" | ||
|rowspan=2| 491. || An expression giving itself ||style="text-align: right;"|<source lang=apl inline></source> | |rowspan=2| 491. || [[Quine|An expression giving itself]] ||style="text-align: right;"|<source lang=apl inline></source> | ||
|- style="background-color: #FFFFFF" | |- style="background-color: #FFFFFF" | ||
|colspan=2 style="background-color: #FFFFFF"|<source lang=apl inline>1⌽22⍴11⍴'''1⌽22⍴11⍴'''</source> | |colspan=2 style="background-color: #FFFFFF"|<source lang=apl inline>1⌽22⍴11⍴'''1⌽22⍴11⍴'''</source> | ||