Adverbs modify verbs to operative iteritvely over nouns. This previous sentence likley will only make sense once you understand it so let’s jump into an example. Imagine you have a list of 4 lists. Using first you will retrieve the first sub-list. In order to retrieve the first element of each sub-list you’ll have to modify the function first with a modifier each.
Adverb f/ over c/ join f\ scan c\ split f' each v' has f': eachp v': bin. f/: eachr ([n;]f)/: c over f\: eachl ([n;]f)\: c scan
Compute f[x;y] such that f@i=f[f@i-1;x@i]. Scan and over are the same functions except that over only returns the last value.
Given a function of two inputs, output for each x according to...
An example
(,\)("a";"b";"c")
a
ab
abc
+\1 20 300
1 21 321
{y+10*x}\1 20 300
1 30 600
{y+10*x}/1 20 300
600
Join the list of strings x using character c between each pair.
"-"/("Some";"random text";"plus";,".")
"Some-random text-plus-."
Split the string x using character c to determine the locations.
" "\"Here is a short sentence." Here is a short sentence.
Apply function f to each value in list x.
*((1 2 3);4;(5 6);7) / first element of the list 1 2 3 *'((1 2 3);4;(5 6);7) / first element of each element 1 4 5 7
Determine if vector v has element x. Simliar to in but arguments reversed.
`a`b`c`a`b'`a 1b `a`b`c`a`b'`d 0b
Apply f[x;y] using the prior value of y, eg. f[y@n;y@n-1]. The first value, n=0, returns f[y@0;x].
,':[`x;(`$"y",'$!5)] y0 x y1 y0 y2 y1 y3 y2 y4 y3 %':[100;100 101.9 105.1 102.3 106.1] / compute returns 1 1.019 1.031403 0.9733587 1.037146 100%':100 101.9 105.1 102.3 106.1 / using infix notation 1 1.019 1.031403 0.9733587 1.037146
Given a sorted (increasing) list x, find the greatest index, i, where y>x[i].
n:exp 0.01*!5;n 1 1.01005 1.020201 1.030455 1.040811 n':1.025 2
Apply f[x;y] to each value of y, i.e. the right one.
(!2)+/:!4 0 1 1 2 2 3 3 4 +/:[!2;!4] 0 1 1 2 2 3 3 4
Apply f[x;y] to each value of x, i.e. the left one.
(!2)+\:!4 0 1 2 3 1 2 3 4 +\:[!2;!4] 0 1 2 3 1 2 3 4
Compute f with initial value x and over list y. f[i] = f[f[i-1];y[i]] except for the case of f[0]=f[x;y[0]]. n over differs from n scan in that it only returns the last value.
f:{(0.1*x)+0.9*y} / ema
0. f\:1+!3
0.9 1.89 2.889
f:{(`$,/$x),(`$,/$y)} / join and collapse
`x f\: `y0`y1`y2
x y0
xy0 y1
xy0y1 y2
`x f/: `y0`y1`y2
xy0y1 y2
Compute f[x], f[f[x]] and continue to call f[previous result] until the output converges to a stationary value or the output produces x.
{x*x}\:.99 / converge to value
0.99 0.9801 0.9605961 ... 1.323698e-18 1.752177e-36 0.
l:3 4 2 1 4 99 / until output is x
l\:0
0 3 1 4
/ the latter example worked out manually
l 0
3
l 3
1
l 1
4
l 4 / f[4]=4 so will stop
4
Same as converge scan but only return last value.
{x*x}\:.99
0.
Convert y (base 10) into base x.
2\:129 10000001b 16\:255 15 15
Convert list y (base x) into base 10.
2/:10101b 21 16/:15 0 15 3855