In my copy of j, only u: 16b2588 (a solid block) and u: 16b258c (a block occupying the left hand side of the character position) display reasonably.
The others are hollow boxes. -- Raul On Mon, Jan 23, 2012 at 4:47 PM, Ian Clark <earthspo...@gmail.com> wrote: > Unicode has a set of characters for building a crude (horizontal) histogram: > > u: 16b2588 NB. U+2588 (9608) 1.0 > u: 16b2589 NB. U+2589 (9609) 0.875 > u: 16b258a NB. U+258a (9610) 0.75 > u: 16b258b NB. U+258b (9611) 0.625 > u: 16b258c NB. U+258c (9612) 0.5 > u: 16b258d NB. U+258d (9613) 0.375 > u: 16b258e NB. U+258e (9614) 0.25 > u: 16b258f NB. U+258f (9615) 0.125 > > They can be used to implement a progress bar, or embed little > histograms in tables of figures. > > Notice a curious thing: these code points divide the basic block > (U+2588) into *eights*, not tenths! Why? > > I imagine it's for ultra-efficient ASM to pick the part-block from the > (base-2) mantissa of a floating-point number. But how do you do it > ultra-efficiently in J? > > See my solution at: http://www.jsoftware.com/jwiki/Essays/Unicode%20Histogram > This uses the verb: fh to pick the code-point for the final > part-block. But I don't believe for a moment it's the best solution. > Anyone got a quicker/slicker fh? > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
