问题
你有一个二维矩阵。 您使用 Loc 函数来定位感兴趣的元素，但 Loc 返回一维矩阵的索引。 如何将一维索引转换回二维索引？
解决方案
您需要使用 Floor 和 Mod 函数，并且需要知道二维矩阵中的列数。 这是三行，隐藏在下面的注释示例中。 pixmat 是一个 2D 矩阵，Loc 正在查找非零像素，即黑色：
blacklocs = (Loc( pixmat ) - 1); // blacklocs 是一维索引矩阵（从 0 开始）
rowlocs = Floor( (blacklocs) / nc ) + 1; // nc 是 2D pixmat 中的列数
collocs = Mod( (blacklocs), nc ) + 1; // rowlocs 和 collocs 是 2D 索引（从 1 开始）
// we need a bitmap to work with; use a text box with a moderate size font
textbox = Text Box( "Gargoyle", <<SETFONTSIZE> .5 are antialiasing fading to white background of the "paper".
// the next statement makes the 0 to .5 values be 1 and the .5 to 1 values be 0
pixmat = pixmat < .75; // cut off edge at some gray value. flip the 0..1 significance.
// the pixels are in a 2D matrix of rows and cols. get the dimensions.
nr = N Rows( pixmat );
nc = N Cols( pixmat );
// now we can find the location of the black pixels (1s) that form the text.
// the LOC function returns a 1D matrix of the offsets of non-zero
// elements in the 2D matrix pixmat. if pixmat was 3 rows of 5 elements each,
// the LOC could return a 1D matrix of 0 to 15 elements with values from 1 to 15.
// JMP uses 1-based indexing. subtract one to get 0-based for future steps.
blacklocs = (Loc( pixmat ) - 1);
// Here's the JSL that convers the 1D answers back to 2D answers.
// use the 0-based blacklocs (previous statement), then add 1 to
// turn the 0-based results back into the 1-based indexes JMP expects.
rowlocs = Floor( (blacklocs) / nc ) + 1;
collocs = Mod( (blacklocs), nc ) + 1;
// Show the results
scale = 16; // try bigger or smaller scale for your screen size
New Window( "Example",
Graph Box(
// make the graph the same aspect ratio as the bitmap
framesize( nc * scale, nr * scale ),
// make the point marker size big; .65 is approximately right
<<MARKER size=""></MARKER></SETFONTSIZE>
Pixels from a bitmap, displayed as markers in a graphPixels from a bitmap, displayed as markers in a graph位图中的像素，在图表中显示为标记
也可以看看： JSL 矩阵是什么？有关 JMP 中矩阵的介绍。
