TM Therneau, JP Sinnwell

27 September, 2022


The plotting function for pedigrees has 5 tasks 1. Gather information and check the data. An important step is the call to align.pedigree. 2. Set up the plot region and size the symbols. The program wants to plot circles and squares, so needs to understand the geometry of the paper, pedigree size, and text size to get the right shape and size symbols. 3. Set up the plot and add the symbols for each subject 4. Add connecting lines between spouses, and children with parents 5. Create an invisible return value containing the locations.

Another task, not yet completely understood, and certainly not implemented, is how we might break a plot across multiple pages.


The dull part is first: check all of the input data for correctness. The [[sex]] variable is taken from the pedigree so we need not check that. The identifier for each subject is by default the [[id]] variable from the pedigree, but users often want to add some extra text. The status variable can be used to put a line through the symbol of those who are deceased, it is an optional part of the pedigree.

The affected status is a 0/1 matrix of any marker data that the user might want to add. It may be attached to the pedigree or added here. It can be a vector of length [[n]] or a matrix with [[n]] rows. If it is not present, the default is to print open symbols without shading or color, which corresponds to a code of 0, while a 1 means to shade the symbol.

If the argment is a matrix, then the shading and/or density value for ith column is taken from the ith element of the angle/density arguments.

For purposes within the plot method, NA values in


Now we need to set the sizes. From align.pedigree we will get the maximum width and depth. There is one plotted row for each row of the returned matrices. The number of columns of the matrices is the max width of the pedigree, so there are unused positions in shorter rows, these can be identifed by having an nid value of 0. Horizontal locations for each point go from 0 to xmax, subjects are at least 1 unit apart; a large number will be exactly one unit part. These locations will be at the top center of each plotted symbol.

We would like to to make the boxes about 2.5 characters wide, which matches most labels, but no more than 0.9 units wide or .5 units high.
We also want to vertical room for the labels. Which should have at least 1/2 of stemp2 space above and stemp2 space below.
The stemp3 variable is the height of labels: users may use multi-line ones. Our constraints then are

  1. (box height + label height)*maxlev \(\le\) height: the boxes and labels have to fit vertically
  2. (box height) * (maxlev + (maxlev-1)/2) \(\le\) height: at least 1/2 a box of space between each row of boxes
  3. (box width) \(\le\) stemp1 in inches
  4. (box width) \(\le\) 0.8 unit in user coordinates, otherwise they appear to touch
  5. User coordinates go from min(xrange)- 1/2 box width to max(xrange) + 1/2 box width.
  6. the box is square (in inches)

The first 3 of these are easy. The fourth comes into play only for very packed pedigrees. Assume that the box were the maximum size of .8 units, i.e., minimal spacing between them. Then xmin -.45 to xmax + .45 covers the plot region, the scaling between user coordinates and inches is (.8 + xmax-xmin) user = figure region inches, and the box is .8*(figure width)/(.8 + xmax-xmin). The transformation from user units to inches horizontally depends on the box size, since I need to allow for 1/2 a box on the left and right.
Vertically the range from 1 to nrow spans the tops of the symbols, which will be the figure region height less (the height of the text for the last row + 1 box); remember that the coordinates point to the top center of the box. We want row 1 to plot at the top, which is done by appropriate setting of the usr parameter.

Subsetting and Sub-Region

This section is still experimental and might change. Also, in the original documentation by TM Therneau, it is within the sizing section above.

Sometimes a pedigree is too large to fit comfortably on one page. The [[subregion]] argument allows one to plot only a portion of the pedigree based on the plot region. Along with other tools to select portions of the pedigree based on relatedness, such as all the descendents of a particular marriage, it gives a tool for addressing this. This breaks our original goal of completely automatic plots, but users keep asking for more.

The argument is [[subregion=c(min x, max x, min depth, max depth)]], and works by editing away portions of the [[plist]] object returned by align.pedigree. First decide what lines to keep. Then take subjects away from each line, update spouses and twins, and fix up parentage for the line below.

Drawing the Tree

  1. First draw and label the boxes. Definition of the drawbox function is deferred until later. symbols code chunk.

  2. Draw in the connections, one by one, beginning with spouses. lines code chunk.

  3. Connect children to parents.

  4. Lines/arcs to connect multiple instances of same subject.

Details on connecting children to parents. First there are lines up from each child, which would be trivial except for twins, triplets, etc. Then we draw the horizontal bar across siblings and finally the connector from the parent. For twins, the _1, y_1)$ and \((x_2, y_2\)) whose height is 1/2 unit above a straight line connection.


Finally we get to the drawbox function itself, which is fairly simple. In 2011 updates, it allows missing, and fixeds up shadings and borders. For affected=0, do not fill. For affected=1, fill with density-lines and angles. For affected=-1 (missing), fill with ? in the midpoint of the polygon, with a size adjusted by how many columns in affected. For all shapes drawn, make the border the color for the person.


There are four sumbols corresponding to the four sex codes: square = male, circle = female, diamond= unknown, and triangle = terminated.
They are shaded according to the value(s) of affected status for each subject, where 0=unfilled and 1=filled, and filling uses the standard arguments of the [[polygon]] function. The nuisance is when the affected status is a matrix, in which case the symbol will be divided up into sections, clockwise starting at the lower left. I asked Beth about this (original author) and there was no particular reason to start at 6 o-clock, but it is now established as history.

The first part of the code is to create the collection of polygons that will make up the symbol. These are then used again and again. The collection is kept as a list with the four elements square, circle, diamond and triangle.

Each of these is in turn a list with ncol(affected) element, and each of those in turn a list of x and y coordinates. There are 3 cases: the affected matrix has only one column, partitioning a circle for multiple columns, and partitioning the other cases for multiple columns.


The circle function is quite simple. The number of segments is arbitrary, 50 seems to be enough to make the eye happy. We draw the ray from 0 to the edge, then a portion of the arc. The polygon function will connect back to the center.


Now for the interesting one — dividing a polygon into ``pie slices’’. In computing this we can’t use the usual \(y= a + bx\) formula for a line, because it doesn’t work for vertical ones (like the sides of the square). Instead we use the alternate formulation in terms of a dummy variable \(z\). \[\begin{eqnarray*} x &=& a + bz \\ y &=& c + dz \\ \end{eqnarray*}\] Furthermore, we choose the constants \(a\), \(b\), \(c\), and \(d\) so that the side of our polygon correspond to \(0 \le z \le 1\). The intersection of a particular ray at angle theta with a particular side will satisfy \[\begin{eqnarray} theta &=& y/x = \frac{a + bz}{c+dz} \nonumber \\ z &=& \frac{a\theta -c}{b - d\theta} \label{eq:z} \\ \end{eqnarray}\]

Equation will lead to a division by zero if the ray from the origin does not intersect a side, e.g., a vertical divider will be parallel to the sides of a square symbol. The only solutions we want have \(0 \le z \le 1\) and are in the forward' part of the ray. This latter %' is true if the inner product \(x \cos(\theta) + y \sin(\theta) >0\). Exactly one of the polygon sides will satisfy both conditions.

Final output

Remind the user of subjects who did not get plotted; these are ususally subjects who are married in but without children. Unless the pedigree contains spousal information the routine does not know who is the spouse. Then restore the plot parameters. This would only not be done if someone wants to further annotate the plot. Last, we give a list of the plot positions for each subject. Someone who is plotted twice will have their first position listed.