n2mfrow Compute Default 'mfrow' From Number of Plots

Description

Easy setup for plotting multiple figures (in a rectangular layout) on one page. This computes a sensible default for par(mfrow).

Usage

n2mfrow(nr.plots, asp = 1)

Arguments

nr.plots	integer; the number of plot figures you'll want to draw.
asp	positive number; the target aspect ratio (columns / rows) in the output. Was implicitly hardwired to 1; because of that and back compatibility, there is a somewhat discontinuous behavior when varying asp around 1, for nr.plots <= 12.

Value

A length-two integer vector (nr, nc) giving the positive number of rows and columns, fulfilling nr * nc >= nr.plots, and currently, for asp = 1, nr >= nc >= 1.

Note

Conceptually, this is a quadratic integer optimization problem, with inequality constraints nr >= 1, nc >= 1, and nr.plots >= nr*nc (and possibly nr >= asp*nc), and two objective functions which would have to be combined via a tuning weight, say w, to, e.g., (nr.plots - nr*nc) + w |nr/nc - asp|.

The current algorithm is simple and not trying to solve one of these optimization problems.

Author(s)

Martin Maechler; suggestion of asp by Michael Chirico.

See Also

par, layout.

Examples

require(graphics)

n2mfrow(8) # 3 x 3

n <- 5 ; x <- seq(-2, 2, length.out = 51)
## suppose now that 'n' is not known {inside function}
op <- par(mfrow = n2mfrow(n))
for (j in 1:n)
   plot(x, x^j, main = substitute(x^ exp, list(exp = j)), type = "l",
   col = "blue")

sapply(1:14, n2mfrow)
sapply(1:14, n2mfrow, asp=16/9)