SIMPLE SOLUTIONS

# FUNCOMBINE(7) - Linux man page online | Overview, conventions, and miscellany

Combining Region and Table Filters.

Chapter
April 14, 2011
funcombine(7) SAORD Documentation funcombine(7)

## NAME

FunCombine - Combining Region and Table Filters

## SYNOPSIS

This document discusses the conventions for combining region and table filters, especially with regards to the comma operator.

## DESCRIPTION

Comma Conventions Filter specifications consist of a series of boolean expressions, separated by commas. These expressions can be table filters, spatial region filters, or combinations thereof. Unfortunately, common usage requires that the comma operator must act differently in dif‐ ferent situations. Therefore, while its use is intuitive in most cases, commas can be a source of confusion. According to long-standing usage in IRAF, when a comma separates two table filters, it takes on the meaning of a boolean and. Thus: foo.fits[pha==1,pi==2] is equivalent to: foo.fits[pha==1 && pi==2] When a comma separates two spatial region filters, however, it has traditionally taken on the meaning of a boolean or. Thus: foo.fits[circle(10,10,3),ellipse(20,20,8,5)] is equivalent to: foo.fits[circle(10,10,3) ⎪⎪ ellipse(20,20,8,5)] (except that in the former case, each region is given a unique id in programs such as funcnts). Region and table filters can be combined: foo.fits[circle(10,10,3),pi=1:5] or even: foo.fits[pha==1&&circle(10,10,3),pi==2&&ellipse(20,20,8,5)] In these cases, it is not obvious whether the command should utilize an or or and opera‐ tor. We therefore arbitrarily chose to implement the following rule: · if both expressions contain a region, the operator used is or. · if one (or both) expression(s) does not contain a region, the operator used is and. This rule handles the cases of pure regions and pure column filters properly. It unam‐ biguously assigns the boolean and to all mixed cases. Thus: foo.fits[circle(10,10,3),pi=1:5] and foo.fits[pi=1:5,circle(10,10,3)] both are equivalent to: foo.fits[circle(10,10,3) && pi=1:5] [NB: This arbitrary rule replaces the previous arbitrary rule (pre-funtools 1.2.3) which stated: · if the 2nd expression contains a region, the operator used is or. · if the 2nd expression does not contain a region, the operator used is and. In that scenario, the or operator was implied by: pha==4,circle 5 5 1 while the and operator was implied by circle 5 5 1,pha==4 Experience showed that this non-commutative treatment of the comma operator was confusing and led to unexpected results.] The comma rule must be considered provisional: comments and complaints are welcome to help clarify the matter. Better still, we recommend that the comma operator be avoided in such cases in favor of an explicit boolean operator.