One-Liner Loop in Functional Style
Xah Lee, 200510
Today we show a example of a loop done as a one-liner of Functional
Programing style.
Suppose you have a list of file full paths of images:
/Users/t/t4/oh/DSCN2059m-s.jpg
/Users/t/t4/oh/DSCN2062m-s.jpg
/Users/t/t4/oh/DSCN2097m-s.jpg
/Users/t/t4/oh/DSCN2099m-s.jpg
/Users/t/Icons_dir/icon_sum.gif
For those ending with -s (indicate a smaller version), you want to
change it to the full version without the -s, if it exists. Here's the
code:
# -*- coding: utf-8 -*-
# python
import re, os.path
imgPaths=[u'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2059m-s.jpg',
u'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2062m-s.jpg',
u'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2097m-s.jpg',
u'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2099m-s.jpg',
u'/Users/t/web/Icons_dir/icon_sum.gif']
# change the image path to the full sized image, if it exists
# that is, if image ends in -s.jpg, find one without the '-s'.
imgPaths2=[]
for myPath in imgPaths:
p=myPath
(dirName, fileName) = os.path.split(myPath)
(fileBaseName, fileExtension)=os.path.splitext(fileName)
if(fileBaseName[-2:] == '-s'):
p2=os.path.join(dirName,fileBaseName[0:-2]) + fileExtension
if os.path.exists(p2): p=p2
imgPaths2.append(p)
print imgPaths2
But how do you do it in a functional programing style? Namely,
something of the from imgPath2=f(imgPath), where the f is some
function. Normally, the f would be a pure function construct made up on
the spot, that is, lambda. But Python's lambda is limited to only a
expression as its body. Nevertheless, one can achieve a functional
style with some workaround, using map twice. Here's the code:
imgPaths3 = map( lambda x: os.path.exists(x[1]) and x[1] or x[0], \
map(lambda x: (x, re.sub( r"^(.+?)-s(\.[^.]+)$",r"\1\2", x)),
imgPaths))
The first map:
newList = map(lambda x: (x, re.sub( r"^(.+?)-s(\.[^.]+)$",r"\1\2", x)),
imgPaths)
generate a list of pairs (x, y), where x is the same element in
imgPaths, and y is one without the -s. Then, a second map:
map( lambda x: os.path.exists(x[1]) and x[1] or x[0], newList,
imgPaths))
checks if the path y exists, if so, use that, else, use the x part. The
function body is essentially of a conditional EXPRESSION, of the from
(test, if true return result1, else return result2). This form of a
test that returns a expression is very important in functional
programing. Because, in functional programing a common pattern is to
sequence functions and passing values. One cannot stop in the middle
and use a block structure. Here's how this form's syntax in several
languages:
test? trueExpression: falseExpression (C, Perl)
If[test, trueExpression, falseExpression] (Mathematica)
(test trueExpression falseExpression) (LISP)
In Python, there's no such form for this, but a semantically equivalent
workaround is to sequence boolean expressions as used in our example.
Namely:
test and trueExpression or falseExpression (Python)
This works because it exploits a particular behavior of how Python
treats boolean sequences. In Python, “x and y” returns y ifx is
true. And, “x or y” returns x if x is true. This behavior is
compatible with the mathematical sense of booleans. For example,
mathematically “x and y” should be true if both are true, yet in
Python if x is true then y is returned, and if y is true then this is
compatible with the math sense, but if y is false then the whole result
is also false, so it also satisfies the math sense. Similar is the case
of “x or y”. The point here is that one of the element is returned,
instead of a real True or False value. Therefore, in a twisted way this
can be used for the If[test, trueExpression, falseExpression] form.
Example.
result1= True and 4 or 5 # returns 4
result2= False and 4 or 5 # returns 5
print result1
print result2
Such language behavior is a result of the laziness of the compiler
implementation, particular seen in unix shells (&& and ||). The problem
with this design is that codes relying on the the returned element of a
boolean sequence does not clearly indicate the programer's intention.
It works by language pecularities, instead of expressed program logic.
For the official doc on evaluating booleans, see:
http://python.org/doc/2.4.2/lib/boolean.html
Here's the Perl code of the same loop block:
# perl
use File::Basename;
@imgPaths=(
'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2059m-s.jpg',
'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2062m-s.jpg',
'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2097m-s.jpg',
'/Users/t/web/Periodic_dosage_dir/lanci/t4/oh/DSCN2099m-s.jpg',
'/Users/t/web/Icons_dir/icon_sum.gif');
# change the image path to the full sized image, if it exists
# that is, if image ends in -s.jpg, find one without the '-s'.
$imgPaths2=();
for $myPath (@imgPaths){
$p=$myPath;
($fileBaseName, $dirName, $fileExtension) = fileparse($myPath,
('\.[^.]+$') );
if (substr($fileBaseName,-2,2) eq '-s') {
$p2= $dirName . '/' .
substr($fileBaseName,0,length($fileBaseName)-2) . $fileExtension;
print $p2, "\n";
if (-e $p2) { $p=$p2}
}
push(@imgPaths2,$p);
}
use Data::Dumper;
print Dumper(\@imgPaths2)
In Perl, this can be written in a functional style one-liner:
@imgPaths3= map{
$y = $_;
$y =~ s/^(.+?)-s(\.[^.]+)$/$1$2/;
-e $y ? $y : $_;
} @imgPaths;
For the official doc of Perl's map, type in command line: “perldoc -f
map”
------
this article is archived at:
http://xahlee.org/perl-python/one-liner_loop.html
Xah
xa*@xahlee.org
∑ http://xahlee.org/
