From 0441cebd72baa0bcefee6a874e834bc852c9ab5a Mon Sep 17 00:00:00 2001
From: Raphael <paloma@css-maps.eu>
Date: Mon, 26 Feb 2018 15:55:50 +0100
Subject: [PATCH] make substitutions more memory efficient

Symmetric substitution cost are looked up on the fly and not 'precomputed'. Symetric parameter now also has an effect on substitutions that are passed as dict.
---
 fizzle.py | 19 ++++++++-----------
 1 file changed, 8 insertions(+), 11 deletions(-)

diff --git a/fizzle.py b/fizzle.py
index 50f60b4..cc4c669 100644
--- a/fizzle.py
+++ b/fizzle.py
@@ -12,9 +12,9 @@ def dl_distance(s1,
                 secondHalfDiscount=False):
     """
     Return DL distance between s1 and s2. Default cost of substitution, insertion, deletion and transposition is 1
-    substitutions is list of tuples of characters (what, substituted by what, cost), 
-    maximal value of substitution is 2 (ie: cost deletion & insertion that would be otherwise used)
-    eg: substitutions=[('a','e',0.4),('i','y',0.3)]
+    substitutions is list of tuples of characters (what, substituted by what, cost) or dictionary 
+    {(what, substituted by what):cost}, maximal value of substitution is 2 (ie: cost deletion & insertion that
+    would be otherwise used) eg: substitutions=[('a','e',0.4),('i','y',0.3)] or {('a','e'):0.4, ('i','y'):0.3}
     symetric=True mean that cost of substituting A with B is same as B with A
     returnMatrix=True: the matrix of distances will be returned, if returnMatrix==False, then only distance will be returned
     printMatrix==True: matrix of distances will be printed
@@ -23,9 +23,6 @@ def dl_distance(s1,
 
     if isinstance(substitutions, list):
         subs_dict = {(from_, to): cost for from_, to, cost in substitutions}
-        if symetric:
-            subs_dict.update({(to, from_)
-                              for from_, to, cost in substitutions})
         substitutions = subs_dict
 
     if nonMatchingEnds:
@@ -39,13 +36,13 @@ def dl_distance(s1,
     for i in range(len(s1)):
         for j in range(len(s2)):
             ch1, ch2 = s1[i], s2[j]
+            cost = 1
             if ch1 == ch2:
                 cost = 0
-            else:
-                if (ch1, ch2) in substitutions:
-                    cost = substitutions[(ch1, ch2)]
-                else:
-                    cost = 1
+            elif (ch1, ch2) in substitutions:
+                cost = substitutions[(ch1, ch2)]
+            elif symetric and (ch2, ch1) in substitutions:
+                cost = substitutions[(ch2, ch1)]
             if secondHalfDiscount and (s1 > half1 or s2 > half2):
                 deletionCost, insertionCost = 0.6, 0.6
             else: