"Nicolas Fleury" <ni******@yahoo.com_remove_the_> wrote in message
news:Lc*******************@nnrp1.uunet.ca...
Hi everyone,
Is there a way to compare recursively two objects (compare their
members recursively)? I'm only interested in equality or non-equality
(no need for lower-than...).
Thx and Regards,
Nicolas
If you're comparing lists, here's an example (although list1==list2 also
works, see the verification in the test() function). If you're comparing
something else, the logic is similar.
This example also does short-circuiting, in that it stops testing for
equality once the first mismatch is found. Also, there is a slight
short-cut in isEqualElement, in comparison of two objects for equal identity
before testing for equal value.
-- Paul
def compareLists(lst1, lst2):
def isEqualElement(e1,e2):
print "Comparing",e1,"and",e2
if e1 is e2:
return True
if type(e1) is type(e2):
if isinstance(e1,list):
return compareLists(e1,e2)
else:
return e1 == e2
else:
return False
# or the same thing as a lambda, just remove '0' from name to
# use the lambda instead
isEqualElement0 = (lambda e1,e2:
(e1 is e2) or
( ( type(e1) is type(e2) ) and
( ( isinstance(e1,list) and compareLists(e1,e2) ) or
(e1 == e2) )
)
)
if len(lst1) != len(lst2):
return False
for e1,e2 in zip(lst1,lst2):
if not isEqualElement(e1,e2):
return False
return True
def test(lst1, lst2):
print "Test:", lst1,"<->",lst2
print compareLists(lst1,lst2)
print lst1 == lst2
print
test([1,2,3], [1,2,3])
test([1,2,3], [1,2])
test([1,'2',3], [1,2,3])
test([1,2,[3,4,5],6], [1,2,[3,4,5],6])
test([1,2,[3,4],6], [1,2,[3,4,5],6])
test([1,2,[7,4,5],6], [1,2,[3,4,5],6])
