Updated REPR
To be able to change what the class is printing, you have to be able to change the repr method. Mainly because what is presented is a bit too constrictive. If instead the class was as follows, then this is possible:
class test():
def __init__(self, a, b, c):
self.a = a
self.b = b
self.c = c
def __repr__(self):
expr = []
if self.a:
expr.append(self.a)
if self.b:
expr.append(self.b)
if self.c:
expr.append(self.c)
return 'test(%s)' % (', '.join(expr))
The biggest hurdle is the fact that we need some control over what is displayed, therefore we need to change the repr method. Alternatively, you might be able to generate the same results by leverage the str method and converting to a string before printing.
Type independent remove duplicates function
Since I didn't know you're exact use case, and wanted to make something relevant to all types of problems, I made the following function. It will allows you to specify an equality condition, what to do when the condition is met, and how to sort the list.
(The default is to check if subsequent values objects are equal, if the are then do not include the second one in the resulting list.)
def remove_dups(old_list,
condition=lambda a, b: a == b,
remove_func=lambda x, y: None,
sort_key=None):
"""
Returns a new list that has duplicates removed from `old_list`
#Arguments
old_list: list to have duplicates discarded
condition: test condition between previous and next obj to test equality.
remove_func: what to add to the list if the condition is met. If `None` is returned
then no value is added to the `new_list`
sort_key: how to sort the `old_list`.
"""
old_list = sorted(old_list, key=sort_key)
comparitor = old_list[0] #used to compare previous and current obj
new_list = [comparitor]
for i, obj in enumerate(old_list[1:]):
#if previous object and current obj are the same
if condition(comparitor, obj):
#run removal function
if remove_func:
new_value = remove_func(comparitor, obj)
else:
new_value = None
else: #then objects are different, add the new one
new_value = obj
if new_value is not None:
new_list.append(new_value)
comparitor = obj
new_value = None
return new_list
Your case (example)
We can use this in your case like follows
test1 = test("1", "abc", "def")
test2 = test("2", "yyy", "xxy")
test3 = test("3", "yyy", "xxy")
objList = [test1, test2, test3]
#in lambda functions `a` corresponds to previous object in list,
#`b` refers to currently observed object
new_list = remove_dups(objList,
condition=lambda a, b: a.b == b.b, #if a.b and b.b the same
remove_func=(lambda a, b: test(b.a, b.b, None)), #then set b.c = None
sort_key=lambda obj: (obj.a, obj.b, obj.c)
)
print(new_list) # [test(1, abc, def), test(2, yyy, xxy), test(3, yyy)]
Edit
If you want to do this without editing the repr method, then you could do some special print statements with the new_list instead of explicitly trying to print the test objects.
Example:
objList = [test1, test2, test3]
#in lambda functions `a` corresponds to previous object in list,
#`b` refers to currently observed object
new_list = remove_dups(objList,
condition=lambda a, b: a.b == b.b, #if a.b and b.b the same
remove_func=(lambda a, b: test(b.a, b.b, None)), #then set b.c = None
sort_key=lambda obj: (obj.a, obj.b, obj.c)
)
exprs = []
for test in new_list:
expr = []
if test.a:
expr.append(test.a)
if test.b:
expr.append(test.b)
if test.c:
expr.append(test.c)
exprs.append('test(%s)' % (', '.join(expr)))
print(exprs) # [test(1, abc, def), test(2, yyy, xxy), test(3, yyy)]
