Python - Why is my list not being modified outside the scope of this method? How do I modify my code so that it is? -

i have following method:

def instances(candidate, candidates):     count = candidates.count(candidate)      # removing candidate candidates. #     list(filter(candidate.__ne__, candidates))     return {candidate: count} 

its intended purpose find how many instances of element in list, delete instances list, , return key:value pair representing element , number of instances.

so if make list , call function, should following:

>>> somelist = [1, 1, 1, 2, 3] >>> print(instances(1, somelist)) {1: 3} >>> print(somelist) [2, 3] 

however, last line, instead:

>>> print(somelist) [1, 1, 1, 2, 3] 

the line list(filter(candidate.__ne__, candidates)) returns correct list want, seems exists in function's scope. if modify line instead read candidates = list(filter(candidate.__ne__, candidates)), reason candidates being interpreted local variable inside scope of function. can't return list filter because need return key:value pair part of program.

i find strange candidates interpreted local variable in line, whereas above it, reference candidates interpreted parameter of function.

can please me understand why happening? , if possible, suggest solution consistent intended purpose? thanks!

i find strange candidates interpreted local variable in line, whereas above it, reference candidates interpreted parameter of function.

python doesn't have references. has variables. foo = bar means "make variable foo point @ whatever bar points at." reassigning variable re-seats pointer. doesn't let change "other" variables point given value. parameters passed using same method: make new pointer object. there no direct support three star programming

modifying list in-place done slicing syntax. this:

candidates[:] = filter(candidate.__ne__, candidates)