Python function is changing the value of passed parameter

Question

Here\'s an example of where I started

mylist = [[\"1\", \"apple\"], [\"2\", \"banana\"], [\"3\", \"carrot\"]]

def testfun(passedvariable):
    for row in passed         


        

Answer 1

mylist outside the function and passedvariable are the same list object. So changing the list is reflected everywhere. The same holds true for copyofdata in the third example. It is no copy, but the same list object again. To make a copy, you have to explicitly copy the list, in your case you even have to copy each element of the list, as they are also list-objects. Now for the second example row = row[:-1]: Here you make a copy of the list, except the last element. So the former row is not changed, but a new list object is bound the the same name row.

Answer 2

Python passes everything by sharing (references passed as value, see call by sharing), however the integrated numeric and string types are immutable, so if you change them the value of the reference is changed instead of the object itself. For mutable types like list, make a copy (e.g. list(passedvariable)). If you are modifying mutable objects within a list (which can only contain references!) you will need to perform a deep copy, to do so use

import copy
copy.deepcopy(passedvariable)

See https://docs.python.org/2/library/copy.html (available since Python 2.6)

Note that since references themselves are passed by value, you cannot change a reference passed as a parameter to point to something else outside of the function (i. e. passedvariable = passedvariable[1:] would not change the value seen outside the function). A common trick is to pass a list with one element and changing that element.