Functional append/extend

Question

The methods append and extend in Python are not functional by nature, they modify the callee and return None.

Is there an alternative way to do what these methods do and get a new list as a returned value?

Consider this example:

def myfun(first, *args):
  for elem in [first].extend(args):
    print elem

Obviously, this won’t work.

Is there a way to construct a new list directly with an expression, instead of being forced to write the following?

def myfun(first, *args):
   all_args = list(first)
   all_args.extend(args)

   for elem in all_args:
     print elem

Thanks.

Asked By: Alexei Sholik

||

Source

Answer 1

You can rewrite that as:

[first] + args

Answered By: tangentstorm

Answer 2

>>> def append(lst, elem):
...     return lst + [elem]
... 
>>> append([1, 2, 3], 4)
[1, 2, 3, 4]
>>> def extend(lst1, lst2):
...   return lst1 + lst2
... 
>>> extend([1, 2], [3, 4])
[1, 2, 3, 4]

Is that what you wanted?

You may also define your own type, which returns the list itself on these operations, in addition to change:

>>> class MyList(list):
...   def append(self, x):
...     super(MyList, self).append(x)
...     return self
...   def extend(self, lst):
...     super(MyList, self).extend(lst)
...     return self
... 
>>> l = MyList([1, 2, 3])
>>> l.append(4)
[1, 2, 3, 4]
>>> l.extend([5, 6, 7])
[1, 2, 3, 4, 5, 6, 7]
>>> l
[1, 2, 3, 4, 5, 6, 7]

Answered By: Michal Chruszcz

Functional append/extend

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