Python and pip, list all versions of a package that’s available

append: Appends object at the end.

x = [1, 2, 3]
x.append([4, 5])
print(x)

gives you: [1, 2, 3, [4, 5]]

extend: Extends list by appending elements from the iterable.

x = [1, 2, 3]
x.extend([4, 5])
print(x)

gives you: [1, 2, 3, 4, 5]

With new_list = my_list, you don't actually have two lists. The assignment just copies the reference to the list, not the actual list, so both new_list and my_list refer to the same list after the assignment.

To actually copy the list, you have various possibilities:

• You can use the builtin list.copy() method (available since Python 3.3):

  new_list = old_list.copy()
  

• You can slice it:

  new_list = old_list[:]
  

  Alex Martelli's opinion (at least back in 2007) about this is, that it is a weird syntax and it does not make sense to use it ever. ;) (In his opinion, the next one is more readable).

• You can use the built in list() function:

  new_list = list(old_list)
  

• You can use generic copy.copy():

  import copy
  new_list = copy.copy(old_list)
  

  This is a little slower than list() because it has to find out the datatype of old_list first.

• If the list contains objects and you want to copy them as well, use generic copy.deepcopy():

  import copy
  new_list = copy.deepcopy(old_list)
  

  Obviously the slowest and most memory-needing method, but sometimes unavoidable.

Example:

import copy

class Foo(object):
    def __init__(self, val):
         self.val = val

    def __repr__(self):
        return 'Foo({!r})'.format(self.val)

foo = Foo(1)

a = ['foo', foo]
b = a.copy()
c = a[:]
d = list(a)
e = copy.copy(a)
f = copy.deepcopy(a)

# edit orignal list and instance 
a.append('baz')
foo.val = 5

print('original: %r\nlist.copy(): %r\nslice: %r\nlist(): %r\ncopy: %r\ndeepcopy: %r'
      % (a, b, c, d, e, f))

Result:

original: ['foo', Foo(5), 'baz']
list.copy(): ['foo', Foo(5)]
slice: ['foo', Foo(5)]
list(): ['foo', Foo(5)]
copy: ['foo', Foo(5)]
deepcopy: ['foo', Foo(1)]