Only to add my 5 pedantic coins to @Warren Weckesser answer. Really from numpy import *
does not overwrite the builtins
sum
function, it only shadows __builtins__.sum
, because from ... import *
statement binds all names defined in the imported module, except those beginning with an underscore, to your current global
namespace. And according to Python's name resolution rule (unofficialy LEGB rule), the global
namespace is looked up before __builtins__
namespace. So if Python finds desired name, in your case sum
, it returns you the binded object and does not look further.
EDIT:
To show you what is going on:
In[1]: print(sum, ' from ', sum.__module__) # here you see the standard `sum` function
Out[1]: <built-in function sum> from builtins
In[2]: from numpy import * # from here it is shadowed
print(sum, ' from ', sum.__module__)
Out[2]: <function sum at 0x00000229B30E2730> from numpy.core.fromnumeric
In[3]: del sum # here you restore things back
print(sum, ' from ', sum.__module__)
Out[3]: <built-in function sum> from builtins
First note: del
does not delete objects, it is a task of garbage collector, it only "dereference" the name-bindings and delete names from current namespace.
Second note: the signature of built-in sum
function is sum(iterable[, start])
:
Sums start
and the items of an iterable
from left to right and returns the total. start
defaults to 0
. The iterable‘s items are normally numbers, and the start value is not allowed to be a string.
I your case print(sum(range(5),-1)
for built-in sum
summation starts with -1. So technically, your phrase the sum over the iterable and then subtracts the second args value from the sum isn't correct. For numbers it's really does not matter to start with or add/subtract later. But for lists it does (silly example only to show the idea):
In[1]: sum([[1], [2], [3]], [4])
Out[1]: [4, 1, 2, 3] # not [1, 2, 3, 4]
Hope this will clarify your thoughts :)