I've seen there are actually two (maybe more) ways to concatenate lists in Python: One way is to use the extend() method:
a = [1, 2]
b = [2, 3]
b.extend(a)
the other to use the plus(+) operator:
b += a
Now I wonder: Which of those two options is the 'pythonic' way to do list concatenation and is there a difference between the two (I've looked up the official Python tutorial but couldn't find anything anything about this topic).
The only difference on a bytecode level is that the
.extend
way involves a function call, which is slightly more expensive in Python than theINPLACE_ADD
.It's really nothing you should be worrying about, unless you're performing this operation billions of times. It is likely, however, that the bottleneck would lie some place else.
According to the Python for Data Analysis.
“Note that list concatenation by addition is a comparatively expensive operation since a new list must be created and the objects copied over. Using extend to append elements to an existing list, especially if you are building up a large list, is usually preferable. ” Thus,
is faster than the concatenative alternative:
I would say that there is some difference when it comes with numpy (I just saw that the question ask about concatenating two lists, not numpy array, but since it might be a issue for beginner, such as me, I hope this can help someone who seek the solution to this post), for ex.
it will return with error
ValueError: operands could not be broadcast together with shapes (0,) (4,4,4)
b.extend(a)
works perfectlyFrom python 3.5.2 source code: No big difference.
According to the Zen of Python:
b += a
is more simple thanb.extend(a)
.The builtins are so highly optimized that there's no real performance difference.
You can't use += for non-local variable (variable which is not local for function and also not global)
It's because for extend case compiler will load the variable
l
usingLOAD_DEREF
instruction, but for += it will useLOAD_FAST
- and you get*UnboundLocalError: local variable 'l' referenced before assignment*