Question: I'm not sure whether it's the same in Python.

Has anyone tried that before?

<http://docs.python.org/library/operator#operator.iadd>

Answer: 
There hardly is a difference in the work python performs for either statement:

```
>>> import dis
>>> def inplace_add():
...     a = 0
...     a += 1
... 
>>> def add_and_assign():
...     a = 0
...     a = a + 1
... 
>>> dis.dis(inplace_add)
  2           0 LOAD_CONST               1 (0)
              3 STORE_FAST               0 (a)

  3           6 LOAD_FAST                0 (a)
              9 LOAD_CONST               2 (1)
             12 INPLACE_ADD         
             13 STORE_FAST               0 (a)
             16 LOAD_CONST               0 (None)
             19 RETURN_VALUE        
>>> dis.dis(add_and_assign)
  2           0 LOAD_CONST               1 (0)
              3 STORE_FAST               0 (a)

  3           6 LOAD_FAST                0 (a)
              9 LOAD_CONST               2 (1)
             12 BINARY_ADD          
             13 STORE_FAST               0 (a)
             16 LOAD_CONST               0 (None)
             19 RETURN_VALUE        

```

The difference is a `INPLACE_ADD` versus a `BINARY_ADD`.

The resulting timings are too close to call which one would be faster:

```
>>> import timeit
>>> timeit.timeit('inplace_add', 'from __main__ import inplace_add', number=10000000)
0.32667088508605957
>>> timeit.timeit('add_and_assign', 'from __main__ import add_and_assign', number=10000000)
0.34172606468200684

```

So, in python, the difference is negligible. Don't worry about it.
I took a slightly different approach using the `cProfile` module:

```
$ python -m cProfile test.py 
     4 function calls in 0.397 seconds

Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.397    0.397 test.py:2(<module>)
        1    0.205    0.205    0.205    0.205 test.py:2(add1)
        1    0.192    0.192    0.192    0.192 test.py:6(add2)
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

aaron@zebrafish:~/pyad$ cat test.py 
def add1(a):
    for x in xrange(10 ** 6):
        a += 1

def add2(a):
    for x in xrange(10 ** 6):
        a = a + 1

add1(0)
add2(0)

```

After about 20 runs I would conclude that add2 (using `a = a + 1`) was very slightly faster, but not in all cases (perhaps try it with a greater number of loops). This is probably not the best heuristic, but I figure a greater number of repetitions with larger and larger numbers should indicate a performance difference.

EDIT - results for 10 \*\* 9 calls:

```
    1  216.119  216.119  216.119  216.119 test.py:2(add1)
    1  195.364  195.364  195.364  195.364 test.py:6(add2)

```