AndrewSmiley - 10 months ago 35

Python Question

I'm not sure if this should go in cs or programmers stack exchange instead, so if it should please let me know. I was speaking with one of my graduate professors about the use of

`continue`

`for`

`continue`

`GO TO`

`continue`

`continue`

I'm also curious about the performance impact. As mentioned above, in other languages,

`continue`

`continue`

Here in the example code (an attempted rework of insertion sort, ignore the overall logic, this stemmed from some experimentation to attempt get better than O(n

`continue`

`def insertion_sort(array):`

sublist = [array[0]]

for i in range(1, len(array)):

if array[i] >= sublist[i-1]:

sublist.append(array[i])

continue #possible performance loss here

if array[i] <= sublist[0]:

sublist.insert(0, array[i])

continue #possible performance loss here

for j in reversed(range(0, i)):

if array[i] >= sublist[j-1] and array [i] <=sublist[j]:

sublist.insert(j, array[i])

break #possible performance loss here

return sublist

Answer

In this case you're using `continue`

as a means of "selection" in the case of mutually-exclusive conditions. Consider the following simpler examples...

```
def f1():
for i in range(10):
if i < 5:
print 'i is less than 5'
if i > 5:
print 'i is greater than 5'
def f2():
for i in range(10):
if i < 5:
print 'i is less than 5'
continue
if i > 5:
print 'i is greater than 5'
```

The two functions will produce identical outputs, but in the first case, given that we know that `i`

cannot be both greater than and less than 5 at the same time, the first function performs an unnecessary comparison, and will likely take longer to execute.

The second example is more commonly expressed as...

```
def f3():
for i in range(10):
if i < 5:
print 'i is less than 5'
elif i > 5:
print 'i is greater than 5'
```

As for which is more performant, you can take a look at the Python `dis`

module, which for `f2()`

yields...

```
0 SETUP_LOOP 63 (to 66)
3 LOAD_GLOBAL 0 (range)
6 LOAD_CONST 1 (10)
9 CALL_FUNCTION 1
12 GET_ITER
>> 13 FOR_ITER 49 (to 65)
16 STORE_FAST 0 (i)
19 LOAD_FAST 0 (i)
22 LOAD_CONST 2 (5)
25 COMPARE_OP 0 (<)
28 POP_JUMP_IF_FALSE 42
31 LOAD_CONST 3 ('i is less than 5')
34 PRINT_ITEM
35 PRINT_NEWLINE
36 JUMP_ABSOLUTE 13
39 JUMP_FORWARD 0 (to 42)
>> 42 LOAD_FAST 0 (i)
45 LOAD_CONST 2 (5)
48 COMPARE_OP 4 (>)
51 POP_JUMP_IF_FALSE 13
54 LOAD_CONST 4 ('i is greater than 5')
57 PRINT_ITEM
58 PRINT_NEWLINE
59 JUMP_ABSOLUTE 13
62 JUMP_ABSOLUTE 13
>> 65 POP_BLOCK
>> 66 LOAD_CONST 0 (None)
69 RETURN_VALUE
```

...and for `f3()`

...

```
0 SETUP_LOOP 60 (to 63)
3 LOAD_GLOBAL 0 (range)
6 LOAD_CONST 1 (10)
9 CALL_FUNCTION 1
12 GET_ITER
>> 13 FOR_ITER 46 (to 62)
16 STORE_FAST 0 (i)
19 LOAD_FAST 0 (i)
22 LOAD_CONST 2 (5)
25 COMPARE_OP 0 (<)
28 POP_JUMP_IF_FALSE 39
31 LOAD_CONST 3 ('i is less than 5')
34 PRINT_ITEM
35 PRINT_NEWLINE
36 JUMP_ABSOLUTE 13
>> 39 LOAD_FAST 0 (i)
42 LOAD_CONST 2 (5)
45 COMPARE_OP 4 (>)
48 POP_JUMP_IF_FALSE 13
51 LOAD_CONST 4 ('i is greater than 5')
54 PRINT_ITEM
55 PRINT_NEWLINE
56 JUMP_ABSOLUTE 13
59 JUMP_ABSOLUTE 13
>> 62 POP_BLOCK
>> 63 LOAD_CONST 0 (None)
66 RETURN_VALUE
```

...the only significant difference being `39 JUMP_FORWARD`

in the first example which can never actually be executed, due to a preceding unconditional jump, but using this to decide which is 'better' is taking optimization to a ridiculous level.

The two examples `f2()`

and `f3()`

only really differ stylistically. Consider...

```
def f4():
while 1:
if expr1:
do_something()
else:
if expr2:
do_something_else()
if expr3:
do_a_different_thing()
def f5():
while 1:
if expr1:
do_something()
continue
if expr2:
do_something_else()
if expr3:
do_a_different_thing()
```

...which are functionally identical, but `f5()`

minimizes the indentation level of nested blocks, which can be more desirable for narrower fixed-width displays. As for which is more 'readable', that's purely subjective.

Back to your code, you could have equally expressed it as...

```
def insertion_sort(array):
sublist = [array[0]]
for i in range(1, len(array)):
if array[i] >= sublist[i-1]:
sublist.append(array[i])
elif array[i] <= sublist[0]:
sublist.insert(0, array[i])
else:
for j in reversed(range(0, i)):
if array[i] >= sublist[j-1] and array [i] <=sublist[j]:
sublist.insert(j, array[i])
break #possible performance loss here
return sublist
```

...which would eliminate the `continue`

s, but as the `dis`

module demonstrates, there is no performance advantage in doing so. As for the `break`

, as far as I can tell, that's required for the algorithm to function correctly.