Re: If one IF is satisfied, skip the rest in the nest...

[John Gordon]

In <50a6789a-3965-430b-9a91-b08adcedf...@googlegroups.com> 
bigred04...@gmail.com writes:

> So here what I have, I have a 3 IF's within the same level.  If one IF is s=
> atisfied, I would like to "skip" the other IFs and continue with my code.

> # 4 second open
> if wb1_sheet1.cell(row=cell + 1, column=2).value == 0 and 
> wb1_sheet1.cell(row=cell + 1, column=3).value == 0 and 
> wb1_sheet1.cell(row=cell + 2, column=2).value == 0 and 
> wb1_sheet1.cell(row=cell + 2, column=3).value == 0 and 
> wb1_sheet1.cell(row=cell + 3, column=2).value == 0 and 
> wb1_sheet1.cell(row=cell + 3, column=3).value == 0:
> open += 3
> open_seconds += 4
> start = wb1_sheet1.cell(row=cell + 4, column=2).coord=
> inate
> break
> # 3 second open
> if wb1_sheet1.cell(row=cell + 1, column=2).value == 0 a=
> nd wb1_sheet1.cell(row=cell + 1, column=3).value == 0 and wb1_sheet=
> 1.cell(row=cell + 2, column=2).value == 0 and wb1_sheet1.cell(row=
> =cell + 2, column=3).value == 0:
> open += 3
> start = wb1_sheet1.cell(row=cell + 3, column=2).coord=
> inate
> open_seconds += 3
> continue
> # 2 second open
> if wb1_sheet1.cell(row=cell + 1, column=2).value == 0 a=
> nd wb1_sheet1   .cell(row=cell + 1, column=3).value == =
> 0:
> open += 3
> start = wb1_sheet1.cell(row=cell + 2, column=2).coord=
> inate
> open_seconds += 2


> if any but the last IF is true, then all IFs will be true...that's my probl=
> em.

It looks like all three of your if statements start out with this condition:

if wb1_sheet1.cell(row=cell + 1, column=2).value == 0 and 
wb1_sheet1.cell(row=cell + 1, column=3).value == 0

So you could reorganize your code by putting an if statement at the top
that only checks this condition.  Then, indented underneath, you can
check for the other conditions.

And be sure to use "else" and "elif" when they are appropriate.

-- 
John Gordon   A is for Amy, who fell down the stairs
gor...@panix.com  B is for Basil, assaulted by bears
-- Edward Gorey, "The Gashlycrumb Tinies"

-- 
https://mail.python.org/mailman/listinfo/python-list

Re: If one IF is satisfied, skip the rest in the nest...

[bigred04bd3]

O...MG I cannot believe I just asked this question ha.  I ended up 
realizing I need to use an elif after I took a break from it and re read 
everything...Brain fart, thanks for replies.

Brice
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: If one IF is satisfied, skip the rest in the nest...

[Denis McMahon]

On Wed, 21 Oct 2015 10:31:04 -0700, bigred04bd3 wrote:

> So here what I have, I have a 3 IF's within the same level.  If one IF
> is satisfied, I would like to "skip" the other IFs and continue with my
> code.

c1 = wb1_sheet1.cell(row=cell + 1, column=2).value == 0 and 
 wb1_sheet1.cell(row=cell + 1, column=3).value == 0

c2 = wb1_sheet1.cell(row=cell + 2, column=2).value == 0 and 
 wb1_sheet1.cell(row=cell + 2, column=3).value == 0

c3 = wb1_sheet1.cell(row=cell + 3, column=2).value == 0 and 
 wb1_sheet1.cell(row=cell + 3, column=3).value == 0

if c1:
if c2:
if c3:
# c1 && c2 && c3
# 4 second open
else:
# c1 && c2
# 3 second open
else:
# only c1
# 2 second open

Each condition only gets evaluated once.

-- 
Denis McMahon, denismfmcma...@gmail.com
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: If one IF is satisfied, skip the rest in the nest...

[Grant Edwards]

On 2015-10-21, Denis McMahon  wrote:
> On Wed, 21 Oct 2015 10:31:04 -0700, bigred04bd3 wrote:
>
>> So here what I have, I have a 3 IF's within the same level.  If one IF
>> is satisfied, I would like to "skip" the other IFs and continue with my
>> code.
>
> c1 = wb1_sheet1.cell(row=cell + 1, column=2).value == 0 and 
>  wb1_sheet1.cell(row=cell + 1, column=3).value == 0
>
> c2 = wb1_sheet1.cell(row=cell + 2, column=2).value == 0 and 
>  wb1_sheet1.cell(row=cell + 2, column=3).value == 0
>
> c3 = wb1_sheet1.cell(row=cell + 3, column=2).value == 0 and 
>  wb1_sheet1.cell(row=cell + 3, column=3).value == 0
>
> if c1:
> if c2:
> if c3:
> # c1 && c2 && c3
> # 4 second open
> else:
> # c1 && c2
> # 3 second open
> else:
> # only c1
> # 2 second open

if c1 && c2 && c3:
pass   # 4 seconds
elif c1 && c2:
pass   # 3 seconds
elif c1:
pass   # 2 seconds

Or if you want to be particulary obtuse:

seconds = {0b111:4, 0b110:3, 0b100:2}.get(c1<<2 | c2<<1 | c3<<0, None)

> Each condition only gets evaluated once.

OK.

-- 
Grant Edwards   grant.b.edwardsYow! You were s'posed
  at   to laugh!
  gmail.com
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: If one IF is satisfied, skip the rest in the nest...

[Denis McMahon]

On Wed, 21 Oct 2015 20:07:21 +, Grant Edwards wrote:

> On 2015-10-21, Denis McMahon  wrote:
>> On Wed, 21 Oct 2015 10:31:04 -0700, bigred04bd3 wrote:
>>
>>> So here what I have, I have a 3 IF's within the same level.  If one IF
>>> is satisfied, I would like to "skip" the other IFs and continue with
>>> my code.
>>
>> c1 = wb1_sheet1.cell(row=cell + 1, column=2).value == 0 and
>>  wb1_sheet1.cell(row=cell + 1, column=3).value == 0
>>
>> c2 = wb1_sheet1.cell(row=cell + 2, column=2).value == 0 and
>>  wb1_sheet1.cell(row=cell + 2, column=3).value == 0
>>
>> c3 = wb1_sheet1.cell(row=cell + 3, column=2).value == 0 and
>>  wb1_sheet1.cell(row=cell + 3, column=3).value == 0
>>
>> if c1:
>> if c2:
>> if c3:
>> # c1 && c2 && c3 # 4 second open
>> else:
>> # c1 && c2 # 3 second open
>> else:
>> # only c1 # 2 second open
> 
> if c1 && c2 && c3:
> pass   # 4 seconds
> elif c1 && c2:
> pass   # 3 seconds
> elif c1:
> pass   # 2 seconds
> 
> Or if you want to be particulary obtuse:
> 
> seconds = {0b111:4, 0b110:3, 0b100:2}.get(c1<<2 | c2<<1 | c3<<0, None)

Not really valid, because #seconds n is simply a marker to indicate which 
branch of the OP's code to execute.

>> Each condition only gets evaluated once.
> 
> OK.

Yes, but in the structure I suggest, you can move the conditions back 
into the if statements and they still only each get evaluated once. Viz 
my alternative to the OP's code:

if wb1_sheet1.cell(row=cell + 1, column=2).value == 0 and wb1_sheet1.cell
(row=cell + 1, column=3).value == 0:

if wb1_sheet1.cell(row=cell + 2, column=2).value == 0 and 
wb1_sheet1.cell(row=cell + 2, column=3).value == 0:

if wb1_sheet1.cell(row=cell + 3, column=2).value == 0 and 
wb1_sheet1.cell(row=cell + 3, column=3).value == 0:

open += 3 
open_seconds += 4 
start = wb1_sheet1.cell(row=cell + 4, column=2).coordinate

else:

open += 3 
open_seconds += 3
start = wb1_sheet1.cell(row=cell + 3, column=2).coordinate 

else:

open += 3 
open_seconds += 2
start = wb1_sheet1.cell(row=cell + 2, column=2).coordinate 

Not trying to be obtuse here, trying to suggest a practical solution.

Of course, the benefit of reducing the number of times each lookup into 
the worksheet is performed by reducing the number of times each 
comparison evaluated is going to depend on the computational and memory 
manipulation cost of doing so (I assume the workbook is loaded in memory, 
so no IO costs), and how frequently this set of comparisons is being 
performed.

-- 
Denis McMahon, denismfmcma...@gmail.com
-- 
https://mail.python.org/mailman/listinfo/python-list

Re: If one IF is satisfied, skip the rest in the nest...

[Ian Kelly]

On Wed, Oct 21, 2015 at 11:31 AM,   wrote:
> So here what I have, I have a 3 IF's within the same level.  If one IF is 
> satisfied, I would like to "skip" the other IFs and continue with my code.

I think you're looking for the elif keyword. An elif branch will only
be considered if the previous branches were false.

if a:
do_something()
elif b:
do_something_else()
elif c:
do_something_different()
-- 
https://mail.python.org/mailman/listinfo/python-list