Checking three conditions (in every way) with if elif stataement-CodePudding

I am trying to check three statements if any of them or two of them is True or False.

But I don't know efficient way to check all three in different ways at same time.

There are three Booleans and I am trying to check like :-

1). If boolean_1 is True and boolean_2 and boolean_3 are False. => do something_1

2). If boolean_2 is True and boolean_1 and boolean_3 are False. => do something_2

3). If boolean_3 is True and boolean_1 and boolean_2 are False. => do something_3

4). If boolean_1 and boolean_2 is True and boolean_3 is False. => do something_4.

5). If boolean_1 and boolean_3 is True and boolean_2 is False. => do something_5.

6). If boolean_2 and boolean_3 is True and boolean_1 is False. => do something_6.

7). If boolean_1 is True and boolean_2 is True and boolean_3 is True. => (and guess what) do something_7.

A way I can do is :-

file.py


if boo_1 == True and bool_2 == False and bool_3 == False:
    do_something_1

elif boo_2 == True and bool_1 == False and bool_3 == False:
    do_something_2

elif boo_3 == True and bool_2 == False and bool_1 == False:
    do_something_3

elif boo_1 == True and bool_2 == True and bool_3 == False:
    do_something_4

elif boo_1 == True and bool_3 == True and bool_2 == False:
    do_something_5

elif boo_2 == True and bool_3 == True and bool_1 == False:
    do_something_6

else:
    do_something_7

I can also use this lengthy if elif method But I don't know why I think that "This is not the efficient way". and I think this will slow. Will this slow ?

Questions

Will this lengthy if elif method be slow ?
Is there a more efficient way instead of this lengthy code ?

Any suggestions would much Appreciated. Thank You

CodePudding user response：

For cleaner code you don't need to check equality vs True/False as they already evaluate as that. Consider changing:

# from this
if boo_1 == True and bool_2 == False, bool_3 == True:
# to this
if boo_1 and not bool_2 and bool_3:

As for efficiency, consider arranging your ifs so it needs to do the least amount of checks as possible to get it's result. Aside from that consider the below for other options.

Using a dict to determine what to do in each case:

switch = {
    (True, True, True): do_thing_1,
    (True, True, False): do_thing_2,
    (True, False, False): do_thing_3,
    (True, False, True): do_thing_4,
    ...
}

Then just access using your variables:

switch[(bool_1, bool_2, bool_3)](*args, **kwargs)

Or for an else clause use dict.get:

switch.get((bool_1, bool_2, bool_3), do_else)(*args, **kwargs)

Using Structural Pattern Matching in python 3.10

match [bool_1, bool_2, bool_3]:
    case [True, False, False]:
        do_thing_1
    case [True, (True | False), False]: # to match a True or False
        do_thing_x
    case _:
        do_else

CodePudding user response：

Well, on such a small scale I doubt this will make any difference, but you could cut it down by if/else in a branched way:

if boo_1:
    if boo_2:
        if boo_3:
            do_something
        else:
            do_something
    else:
        if boo_3:
            do_something
        else:
            do_something
else:
    if boo_2:
        if boo_3:
            do_something
        else:
            do_something
    else:
        if boo_3:
            do_something
        else:
            do_something

This way you can in best case cut down the number checks down. In your best case, it checks already 3 times, in worst case 18 times. Just to get the conclusion that all have been False. With each if/elif statement it must check for 3 conditions. If you do it in a branched way, it must check in best case 3 conditions, as well as in worst case. The difference is not big, but there is a difference. I quick wrote a program which got to assign to boo_1 to boo_3 a random 0 or 1 and check it through the branched if/else statements and your if/elif/else statement.

After 1mio runs, the branched function had an average duration of 2.8696 sec while your construct had 3.0864 sec.

CodePudding user response：

With numpy and dict you can minimize rows

import numpy as np

bool_0 = False
bool_1 = True
bool_2 = True

t = [bool_0,bool_1,bool_2] # this is list of true/False index are [0,1,2]

switch = {0:"--0--",1:"--1--",2:"--2--",'01':"--01--",'12':'--12--','03':'--03--'} # these are conditions to follow

if bool_0 bool_1 bool_2 == 1:
    print(switch[np.where(t)[0][0]])
elif bool_0 bool_1 bool_2 == 2:
    print(switch[str(np.where(t)[0][0]) str(np.where(t)[0][1])])
else:
    print('Either all True or all False')

you need to change conditions inside switch dict keeping keys as it is