better writing style for the following code

I was wondering if someone could tell me the pythonic way to check out the following.

I have a 6 bit binary number and want to check with its decimal values. Using mathematical function is one way but still it would require that I write around 2**6 if constructs.

So I wanted to know if there's an easier statement to write it.

Also assume that lets say it's not binary then what's the better way to check for 2**6 values in python.

if(a==1):
    ....
else:
  开发者_如何转开发  if(a==2)
.....

One way is saving it in a list and checking it with the indexes but still that would require that many if-else I guess.....

Thanks ....

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Use a dictionary mapping values into outcomes (which can be functions in Python).

For example:

d = {}
d[0] = ....
d[1] = ....
d[2] = ....

outcome = d[a]

Naturally, how this works depends on your ...., but this construct can be very flexible. The most important feature of this approach is that this dictionary can be populated programmatically, and you don't need to write a lot of manual assignments. It's of course also much more efficient than going over many values with nested if statements (or elsif)

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To add to the responses of the others, you should read about the recommended Python style in PEP 8.

With your if version, the brackets are undesirable and spacing is desirable:

if a == 1:
    pass
elif a == 2:
    pass
elif a == 3:
    pass
else:
    pass

---

I would use a decorator to map into a dictionary based dispatch:

_dispatch_table = {}

def dispatch_on(*values):
    def dec(f):
        _dispatch_table.update((v, f) for v in values)
        return f
    return dec

@dispatch_on(0, 2, 47)
def one():
    foo()
    bar()

@dispatch_on(2, 23, 89)
def two():
    bar()
    baz()

x = some_number
_dispatch_table[x]()    

---

Personally I prefer a if/elif if I am understanding your ...

so your:

if(a==1):
    ....
else:
    if(a==2)
.....

Becomes this if you use if elif ladder:

if a==1:
    ....
elif a==2:
    .....
else:
    default

You can also use Python's version of a conditional expression for simple ladders:

def one():
   print("option 1 it is\n")

def two():
   print("option 2 it is\n")

def three():
   print("not one or two\n")

one() if a==1 else two() if a==2 else three()

Or even dictionaries:

def one():
   print("option 1 it is\n")

def two():
   print("option 2 it is\n")

def three():
   print("not one or two\n")

options = {1:one,
           2:two,
           3:three,
}

options[2]()

There is a great discussion on Python forms of switch-case in this SO post.

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Based on the somewhat vague information in the question and what I've been able to gather from the OP's comments, here's my guess:

def func1(): pass
def func2(): pass
def func3(): pass
#     ...
def func62(): pass
def func63(): pass

if 0 < a < 64:
    globals()['func'+str(a)]()
else:
    print 'a is out of range'