Member Avatar for Thisisnotanid

I was wondering, is it possible to do something like the following? The given code obviously doesn't work.

l = 5

(l == 5)*(k = 7) #Returns error message.

The reason is, I'd rather avoid doing something like the following for the project I'm currently working on.

if stmt1 == True or stmt2 == True:


  if stmt1 == True:

    k = 7

  if k == 7:

    # Evaluate suite

  else:

    # Evaluate different suite
Edited by Thisisnotanid because: Additional info
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Member Avatar for Gribouillis

No it's not possible to have assignments in if conditions. The reason is that assignment is a statement and cannot be part of an expression (the same applies to += -=, etc), while the 'if' statement expects an expression. The potential problem with your code above is that it evaluates stmt1 several times. You could write it this way

c = expr1
if c or expr2:
    if c:
        k = 7
    if k == 7:
        # etc
    else:
        # etc

Notice that '== True' is usually not needed. I know it looks restricting if you're used to C code, but it is one of the features that make python code clear.

Edited by Gribouillis because: n/a
Member Avatar for TrustyTony

Something along these lines?

>>> ctrl = 5
>>> k = 7 if ctrl == 5 else None
>>> k
7
Member Avatar for Thisisnotanid

Thanks fellas, that was helpful. Gribouillis, It turns out that the method you outlined is actually equivalent to what I am trying to do, so I can make use of that. PyTony, I wanted to have the assignment nested in an if statement; I was hoping to avoid having to do precisely what you outlined.

Member Avatar for Thisisnotanid

It turns out that something like what's being discussed is possible. It's not equivalent, but if you design with this catch in mind, you can make your programs behave in a way which mimics the desired behavior:

l = 5

k = 7*(l == 5)

Note, that while implementing boolean algebra in such a way can come handy, it can not always replace logic. For example, if '*' in the following is not replaced with 'and', the code will return an error message.

list = range(10) # Some list

a = len(list)

b = a + 5

for l in xrange(b):

  if (l < a)*(not list[l]%2):

    print list[l]
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