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Home/ Questions/Q 816789
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T01:53:32+00:00 2026-05-15T01:53:32+00:00

I noticed a problem when I was trying to use del in a lambda

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I noticed a problem when I was trying to use del in a lambda to thin out a list of threads to just those running:

map(lambda x: del(x) if not x.isAlive() else x, self.threads)

Ignore for a second that this doesn’t do anything, I’m just fooling around with map, reduce, and lambda.

This fails with a syntax error at del(x). With some messing around, I think the problem is del() doesn’t return a value. For example, this fails with the same error:

b = 5
x = del(b)

This doesn’t, however:

def rmThis(x): del(x)

Which means I’m using this workaround:

map(lambda x: rmThis(x) if not x.isAlive() else x, self.threads)

So is the limitation just because del() doesn’t return a value? Why not?

I’m using python 2.6.2

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  1. Editorial Team
    Editorial Team
    2026-05-15T01:53:33+00:00Added an answer on May 15, 2026 at 1:53 am

    Two problems. The first one is more subtle so I’ll explain that first.

    The issue is that del removes a variable binding. Passing it a value will not serve your purpose.

    Here’s an illustration

    >>> a = 5
    >>> del(a)
    >>> a
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
    NameError: name 'a' is not defined
    >>> def mydel(x): del(x)
    ... 
    >>> a = 5
    >>> mydel(a)
    >>> a
    5
    >>> 
    

    As you can see. In the first case, the variable a has been removed from the current namespace. There’s no way to refer to the object it pointed to anymore (assuming nothing else points to the same thing).

    In the second case, you are not removing a from the namespace. You are deleting the binding of x in the function namespace the effect of which is that you won’t be able to use x as as an rvalue anymore in the function (ie. it’s an undefined variable).

    The second problem is the SyntaxError which is simpler. Python lambda functions can only have expressions in them and not statements. del in not an expression (i.e., not a function call) – it’s a statement (the del_stmt) and so it can’t appear in the body of the lambda. You’d see the same issue if you tried putting a print in the body of the lambda.

    >>> lambda x: print x
      File "<stdin>", line 1
        lambda x: print x
                      ^
    SyntaxError: invalid syntax
    

    This also accounts for why the x=del(a) fails. The statement syntax is invalid. It’s not a function that can be called.

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