when i run the code i get a memory location as an output
<function at 0x00000200FF8A5E50>
val = (lambda colmns:(colmns.max + colmns.min)/2
print(val)
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3You need to actually use the `lambda`. If you try to print it directly, you will get the address of where the lambda code lives in memory. – wkl Jul 21 '22 at 22:54
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Right. Assuming you have an array, do `print(val(array))`. – Tim Roberts Jul 21 '22 at 22:57
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`val` is a function object, it isn't a "memory address", when you print a function object, that is how it is represented, just like any function, try `def foo(): return 42` then `print(foo)` – juanpa.arrivillaga Jul 21 '22 at 23:05
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Why did you *expect* a different output? – juanpa.arrivillaga Jul 21 '22 at 23:05
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A lambda is just an anonymous function object and that's why it prints that way. In fact, as soon as you assign it to a variable... then why have a lambda in the first place? Might as well do
def foo(colmns):
return (colmns.max + colmns.min)/2
And then it wouldn't be too surprising that print(foo) is <function 0x...>.
tdelaney
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@hghebrem - what are you talking about? The thing you seemed to have missed is that lambda creates a function object, and that's what I illustrated with the `def`. But I still needed to wrap that back specifically to the question about where "
" comes from. When you see it as a `def` I didn't think there would be any question about why is printed as a "function". – tdelaney Jul 22 '22 at 18:12