I know lambda doesn't have a return expression. Normally
def one_return(a):
#logic is here
c = a + 1
return c
can be written:
lambda a : a + 1
How about write this one in a lambda function:
def two_returns(a, b):
# logic is here
c = a + 1
d = b * 1
return c, d
Yes, it's possible. Because an expression such as this at the end of a function:
return a, b
Is equivalent to this:
return (a, b)
And there, you're really returning a single value: a tuple which happens to have two elements. So it's ok to have a lambda return a tuple, because it's a single value:
lambda a, b: (a, b) # here the return is implicit
, has lower precedence than lambdaSure:
lambda a, b: (a + 1, b * 1)
what about:
lambda a,b: (a+1,b*1)
Print the table of 2 and 3 with a single range iteration.
>>> list(map(lambda n: n*2, range(1,11)))
[2, 4, 6, 8, 10, 12, 14, 16, 18, 20]
>>> list(map(lambda n: (n*2, n*3) , range(1,11)))
[(2, 3), (4, 6), (6, 9), (8, 12), (10, 15), (12, 18), (14, 21), (16, 24), (18, 27), (20, 30)]
not sure if what worked for me is a good practice, but I needed a conditional lambda, which worked this way:
If variable A is 'Not A', I want to set variable A to 'A' and variable B to 'B'
else (if variable A is 'A'), I want to set variable A to 'Not A' and variable B to 'Not B'
lambda: (variable_a='A', variable_b='B') if variable_a=='Not A' else (variable_a='Not A', variable_b='Not B')
,to tuple all over the place. It obfuscates what's actually going on.,is not "promoted to tuple", that's literally the syntax for tuple creation. And it's perfectly clear IMHO.