I want to return a comparison function from a lambda function and use it depending on the type of the input. Is there a way to do it?
def p12(p):
f = lambda p: compare_nr if type(p) == int else compare_str
return f
if __name__ == '__main__':
cmp1 = p12(0)
cmp1(2,3)
cmp2 = p12('')
cmp2('Mississippi','fall')
There is a way to do it (in Python 3) without using a lambda: functools.singledispatch. With this, you can create a single-dispatch generic function:
A form of generic function dispatch where the implementation is chosen based on the type of a single argument.
[Emphasis mine]
It adds overloaded implementations to the function which are called depending on the type of the first argument, and fortunately, your p12 function takes only one argument.
from functools import singledispatch
@singledispatch
def p12(p):
...
@p12.register(int)
def _(p):
'''compare integers'''
...
@p12.register(str)
def _(p):
'''compare strings'''
...
p12(0) # calls integer dispatch
p12('') # calls string dispatch
And you can still add more candidate functions for other Python types to overload p12.
The problem encountered was that I returned the actual lambda function instead of the defined one.
def p12(p):
f = lambda p: compare_nr if type(p) == int else compare_str
#here was the problem
#return f
return f(p)
if __name__ == '__main__':
cmp1 = p12(0)
cmp1(2,3)
cmp2 = p12('')
cmp2('Mississippi','fall')
You could alternatively replace lambda with a higher order function and do the same.
def p12(p):
if type(p) == int :
return compare_nr
else:
return compare_str
if __name__ == '__main__':
cmp1 = p12(0)
cmp1(2,3)
cmp2 = p12('')
cmp2('Mississippi','fall')
lambdashouldn't be used if you are going to name it anyway... it defeats the entire purpose of an anonymous function.lambda.