Is it possible to replace any lambda function with normal function?
e.g. lst1 = [lambda x:x*i for i in range(5)]
lst2 = [j(2) for j in lst1]
Here can we use normal function in stead of lambda?
If it's possible then please tell how to do the same.
Thanks!
Yes you can using functools.partial:
from functools import partial
def fun(x, i):
return x * i
lst1 = [partial(fun, i=i) for i in range(5)]
lst2 = [j(2) for j in lst1]
lst2 does print [8, 8, 8, 8, 8] however because i is 4 at the end of the loop but I don't think that's what he wants.
[8, 8, 8, 8, 8] you could use [8] * 5.yes - lambda is actually "make function"; you will need to give it a name
lst1 = [lambda x:x*i for i in range(5)]
def replace_lambda(x):
return x * x
lst2 = [replace_lambda for i in range(5)]
print lst1
print lst2
for idx, func in enumerate(lst1):
print func(idx)
for idx, func in enumerate(lst2):
print func(idx)
result:
[<function <lambda>>, <function <lambda>>, <function <lambda>>, <function <lambda>>, <function <lambda>>]
[<function replace_lambda>, <function replace_lambda>, <function replace_lambda>, <function replace_lambda>, <function replace_lambda>]
0
1
4
9
16
0
1
4
9
16
[lambda x:x*i for i in range(5)]you will run into classic What do (lambda) function closures capture in Python? problem.