CODEWITHSUNDEEP
pythonlambdaiterator
Ruidongd
Ruidongd

Reputation: 41

How to use iteration of lambda in Python

Thank you guys. I find I cannot post my question on internet.

However, it is really helpful to me.

Upvotes: 0

Views: 289

Answers (4)

chepner
chepner

Reputation: 531265

While this can be implemented concisely using recursion, an iterative solution is more efficient and won't break for large values of n.

def compose(f, n):
    def iterated_f(x):
        rv = x
        while n > 0:
            rv = f(rv)
        return rv
    return iterated_f

Upvotes: 0

saulspatz
saulspatz

Reputation: 5261

Compose will need 3 arguments: the function to compose, the number of times to compose it, and the initial argument. So if you want to evaluate f(f(f(7))) you would call compose(f, 3, 7).

def compose(f, n, x):
   if n == 0: return x
   return compose(f, n-1, f(x))

It's occurred to me that maybe this isn't exactly what you want. Perhaps you want to be able to write g = compose(f, 3) and then call g(7). This requires only a minor modification:

def compose(g, m):
    def composer(f, n, x):
       if n == 0: return x
       return compose(f, n-1, f(x))
    return lambda x: composer(g, m, x)

Upvotes: 5

LetzerWille
LetzerWille

Reputation: 5658

use saulspatz's compose and a lambda 

cubed = lambda x: x**3

def compose(f, n, x):
   if n == 0: return x
   return compose(f, n-1, f(x))

print(compose(cubed,2,2))

512

Upvotes: 0

Daniel Castro
Daniel Castro

Reputation: 1290

I am not sure if I understand correctly, but I believe you expect compose() to return a function. In that case, this would work:

def compose(f, n):
   def composed_f(x):
      result = x

      for i in range(n):
         result = f(result)
      return result

   return composed_f

So, if you write:

a = compose((lambda x:x*2), 5)
print a(1)

You will get 32

Upvotes: 1

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