Strong numbers in python

Question

A number is said to be a strong number if the sum of the factorials of the individual digits is equivalent to the number itself. For instance: 145 = 1! + 4! +5! 

I composed the accompanying code in python for this:

import math

def strong_num():

    return [x for x in range(1,1000) if x==int(reduce(lambda p,q:math.factorial(int(p))+math.factorial(int(q)),str(x)))]

print strong_num()

I don't know where this code went wrong??

Answer 1

Your reduce input isn't right, at that point you shouldn't figure the factorial of p. Truth be told, it is simpler to simply utilize the sum 

return [x for x in range(1, 1000) 

          if x == sum(math.factorial(int(q)) for q in str(x))]

functions.reduce function will be considered as:

reduce(f, [a, b, c, d, ...]) == f(f(f(a, b), c), d) ...

In this way, for example, if that x == 145, at that point your reduce part will compute

   int(reduce(lambda p, q: factorial(int(p)) + factorial(int(q)), str(x)))

== int(reduce(lambda p, q: factorial(int(p)) + factorial(int(q)), "145"))

== int(factorial(factorial(1) + factorial(4)) + factorial(5))

== int(factorial(1 + 24) + 120)

== int(15511210043330985984000000 + 120)

== 15511210043330985984000120

The interpreter doesn't complete likely because of expecting to process the factorial of a very huge number (consider (2 × 9!)!...) 

In the event that you actually need to keep the reduce, you should transform it to:

 reduce(lambda p,q: p + math.factorial(int(q)),  str(x),  0)

#                   ^                                     ^

#                   No need to factorial                  Add initializer too

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