在Python中，当执行`if`语句时，如何从函数返回值？

Question

In this function, I want to return a value when the if statement is executed.

Since Python always returns something, it returns None .

def Persistence(number, counter):
    numArr = [int(i) for i in str(number)]
    result = 1
    data = None
    for j in numArr:
        result *= j
    if len(str(number)) == 1:
        data = str(f'Done. {counter} steps taken')
        print(data)
        # return data
    else:
        counter = counter + 1
        Persistence(result, counter)
    # return data

print(Persistence(333,0))

Maybe I put the return keyword in the wrong place (I checked by putting it in two different places, marked as a comment) or something else, but I couldn't figure it out.

Please help me out. Also, if there is another way to count the recursion steps besides my technique, please let me know.

Answer 1

Maybe try this :

def Persistence(number, counter):
    numArr = [int(i) for i in str(number)]
    result = 1

    for j in numArr:
        result *= j
    if len(str(number)) == 1:
        return str(f'Done. {counter} steps taken')

    counter = counter + 1
    return Persistence(result, counter)

print(Persistence(333,0))

Hope it helps

Answer 2

The issue is that you're not setting the value from the else call to persistence to anything. The following code return the data value for me:

def Persistence(number, counter):
    numArr = [int(i) for i in str(number)]
    result = 1
    data = None
    for j in numArr:
        result *= j
    if len(str(number)) == 1:
        data = str(f'Done. {counter} steps taken')
        print(data)
        return data
    else:
        counter = counter + 1
        data = Persistence(result, counter)
    return data

x = Persistence(333, 0)

Then if we print x:

print(x)
# Done. 3 steps taken

Answer 3

Your logic to count the recursion steps is basically correct, you just need to place the return statements for both the: 1)Base case 2)The recursive call itself

The following modification to your code will do the trick of what you are asking:

def Persistence(number, counter):
    numArr = [int(i) for i in str(number)]
    result = 1
    data = None
    for j in numArr:
        result *= j
    if len(str(number)) == 1:
        data = str(counter)
        return data
    else:
        counter = counter + 1
        return Persistence(result, counter)


print(Persistence(333,0))

The above code will return the output of:

3

Please note that the reason you were getting "None" as the output in your original code is because you were not making a return at the actual recursive call itself: **return** Persistence(result, counter)

So when you ran print(Persistence(333,0)) it was returning nothing resulting in the None .

Answer 4

This question is better to learn about recursion than you may think. But we won't muddy the waters by mixing recursion (functional style) with statements and side effects (imperative style).

It looks like you're trying to calculate multiplicative root and persistence. Instead of putting all concerns of the computation into a single function, break it down into sensible parts -

def digits (n = 0):
  if n < 10:
    return [ n ]
  else:
    return digits (n // 10) + [ n % 10 ]

def product (n = 0, *more):
  if not more:
    return n
  else:
    return n * product (*more)

def mult_root (n = 0):
  if n < 10:
    return [ n ]
  else:
    return [ n ] + mult_root (product (*digits (n)))

def mult_persistence (n = 0):
  return len (mult_root (n)) - 1

print (mult_persistence (333))
# 3

print (mult_root (333))
# [ 333, 27, 14, 4 ]