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Python: Accept a positive number and subtract from this number the sum of its digits

Python Basic - 1: Exercise-23 with Solution

Write a Python program that accept a positive number and subtract from this number the sum of its digits and so on. Continues this operation until the number is positive.

Sample Solution:

Python Code:

def repeat_times(n):
  n_str = str(n)
  while (n > 0):
    n -= sum([int(i) for i in list(n_str)])
    n_str = list(str(n))
  return n
print(repeat_times(9))
print(repeat_times(20))
print(repeat_times(110))
print(repeat_times(5674))

Sample Output:

0
0
0
0

Flowchart:

Flowchart: Python - Accept a positive number and subtract from this number the sum of its digits

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Python Code Editor:

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Previous: Write a Python program to create a sequence where the first four members of the sequence are equal to one, and each successive term of the sequence is equal to the sum of the four previous ones. Find the Nth member of the sequence.
Next: Write a Python program to find the number of divisors of a given integer is even or odd.

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Python: Tips of the Day

How to make a flat list out of list of lists?

Given a list of lists l

flat_list = [item for sublist in l for item in sublist]

which means:

flat_list = []
for sublist in l:
    for item in sublist:
        flat_list.append(item)

is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:

flatten = lambda l: [item for sublist in l for item in sublist]

As evidence, you can use the timeit module in the standard library:

$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Ref: https://bit.ly/3dKsNTR