w3resource

Python: Function that takes a sequence of numbers and determines whether all are different from each other

Python Basic - 1: Exercise-1 with Solution

Write a Python function that takes a sequence of numbers and determines whether all the numbers are different from each other.

Pictorial Presentation:

Python: Function that takes a sequence of numbers and determines whether all are different from each other

Sample Solution:

Python Code :

def test_distinct(data):
  if len(data) == len(set(data)):
    return True
  else:
    return False;
print(test_distinct([1,5,7,9]))
print(test_distinct([2,4,5,5,7,9]))

Sample Output:

True
False

Flowchart:

Flowchart: Python - Function that takes a sequence of numbers and determines whether all are different from each other

Visualize Python code execution:

The following tool visualize what the computer is doing step-by-step as it executes the said program:

Python Code Editor :

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Previous: Basic - Part-I.
Next: Write a Python program to create all possible strings by using 'a', 'e', 'i', 'o', 'u'. Use the characters exactly once.

What is the difficulty level of this exercise?

Test your Programming skills with w3resource's quiz.



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