w3resource

Python: Count the number of equal numbers from three given integers

Python Basic - 1: Exercise-86 with Solution

Write a Python program to count the number of equal numbers from three given integers.

Sample Solution:

Python Code:

def test_three_equal(x, y, z):
  result= set([x, y, z])
  if len(result)==3:
    return 0
  else:
    return (4 - len(result))

print(test_three_equal(1, 1, 1))
print(test_three_equal(1, 2, 2))
print(test_three_equal(-1, -2, -3))
print(test_three_equal(-1, -1, -1))

Sample Output:

3
2
0
3

Pictorial Presentation:

Python: Count the number of equal numbers from three given integers.
Python: Count the number of equal numbers from three given integers.

Flowchart:

Flowchart: Python - Count the number of equal numbers from three given integers.

Python Code Editor:

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

Previous: Write a Python program to check whether a given string is an "isogram" or not.
Next: Write a Python program to check whether a given employee code is exactly 8 digits or 12 digits.

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