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Python: Count the number of prime numbers less than a given non-negative number

Python Basic - 1: Exercise-68 with Solution

Write a Python program to count the number of prime numbers less than a given non-negative number.

Sample Solution:

Python Code:

def count_Primes_nums(n):
    ctr = 0
    
    for num in range(n):
        if num <= 1:
            continue
        for i in range(2, num):
            if (num % i) == 0:
                break
        else:
            ctr += 1

    return ctr

print(count_Primes_nums(10))
print(count_Primes_nums(100))

Sample Output:

4
25

Pictorial Presentation:

Python: Count the number of prime numbers less than a given non-negative number

Flowchart:

Flowchart: Python - Count the number of prime numbers less than a given non-negative number

Python Code Editor:

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Python: Tips of the Day

Iterating over dictionaries using 'for' loops:

I am a bit puzzled by the following code: d = {'x': 1, 'y': 2, 'z': 3} for key in d: print key, 'corresponds to', d[key] What I don't understand is the key portion. How does Python recognize ...

key is just a variable name.

for key in d:

For Python 3.x:

for key, value in d.items():

For Python 2.x:

for key, value in d.iteritems():

To test for yourself, change the word key to poop.

In Python 3.x, iteritems() was replaced with simply items(), which returns a set-like view backed by the dict, like iteritems() but even better. This is also available in 2.7 as viewitems().

The operation items() will work for both 2 and 3, but in 2 it will return a list of the dictionary's (key, value) pairs, which will not reflect changes to the dict that happen after the items() call. If you want the 2.x behavior in 3.x, you can call list(d.items()).

Ref: https://bit.ly/37dm0Qo