Learn Python Fundamental in 30 Days — Day 9(while/for loop)

With while loop we can make block of code executed as long as while statement is True

test = 0
while ( test < 5):
    print("hello")
    test = test + 1
hello
hello
hello
hello
hello

Here as soon as the execution reaches the end of a while statement’s block, it jumps back to the start to re-check the condition. We are initializing test=0 and as long as the value of a test is less than 5, While checking the condition and prints the hello.

When execution runs through the loop we call it iteration we say this while loop iterates 5 times.

But let’s take a look at other two cases

• If forget to initialize the value, I will receive an error

while n < 5:
    print(n)
    n = n+1

Output

Traceback (most recent call last):
 File “/Users/plakhera/Documents/python_testing/test.py”, line 1, in <module>
 while n < 5:
NameError: name ’n’ is not defined
Process finished with exit code 1

• If I forget to increment the value, then it will become an infinite loop

n = 0
while n < 5:
    print(n)

Infinite Loop

>>> while True:
… print(“hello”)
…
hello
hello
hello

Since this condition is always True this will cause the loop to be executed forever. To come out of infinite loop use Ctrl+c

Break: Cause statement to jump out of loop, without re-checking the condition

test = 0
while (test < 10):
    print(test)
    test = test + 1
    if test == 5:
        break

Output

0
1
2
3
4

Continue

In the case of continue statement when the program execution reach the continue statement, program execution immediately jump back to the start off while loop and re-check the condition, and that is the reason test==5 never executed

test = 0
while (test < 6):
    test = test + 1
    if test == 5:
        continue
    print("Current Value of test: " + str(test))

Output

Current Value of test: 1
Current Value of test: 2
Current Value of test: 3
Current Value of test: 4
Current Value of test: 6

For Loop

For loop is used when we want to iterates a specific number of times.

for i in range(5):
    print(i)

Output

0
1
2
3
4

Sum of first 100 number

total = 0
#Because it goes one less
for i in range(101):
    total = total + i
print(total)

Output

5050

OR one more favorite interview question, print all the even number between 0 to n number

for i in range(0,11,2):
    print(i)

Output

0
2
4
6
8
10

Let summarize

Let’s try to solve few more problem

In this problem we need to count number of vowels in a string

s = 'welcome to the world of python'
count = 0
for i in s:
    if (i == "a" or i == "e" or i == "i" or i == "o" or i == "u"):
        count += 1
print(count)

Is there is any better way to write the same code?

s = 'welcome to the world of python'
count = 0
vowels = ["a","e","i","o","u"]
for i in s:
    if i in vowels:
        count += 1
print(count)

But this code will fail if we have Vowels like AEIOU

s = 'welcOmE to the world of python'

It will return only 6 vowels

There are multiple ways to deal with this problem but the easiest way right now is the use of lower()

s = 'welcOmE to the world of python'
s = s.lower()
count = 0
vowels = ["a","e","i","o","u"]
for i in s:
    if i in vowels:
        count += 1
print(count)

Later on when we jump to regular expression we will see we can solve the same problem using regular expression

>>> import re
>>> s = ‘welcOmE to the world of python’
>>> vowels = re.findall(‘[aeiou]’,s,re.IGNORECASE)
>>> print(len(vowels))
8

There might be a situation where we might need to know how many times the gives element is repeated in a string. There are number of ways to do that but the easiest way is to use collections module

>>> from collections import Counter
>>> s
‘welcOmE to the world of python’
>>> a = Counter(s)
>>> print(a)
Counter({‘ ‘: 5, ‘o’: 4, ‘t’: 3, ‘w’: 2, ‘e’: 2, ‘l’: 2, ‘h’: 2, ‘c’: 1, ‘O’: 1, ‘m’: 1, ‘E’: 1, ‘r’: 1, ‘d’: 1, ‘f’: 1, ‘p’: 1, ‘y’: 1, ’n’: 1})

Let’s take one more case in which I need to find out how many times prash repeated in this string

s = 'hello my name is prash and the last name is prash too'
count = 0
for i in range(len(s)):
    if s[i:i+5] == "prash":
        count += 1
print(count)
2

So this end of Day9, In case if you are facing any issue, this is the link to Python Slack channel https://devops-myworld.slack.com
Please send me your details
* First name
* Last name
* Email address
to devops.everyday.challenge@gmail.com, so that I will add you to this slack channel