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🔧 Python Day-26 List comprehension-Exercises


Nachrichtenbereich: 🔧 Programmierung
🔗 Quelle: dev.to

List Comprehension

List comprehension offers a shorter syntax when you want to create a new list based on the values of an existing list. (refer-https://www.w3schools.com/python/python_lists_comprehension.asp)

Example:1
Method:1

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]
newlist = []

for x in fruits:
  if "a" in x:
    newlist.append(x)

print(newlist)

Method:2(comprehensive)

fruits = ["apple", "banana", "cherry", "kiwi", "mango"]

newlist = [x for x in fruits if "a" in x]
print(newlist)

Output:

['apple', 'banana', 'mango']

Example:2

l = [10,20,30,40]
newlist = []
#using normal loop
for num in l:
    newlist.append(num**2)
print(newlist)

#using loop in comprehensive way
newlist = [num**2 for num in l]
print(newlist)

Output:

[100, 400, 900, 1600]
[100, 400, 900, 1600]

Exercise:
1.Find similar numbers from 2 lists and different numbers from the same 2 lists.
l1 = [10,20,30,40]
l2 = [30,40,50,60]
Get this output:
a) 30,40

#30,40
l1 = [10,20,30,40]
l2 = [30,40,50,60]
#normal method

for num in l1:
    for no in l2:
        if num== no:
            print(num,end=' ')
#comprehensive

print([num for num in l1 for no in l2 if num==no])

output:

[30, 40]

b) 10,20,50,60

l1 = [10,20,30,40]
l2 = [30,40,50,60]
#comprehensive
output = [num for num in l1 if num not in l2]

output = output + [num for num in l2 if num not in l1]
print(output)

#normal method
for num in l1:
    if num not in l2:
        print(num,end=' ')

for num in l2:
    if num not in l1:
        print(num,end=' ')

Output:

[10, 20, 50, 60]
10 20 50 60 

2. Find program for the given output in comprehensive approach
l1 = [1,2,3]
l2 = [5,6,7]
Output:[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7)]

l1 = [1,2,3]
l2 = [5,6,7]

l = [(i,j) for i in l1 for j in l2 if i!=j]
print(l)

Output:

[(1, 5), (1, 6), (1, 7), (2, 5), (2, 6), (2, 7), (3, 5), (3, 6), (3, 7)]

3. Find program for the given output:
s = "a1b2c3"
output: abc123

Method:1

s = "a1b2c3"

alpha_list = []
num_list = []

for letter in s:
    if letter.isalpha():
        alpha_list.append(letter)
    else:
        num_list.append(letter)

print("".join(alpha_list+num_list))

Method:2

s = "a1b2c3"
letter=''.join([i for i in s if i.isalpha()])
no=''.join([i for i in s if i.isdigit()])

print(letter+no)

Output:

abc123

4. Find program for the given output:

s = "a4k3b2"
output: aeknbd

s = "a4k3b2"
i = 0 
while i<len(s):
    first = s[i]
    second = int(s[i+1])
    print(first, chr(ord(first)+second),sep='',end='')
    i+=2

Output:

aeknbd

Explanation:

-->The ASCII value of first is obtained using ord(first), and second is added to it to find the new character.
-->ord() used to find ASCII value.
-->chr() converts ASCII value-->character.

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