# Reversing a string using slicing

my_string = "ABCDE"
reversed_string = my_string[::-1]

print(reversed_string)

# Output
# EDCBA

my_string = "my name is chaitanya baweja"

# using the title() function of string class
new_string = my_string.title()

print(new_string)

# Output
# My Name Is Chaitanya Baweja

my_string = "aavvccccddddeee"

# converting the string to a set
temp_set = set(my_string)

# stitching set into a string using join
new_string = ''.join(temp_set)

print(new_string)

# output
# cdvae

n = 3 # number of repetitions

my_string = "abcd"
my_list = [1,2,3]

print(my_string*n)
# abcdabcdabcd

print(my_list*n)
# [1,2,3,1,2,3,1,2,3]

# Multiplying each element in a list by 2

original_list = [1,2,3,4]

new_list = [2*x for x in original_list]

print(new_list)
# [2,4,6,8]

a = 1
b = 2

a, b = b, a

print(a) # 2
print(b) # 1

string_1 = "My name is Chaitanya Baweja"
string_2 = "sample/ string 2"

# default separator ' '
print(string_1.split())
# ['My', 'name', 'is', 'Chaitanya', 'Baweja']

# defining separator as '/'
print(string_2.split('/'))
# ['sample', ' string 2']

list_of_strings = ['My', 'name', 'is', 'Chaitanya', 'Baweja']

# Using join with the comma separator
print(','.join(list_of_strings))

# Output
# My,name,is,Chaitanya,Baweja

my_string = "abcba"

if my_string == my_string[::-1]:
    print("palindrome")
else:
    print("not palindrome")

# Output
# palindrome

# finding frequency of each element in a list
from collections import Counter

my_list = ['a','a','b','b','b','c','d','d','d','d','d']
count = Counter(my_list) # defining a counter object

print(count) # Of all elements
# Counter({'d': 5, 'b': 3, 'a': 2, 'c': 1})

print(count['b']) # of individual element
# 3

print(count.most_common(1)) # most frequent element
# [('d', 5)]

from collections import Counter

str_1, str_2, str_3 = "acbde", "abced", "abcda"
cnt_1, cnt_2, cnt_3  = Counter(str_1), Counter(str_2), Counter(str_3)

if cnt_1 == cnt_2:
    print('1 and 2 anagram')
if cnt_1 == cnt_3:
    print('1 and 3 anagram')

# output
# 1 and 2 anagram

a, b = 1,0

try:
    print(a/b)
    # exception raised when b is 0
except ZeroDivisionError:
    print("division by zero")
else:
    print("no exceptions raised")
finally:
    print("Run this always")

# output
# division by zero
# Run this always

my_list = ['a', 'b', 'c', 'd', 'e']

for index, value in enumerate(my_list):
    print('{0}: {1}'.format(index, value))

# 0: a
# 1: b
# 2: c
# 3: d
# 4: e

import sys

num = 21

print(sys.getsizeof(num))

# In Python 2, 24
# In Python 3, 28

dict_1 = {'apple': 9, 'banana': 6}
dict_2 = {'banana': 4, 'orange': 8}

combined_dict = {**dict_1, **dict_2}

print(combined_dict)
# Output
# {'apple': 9, 'banana': 4, 'orange': 8}

import time

start_time = time.time()
# Code to check follows
for i in range(10**5):
    a, b = 1,2
    c = a+ b
# Code to check ends
end_time = time.time()
time_taken_in_micro = (end_time- start_time)*(10**6)

print(time_taken_in_micro)

# output
# 18770.217895507812

from iteration_utilities import deepflatten

# if you only have one depth nested_list, use this
def flatten(l):
  return [item for sublist in l for item in sublist]

l = [[1,2,3],[3]]
print(flatten(l))
# [1, 2, 3, 3]

# if you don't know how deep the list is nested
l = [[1,2,3],[4,[5],[6,7]],[8,[9,[10]]]]

print(list(deepflatten(l, depth=3)))
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

import random

my_list = ['a', 'b', 'c', 'd', 'e']
num_samples = 2

samples = random.sample(my_list,num_samples)
print(samples)
# [ 'a', 'e'] this will have any 2 random values

num = 123456

# using map
list_of_digits = list(map(int, str(num)))

print(list_of_digits)
# [1, 2, 3, 4, 5, 6]

# using list comprehension
list_of_digits = [int(x) for x in str(num)]

print(list_of_digits)
# [1, 2, 3, 4, 5, 6]

def unique(l):
    if len(l)==len(set(l)):
        print("All elements are unique")
    else:
        print("List has duplicates")

unique([1,2,3,4])
# All elements are unique

unique([1,1,2,3])
# List has duplicates