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Assigning types to variables as values

a_var = str

random choice

#from random import choice

a, b, c = float, int, str
for i in range(5):
    j = choice([a,b,c])(i)
    print(j, type(j))

Only the first clause of generators is evaluated immediately

gen_fails = (i for i in 1/0)

lazy evaluation

gen_succeeds = (i for i in range(5) for j in 1/0)
print('But obviously fails when we iterate ...')
for i in gen_succeeds:

Usge of *args

def a_func(*args):
    print('type of args:', type(args))
    print('args contents:', args)
    print('1st argument:', args[0])
a_func(0, 1, 'a', 'b', 'c')

usage of kwargs

def b_func(**kwargs):
    print('type of kwargs:', type(kwargs))
    print('kwargs contents: ', kwargs)
    print('value of argument a:', kwargs['a'])

b_func(a=1, b=2, c=3, d=4)

Unpacking of iterables

val1, *vals = [1, 2, 3, 4, 5]
print('val1:', val1)
print('vals:', vals)

if else for

for i in range(5):
    if i == 1:
        print('in for')
    print('in else')
print('after for-loop')

usage of break

for i in range(5):
    if i == 1:
    print('in else')
print('after for-loop')

conditional usecase

a_list = [1,2]
if a_list[0] == 1:
    print('Hello, World!')
    print('Bye, World!')

### Usage of while
i = 0
while i < 2:
    i += 1
    print('in else')

Interning of string

hello1 = 'Hello'
hello2 = 'Hell' + 'o'
hello3 = 'Hell'
hello3 = hello3 + 'o'
print('hello1 is hello2:', hello1 is hello2)
print('hello1 is hello3:', hello1 is hello3)