if 1:
print('The expression is true')if 103.51:
print('The expression is true')if -1:
print('The expression is true')a = True
if a:
print('The expression is true')b = -73.445
if b:
print('The expression is true')if 5 < 5.1:
print('The expression is true')if 5 <= 5:
print('The expression is true')if 5.1 > 5.1:
print('The expression is true')if 5.1 >= 5.1:
print('The expression is true')# "!=" is "not-equals"
if 5 != 5.1:
print('The expression is true')if 5 < 6 and 10 <= 9:
print("i don't think this is true")
if not False or True:
print("it is true?")
else:
print("it wasn't true?")it is true?
def check_number(a):
if a <= 5:
print(a, "is less than or equal to 5")
elif a < 10:
# note we don't need to test if it's greater than 5 since the first if already took
# care of that for us
print(a, "is between 5 and 10")
else:
print(a, "is greater than or equal to 10")
# Testing your function
check_number(1)
check_number(5)
check_number(7)
check_number(10)
check_number(15)The first code below is the bad one, with nested ifs.# our function
def connected(subscription,paid,computer,on):
if subscription:
if paid:
if computer:
if on:
return print('Hello world')
else:
return print('computer not on')
else:
return print('no computer')
else:
return print('not paid')
else:
return print('no subscription')
#test
connected(1,0,1,1)
connected(1,1,1,1)not paid
Hello world
Improved way of writing, without nesting:
# our function
def connected(subscription,paid,computer,on):
if not subscription:
return print('no subscription')
if not paid:
return print('not paid')
if not computer:
return print('no computer')
if not on:
return print('computer not on')
else:
return print('Hello world')
#test
connected(1,0,1,1)
connected(1,1,1,1)not paid
Hello world
def factorial(a):
# f = our factorial result
f = 1
i = 2
while(i<=a):
f *= i
i += 1
return f
print(factorial(4))
print(factorial(2))# Your code here
s = 0
for i in range(11):
s += np.sin(i)**2
print("Sum is", s)s = 0
# Note the range should end at 101!
# It should also start at 1: range(101) will also execute
# the loop with i=0 (which won't matter here...)
for i in range(1,101):
s += i
print(s)maxnum=1e6
i=0
while i <= maxnum:
if i*(i-10)==257024:
break
i+=1
print(i)def leap_year(year):
if year % 4 == 0:
if year % 100 == 0:
if year % 400 == 0:
return True
return False
return True
return False
years = [1, 4, 100, 400, 2000, 2012, 2020, 2021, 2024, 2100]
for year in years:
print(year, leap_year(year))1 False
4 True
100 False
400 True
2000 True
2012 True
2020 True
2021 False
2024 True
2100 False
## Solution
import numpy as np
y = np.array([2, 2, 3, 4, 5, 4, 3, 8, 6, 4])
for i in range(1, len(y)-1):
if y[i-1] < y[i] and y[i+1] < y[i]:
print('index:', i, '\t value:', y[i])
password = 'practicum123'
tries = 0
from time import sleep
while True:
input_pw = input('What is the password?')
if input_pw == password:
break
if (tries+1)%3==0:
sleep(60+tries**3)
tries += 1