Conditional judgment
The reason why a computer can do many automated tasks is because it can make conditional judgments on its own.
For example, enter the user's age, and print different content according to the age. In the Python program, use the ifstatement to achieve:
age = 20
if age >= 18:
print('your age is', age)
print('adult')
According to Python's indentation rules, if the ifstatement is judged to be true True, the indented two-line print statement is executed, otherwise, nothing is done.
You can also ifadd a elsestatement, which means that if ifthe judgment is yes False, do not execute ifthe content, and elseexecute it:
age = 3
if age >= 18:
print('your age is', age)
print('adult')
else:
print('your age is', age)
print('teenager')
Be careful not to omit the colon :.
Of course, the above judgment is very rough, and it can be used eliffor a more detailed judgment:
age = 3
if age >= 18:
print('adult')
elif age >= 6:
print('teenager')
else:
print('kid')
elifYes else if, there can be more than one abbreviation, elifso ifthe full form of the statement is:
if <条件判断1>:
<执行1>
elif <条件判断2>:
<执行2>
elif <条件判断3>:
<执行3>
else:
<执行4>
ifStatement execution has a characteristic. It is judged from top to bottom. If it is in a certain judgment, after the statement corresponding to the judgment is executed, the remaining sum Trueis ignored :elifelse
age = 20
if age >= 6:
print('teenager')
elif age >= 18:
print('adult')
else:
print('kid')
ifJudgment conditions can also be abbreviated, such as writing:
if x:
print('True')
As long as it xis a non-zero value, a non-empty string, a non-empty list, etc., it is judged as True, otherwise it is False.
input
By input()reading the user's input, you can enter your own input, and the program runs more interestingly:
birth = input('birth: ')
if birth < 2000:
print('00前')
else:
print('00后')
Input 1982, the result is an error:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unorderable types: str() > int()
This is because input()the returned data type strcan strnot be directly compared to an integer, and must be strconverted to an integer first. Python provides int()functions to do this:
s = input('birth: ')
birth = int(s)
if birth < 2000:
print('00前')
else:
print('00后')
Run it again to get the correct result. But what if the input abc? Again you get an error message:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: 'abc'
It turns out that int()when the function finds that a string is not a valid number, it will report an error and the program will exit.
Code:
print('------------------------------------------------------')
age = 3
if age >= 18:
print('your age is',age)
print('adult')
elif age >= 6:
print('teenager')
else:
print('kid')
print('------------------------------------------------------')
s = input('birth:\n')
birth = int(s)
if birth < 2000:
print('before 00')
else:
print('after 00')
print('------------------------------------------------------')
x = input('please enter ...\n')
if x:
print('True')
else:
print('False')
TestCode:
# -*- coding:utf-8 -*-
'''
Xiao Ming is 1.75 in height and 80.5 kg in weight. Please help Xiaoming calculate his BMI index according to the BMI formula (weight divided by the square of height), and according to the BMI index:
Below 18.5: too light
18.5-25: Normal
25-28: Overweight
28-32: Obesity
Above 32: Severely obese
height = 1.75
weight = 80.5
'''
height = input('Please input height:')
weight = input('Please enter your weight:')
h = float(height)
w = float(weight)
BIM = w / (h * h)
print('Your BIM value is %.2f'%BIM,'\nbelongs to:')
if BIM < 18.5:
print('too light')
elif 18.5 < BIM < 25:
print('normal')
elif 25 < BIM < 28:
print('too heavy')
elif 28 < BIM < 32:
print('fat')
elif BIM > 32:
print('Severely obese')
