"Python中串联字符串的操作符有两个,一个是"+", 一个是",", "+"不能用来串联非字符串,而","可以!而且","还会输出一个空格"
2016-12-18
def average(*args):
n=0.0
sum=0.0
if args:
for x in args:
n=n+1
sum=sum+x
return sum/n
else:
return 0.0
print average()
print average(1, 2)
print average(1, 2, 2, 3, 4)
n=0.0
sum=0.0
if args:
for x in args:
n=n+1
sum=sum+x
return sum/n
else:
return 0.0
print average()
print average(1, 2)
print average(1, 2, 2, 3, 4)
2016-12-17
def greet(a='world' ):
print "Hello"",",a,"."
greet()
greet('Bart')
print "Hello"",",a,"."
greet()
greet('Bart')
2016-12-17
def move(n, a, b, c):
if n==1:
print a,"-->",c
return
else:
move(n-1,a,c,b) #将n-1个a先移动到b
move(1,a,b,c) #将a最后的一个移动到c
move(n-1,b,a,c) #再把b中的n-1移动到c
move(4, 'A', 'B', 'C')
虽然我知道这道题怎么做了,但是感觉这样的数学方法充满了哲学趣味,虽然说不上来
if n==1:
print a,"-->",c
return
else:
move(n-1,a,c,b) #将n-1个a先移动到b
move(1,a,b,c) #将a最后的一个移动到c
move(n-1,b,a,c) #再把b中的n-1移动到c
move(4, 'A', 'B', 'C')
虽然我知道这道题怎么做了,但是感觉这样的数学方法充满了哲学趣味,虽然说不上来
2016-12-17
...
def generate_tr(name, score):
if score < 60
return '<tr><td>%s</td><td>%s</td></tr>' % (name, score)
return '<tr><td>%s</td><td style="color:red">%s</td></tr>' % (name, score)
tds = [generate_tr(name, score) for name, score in d.iteritems()]
.......
def generate_tr(name, score):
if score < 60
return '<tr><td>%s</td><td>%s</td></tr>' % (name, score)
return '<tr><td>%s</td><td style="color:red">%s</td></tr>' % (name, score)
tds = [generate_tr(name, score) for name, score in d.iteritems()]
.......
2016-12-17
import math
def quadratic_equation(a, b, c):
M=(b**2 - 4*a*c)
if M<0:
return none
else:
nx=(-b + math.sqrt(M))/(2*a)
ny=(-b - math.sqrt(M))/(2*a)
return nx,ny
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
注意缩进!!!
def quadratic_equation(a, b, c):
M=(b**2 - 4*a*c)
if M<0:
return none
else:
nx=(-b + math.sqrt(M))/(2*a)
ny=(-b - math.sqrt(M))/(2*a)
return nx,ny
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
注意缩进!!!
2016-12-17
import math
def quadratic_equation(a, b, c):
M=(b**2 - 4*a*c)
if M<0:
return none
else:
nx=(-b + math.sqrt(b**2 - 4*a*c))/(2*a)
ny=(-b - math.sqrt(b**2 - 4*a*c))/(2*a)
return nx,ny
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
注意缩进!!
def quadratic_equation(a, b, c):
M=(b**2 - 4*a*c)
if M<0:
return none
else:
nx=(-b + math.sqrt(b**2 - 4*a*c))/(2*a)
ny=(-b - math.sqrt(b**2 - 4*a*c))/(2*a)
return nx,ny
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
注意缩进!!
2016-12-17
如果是r'''...'''的形式,...代表的内容最后是'的话不能识别
2016-12-17