Python: 定义一个接受很多点列表的函数
Python: Define a function that takes a list of many points
当我这样写我的函数时,它运行得很好:
a = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
def func(a)
#function calculation
return Min dist
但是这个函数不太可行,因为它只会计算代码中给定的'a'。有没有办法让函数将“a”识别为包含用户输入的许多点的列表?例如,我希望函数像这样工作:
def func(a)
#calculation
return Min dist
在控制台中:
[In:] a = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
[O:] 3.436788
是的,你已经可以了:
你的模块
def func(a)
#function calculation
return Min dist
你的主要
a = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
b = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
c = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
func(a)
func(b)
func(c)
当我这样写我的函数时,它运行得很好:
a = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
def func(a)
#function calculation
return Min dist
但是这个函数不太可行,因为它只会计算代码中给定的'a'。有没有办法让函数将“a”识别为包含用户输入的许多点的列表?例如,我希望函数像这样工作:
def func(a)
#calculation
return Min dist
在控制台中:
[In:] a = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
[O:] 3.436788
是的,你已经可以了:
你的模块
def func(a)
#function calculation
return Min dist
你的主要
a = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
b = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
c = [(9,0), (6,6), (1,15), (243, 635), (9, 7), (7,8), (2,4)]
func(a)
func(b)
func(c)