[英]How to pass a function as a function parameter in Python
This is what I currently have and it works fine:这是我目前拥有的,并且工作正常:
def iterate(seed, num):
x = seed
orbit = [x]
for i in range(num):
x = 2 * x * (1 - x)
orbit.append(x)
return orbit
Now if I want to change the iterating equation on line 5 to, say, x = x ** 2 - 3, I'll have to create a new function with all the same code except line 5. How do I create a more general function that can have a function as a parameter?现在,如果我想将第 5 行的迭代方程更改为 x = x ** 2 - 3,我将不得不使用除第 5 行之外的所有相同代码创建一个新函数。如何创建一个更通用的函数可以将函数作为参数的函数?
Functions are first-class citizens in Python.函数是 Python 中的一等公民。 you can pass a function as a parameter:您可以将函数作为参数传递:
def iterate(seed, num, fct):
# ^^^
x = seed
orbit = [x]
for i in range(num):
x = fct(x)
# ^^^
orbit.append(x)
return orbit
In your code, you will pass the function you need as the third argument:在您的代码中,您将传递您需要的函数作为第三个参数:
def f(x):
return 2*x*(1-x)
iterate(seed, num, f)
# ^
Or或者
def g(x):
return 3*x*(2-x)
iterate(seed, num, g)
# ^
Or ...或者 ...
If you don't want to name a new function each time, you will have the option to pass an anonymous function (ie: lambda ) instead:如果您不想每次都命名一个新函数,您可以选择传递一个匿名函数(即: lambda ):
iterate(seed, num, lambda x: 3*x*(4-x))
Just pass the function as a parameter.只需将函数作为参数传递即可。 For instance:例如:
def iterate(seed, num, func=lambda x: 2*x*(1-x)):
x = seed
orbit = [x]
for i in range(num):
x = func(x)
orbit.append(x)
return orbit
You can then either use it as you currently do or pass a function (that takes a single argument) eg:然后,您可以像当前一样使用它,也可以传递一个函数(采用单个参数),例如:
iterate(3, 12, lambda x: x**2-3)
You can also pass existing (non lambda functions) in the same way:您还可以以相同的方式传递现有(非 lambda 函数):
def newFunc(x):
return x**2 - 3
iterate(3, 12, newFunc)
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