在函数中使用 lambda

Question

I'm learning about lambda functions in Python through tutorials online. I understand how it works but I came across an example that puzzles me (on this page ):

def myfunc(n):
    return lambda a : a * n

mydoubler = myfunc(2)
print(mydoubler(11))

I don't understand how mydoubler function works here. How does it take 11 as an argument when we didn't define it before?

Answer 1

mydoubler is what myfunc(2) returns. It returns a lambda that accepts a single argument, a .

When you call on a function like this: myfunction(this_argument) , it is going to resolve to what is returned in that spot. So this is effectively the same as writing mydoubler = lambda a : a * 2

Answer 2

A lambda function is a small anonymous function. In your example

myfunc(2) # lambda a: a * 2

You translate it as apply a function on each input element. It is quite obvious when an input is just a scalar, for example

mydoubler(11) #outputs 11 * 2 = 22

But what do you expect when the input is an array or a string ?

mydoubler([1,1,1]) #outputs [1,1,1] * 2 = [1,1,1,1,1,1]
mydoubler("str") #outputs "str" * 2 = "strstr"

Answer 3

Your example has two functions: the outer function myfunc and the inner function lambda . Normally you can call a lambda function directly:

n = 2
print((lambda a: a * n)(11))
# 22

Or you can assign some variable to this function and call it through this variable:

inner = lambda a: a * n
print(inner(11))
# 22

You can also define some outer function, which will return the inner lambda function:

def myfunc():
    n = 2
    return lambda a: a * n

mydoubler = myfunc()
print(mydoubler(11))
# 22

What is equivalent to:

mydoubler = lambda a: a * 2
print(mydoubler(11))
# 22

In the example above the variable n was declared inside myfunc and in your case n is the parameter of myfunc , which is passed to the lambda function. The function myfunc returns the lambda function with n equal to the argument, which you pass to myfunc by the function call. So the function call myfunc(2) returns the fuction lambda a: a * 2 .

Answer 4

as Python documentation says, lambda is only anonymous function

Lambda expressions (sometimes called lambda forms) are used to create anonymous functions. The expression lambda parameters: expression yields a function object.

you can see it in here

what's going on in your snippet of code is that your myfunc function use n as a constant to new anonymous function that receive one parameter called a and return the multiplication of a with the n . In your calling n value is 2 , result by your call myfunc(2) . when you call mydoubler(11) you call your new anonymous function when a has value 11

Answer 5

• As per the lambda documentation in https://pythonreference.readthedocs.io/en/latest/docs/operators/lambda.html

  lambda returns an anonymous function.

• In the above-mentioned example, lambda function is lambda a : a * n and the lambda itself returns some anonymous function which must be something like

  def mydoubler(a, n): return a*n

• But here, as we have already passed n=2 while calling myfun(), hence doing mydoubler(a) just returns a*2 (as here n=2) and hence the result.

Answer 6

Here mydoubler=myfunc(2) #myfunc(2) returns lambda function to mydoubler

 mydoubler=lambda a:a*2
 # it is similar to x=lambda a:a*2

 print(mydoubler(11)
 # it is similar to print(x(11))

Answer 7

当您在将 mydoubler 分配给它后调用该函数时，它返回一个函数，即 lambda FUNCTION，因此，mydoubler 本身成为一个函数，可以传入参数。

Answer 8

Had same question and landed here by google search. It seems hard at the beginning but crystal clear once understood.

myfunc defines a lambda function with parameter n which is unknown yet.

mydoubler defines a lambda function using myfunc with by giving value to that parameter n=2.

Keep in mind mydoubler is still a lambda function, so it takes variable(s) to work, here within the print function a=11.

Answer 9

How do we write lambda func in simple 1 line ie, consider:

x = lambda a: a*3
print(x(2))

the output we expect is 2*3 =6 .

Here 2nd line print(x(2)) is nothing but value of a to-be-considered if you write print(x) it will show error as <function myfunc.<locals>.<lambda> at 0x000001F2D170AD40> or something.

same thing in

def myfunc(n):
    return lambda a : a * n

mydoubler is nothing but (x) = myfunc(2) is nothing but (lambda a:a*2) func, just name used are different. print(mydoubler(11)) is similar to print(x(valueof a))

Answer 10

To understand this concept the above code can be simplified as given below:

def myfunc(n):
    return lambda a : a*n
mydoubler = myfunc(2)
mydoubler(11)

This is only because lambda is itself a function . myfunc(2) is called and stored in mydoubler . Also myfunc(2) shares parameter ie 2 with lambda. Lambda and mydoubler gets connected through myfunc(2). When mydoubler is called and passed with equal parameter to lambda,Lambda function will be automatically called.

Lambda function are the function which are defined without a name . There for it is also called anonymous function.