lambda 函数访问外部变量 [英] lambda function accessing outside variable

问题描述

我想尝试使用匿名函数，所以我决定制作一个简单的素数查找器.这是:

I wanted to play around with anonymous functions so I decided to make a simple prime finder. Here it is:

tests = []
end = int(1e2)
i = 3
while i <= end:
    a = map(lambda f:f(i),tests)
    if True not in a:
        tests.append(lambda x:x%i==0)
        print i
    print tests
    print "Test: "+str(i)
    print str(a)
    i+=2

然而，我发现 lambda x:x%i==0 中的 i 每次都会被访问，而我希望它是一个文字数字.我怎样才能让它变成 lambda x:x%3==0 呢?

What I find however, is that the i in the lambda x:x%i==0 is accessed each time, while i want it to be a literal number. how can I get it to become lambda x:x%3==0 instead?

推荐答案

您可以捕获"创建 lambda 时的 i

You can "capture" the i when creating the lambda

lambda x, i=i: x%i==0

这会将 lambda 上下文中的 i 设置为等于创建它时的任何 i.如果你愿意，你也可以说 lambda x, n=i: x%n==0，它不是完全捕获，但它可以满足你的需要.

This will set the i in the lambda's context equal to whatever i was when it was created. you could also say lambda x, n=i: x%n==0 if you wanted, it's not exactly capture, but it gets you what you need.

这是一个类似于以下定义函数的查找问题:

It's an issue of lookup that's analogous to the following with defined functions:

i = "original"

def print_i1():
    print(i) # prints "changed" when called below

def print_i2(s=i): # default set at function creation, not call
    print(s) # prints "original" when called below


i = "changed"
print_i1()
print_i2()

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