将Python函数作为参数传递而不执行它? [英] Passing Python Function as argument without executing it?

问题描述

我有这个功能:

def a(one, two, the_argument_function):
    if one in two:
        return the_argument_function

我的the_argument_function看起来像这样:

my the_argument_function looks something like this:

def b(do_this, do_that):
    print "hi."

以上两项均导入到文件"main_functions.py"中，以使我的最终代码如下所示:

Both of the above are imported to a file "main_functions.py" for my ultimate code to look like this:

print function_from_main(package1.a, argument, package2.b(do_this, do_that)

"a"函数中的如果是二分之一"有效，但是当传递给"function_from_main"时，"b"函数仍会执行，而无需等待"a"中的检查是否应该执行.

The "if one in two" from "a"function works but "b"function still executes when being passed to "function_from_main" without waiting the check from "a" to see if it actually should execute.

我该怎么办?

推荐答案

package2.b(do_this，do_that)是一个函数调用(函数名称后加括号).相反，您应该只传递函数名称 package2.b 函数 a

package2.b(do_this, do_that) is a function call (a function name followed by parenthesis). Instead you should be passing only the function name package2.b the function a

您还需要修改函数 a ，以便在满足条件时调用该函数

You will also need to modify function a such that function be is called when the condition is satisfied

# function a definition 
def a(one, two, the_argument_function, argument_dict):
    if one in two:
        return the_argument_function(**argument_dict)

def b(do_this, do_that):
    print "hi."

# function call for a
a(one, two, b, {'do_this': some_value, 'do_that': some_other_value}) 

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