Conditional when calling functions sequentially in python-CodePudding

I have four functions and each one returns a list as a result. I want whichever of these functions that I call sequentially brings a list, not to look at the others. Thank you for your help.

def main(self):
    print(self.df_list)
    
    funcs = [(Calculate.free_two_kdv(self.df_list)),(Calculate.free_kdv(self.df_list)),(Calculate.zero_kdv(Calculate.find_largest_duplicated_value,self.df_list)),(Calculate.zero_with_other_kdv(Calculate.find_largest_duplicated_value,self.df_list))]
    for func in funcs:
        try:
            
            func()
            
        except ValueError:
            break

In short, I don't want it to go to other functions at all, whichever brings results.

CodePudding user response：

You are calling your functions immediately and storing the result in the list. Define a new function which makes the desired function call.

def main(self):
    print(self.df_list)
    
    funcs = [lambda: Calculate.free_two_kdv(self.df_list),
             lambda: Calculate.free_kdv(self.df_list),
             lambda: Calculate.zero_kdv(Calculate.find_largest_duplicated_value, self.df_list),
             lambda: Calculate.zero_with_other_kdv(Calculate.find_largest_duplicated_value, self.df_list)
    ]

    for func in funcs:
        try:
            return func()
        except ValueError:
            pass

It's not clear if you want Calculate.find_largest_duplciated_value as an argument to the first function, or if you want function composition (i.e., f(g, x) vs f(g(x)). If you want composition, it's

    funcs = [lambda: Calculate.free_two_kdv(self.df_list),
             lambda: Calculate.free_kdv(self.df_list),
             lambda: Calculate.zero_kdv(Calculate.find_largest_duplicated_value(self.df_list)),
             lambda: Calculate.zero_with_other_kdv(Calculate.find_largest_duplicated_value(self.df_list))
    ]

CodePudding user response：

Assuming that the func in funcs return either a list or None:

for func in funcs:
    if func(): break

for func in funcs:
    flist = func()
    if flist: break

The second receives the return value from the function for further work. try..except in this case has a greater performance cost than a straightforward boolean condition.