How to "double pop"?-CodePudding

I'm sorry if my title is a bit unclear, I wasn't sure how to express my problem better. I will give a quick code snippet to illustrate what I am trying to do:

for data_file in data_files:
    analyze(data_file)

def analyze(data_file):
    for _ in np.arange(n_steps):
        foo(data_file)
    do some other things

def foo(data_file):
    do some things
    if (value > 1): 
        do some things
        double_return()
    else:
        continue doing other things

What I want the function double_return() to do then, is to return back to the larger loop:

for data_file in data_files:

(Instead of return, which would continue to the next step in for _ in n_steps). However, I am not sure what command to use here and haven't been able to find it.

CodePudding user response：

foo() could return True, if it completes, or False if you want to “leave early”:

for data_file in data_files:
   analyze(data_file)

def analyze(data_file):
   for _ in np.arange(n_steps):
       if not foo(data_file):
          return
   do some other things

def foo(data_file):
   do some things
   if (value > 1):
       do some things
       return False
   else:
       continue doing other things

   return True

Edit:
To make the meaning clearer, you could define foo_completed=True, and foo_not_completed=False, and use these constants instead of True/False.

CodePudding user response：

What you can do is to return a truthy value (assuming all the other routes out of foo() are the default return None) and test for it:

def analyze(data_file):
    for _ in np.arange(n_steps):
        if foo(data_file):
            return
    do some other things

def foo(data_file):
    do some things
    if (value > 1): 
        do some things
        return True
    else:
        continue doing other things

for data_file in data_files:
    analyze(data_file)

It rather obscures the meaning, but it will work in a testable way.