返回函数值到范围参数步骤

Question

Hi can some one point me some guidance, i pretend to pass the value of function schema to a variable that will be used as the argument step inside the range module. i hope its clear .. sorry the newbie guys. after some trials i got the edit fine.

import os
import sys


def schema(x):
    if x == 'M':
        step=1
    else:
        step=2
        return step

def main():
    left_start = 1
    left_end = 9
    sch = schema('M')

    range1= range(left_start,left_end,sch)
    if 2 in range1:
        print "In range"
    else:
        print "Missing in range"

if __name__ == '__main__':
    main()

Answer 1

Your code has a problem with indentation. The following:

def schema(x):
    if x == 'M':
        step=1
    else:
        step=2
        return step

should read

def schema(x):
    if x == 'M':
        step=1
    else:
        step=2
    return step

Otherwise, the function returns None when x == 'M' .

A more concise way to write that function is:

def schema(x):
    return 1 if x == 'M' else 2

Answer 2

It looks like you want to do the following:

def schema(x):
    if x == 'M':
        step=1
    else:
        step=2
    return step

Notice how the return step is indented to the same level as the if statement. In your code, the return step will only run in the else case. In the case where x == 'M' , your code would return None from schema() .

My suggested code will run the return step in either case.

Answer 3

The problem in indentation your code:

def schema(x):
    if x == 'M':
        step=1
    else:
        step=2
        return step

it should as:

def schema(x):
    if x == 'M':
        step=1
    else:
        step=2
    return step

in your code the step is returned IF and only if x!='M'.