How to avoid write similar code multiple times in this situation?-CodePudding

I have a code like that,

i = 0
l1 = ['a','b','c']
while(i < len(l1)):
    if i   2 < len(l1):
        if  l1[i   2] == 'c':
            print("ok")
        else:
            print("None,Error!")
    else:
        print("None,Error!")
    i  = 1

As you can see, in the else part, print("None,Error!") are used in two times. In practice, they are very long but totally same so I want to merge them to make them simple.However, I need to check if i 2 is out of list bound, so I cannot rewrite it to

if i 2 < len(l1) and l1[i   2] == 'c':

Is there any idea can solve this problem?

CodePudding user response：

As Phu Ngo mentions, we can take advantage of short circuiting to write

i = 0
l1 = ['a','b','c']
while(i < len(l1)):
    if i   2 < len(l1) and l1[i   2]=='c':
        print("ok")
    else:
        print("None,Error!")
    i =1

If i 2<len(l1) is false, then the expression l1[i 2]=='c' is never evaluated, which means that no error is raised.

Another solution that can be considered here is a try/except clause.

i = 0
l1 = ['a','b','c']
while(i < len(l1)):
    try:
        if l1[i 2]=='c':
            print("ok")
        else:
            raise ValueError
    except Exception:
        print("None,Error!")
    i =1

Both result in the printouts

ok
None,Error!
None,Error!

CodePudding user response：

Unless your data is large, ditch the algorithmic details and just arrange the data in a convenient way:

xs = ['a', 'b', 'c']

for x1, x2 in zip(xs, xs[2:]   [None, None]):
    print("ok" if x2 == 'c' else "None,Error!")

Yet another illustration of the general point: one rarely needs to iterate over a collection via indexes.