Improve company productivity with a Business Account.Sign Up

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 553
  • Last Modified:

If Condition with Strings

Hi,

Very short question.

I want to check this condition.

list = ['3', '5', '8']

if list[0] != NA & list[0] != NA & list[0] != NA:
       print 'No NAs'
else:
       print 'there is at least one NA'

But i get the following error:
TypeError: unsupported operand type(s) for &: 'str' and 'str'

What's the shortest way to do what i want to do without nested if else???! don't wanna use nested if else.
0
dfernan
Asked:
dfernan
3 Solutions
 
Infinity08Commented:
& is the bitwise AND operator. I guess you intended to use the logical AND operator :

if list[0] != NA and list[1] != NA and list[2] != NA:
       print 'No NAs'
else:
       print 'there is at least one NA'

Open in new window

0
 
LunarNRGCommented:
You should avoid rebinding built-in names. In other words, don't use 'list' as a variable name.

Here's another option ...


l = ['3', '5', '8']
NA = 'NA'
 
if NA not in l:
   print 'No NAs'
else:
   print 'there is at least one NA'
 
# ... or ...
 
if NA in l:
   print 'there is at least one NA'   
else:
    print 'No NAs'
    

Open in new window

0
 
peprCommented:
l (small L) is also not very practical identifier. It can be considered I or 1 depending on the used font. The lst identifier is quite usual in the cases. Otherwise, LunarNRG's solution is fine. Notice also that there is less need to assign

NA = 'NA'

as the 'NA' is used on a single place.
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

Join & Write a Comment

Featured Post

What Kind of Coding Program is Right for You?

There are many ways to learn to code these days. From coding bootcamps like Flatiron School to online courses to totally free beginner resources. The best way to learn to code depends on many factors, but the most important one is you. See what course is best for you.

Tackle projects and never again get stuck behind a technical roadblock.
Join Now