how to check if a list contains a specific string or not in python

Question

I want to search if a list contains a string or not. The following code does not compile.

list= []         
list.append("item1")         
list.append("item2")         
list.append("item3")     

if 'item1' in list
    print "yes"  
else  
    print "no"

Error:

 File "<string>", line 1, in <module>  
  File "/usr/lib/python2.7/py_compile.py", line 117, in compile  
    raise py_exc  
py_compile.PyCompileError:   File "prog.py", line 6  
    if 'item1' in command  
                          ^  
SyntaxError: invalid syntax

since list is a builtin, you should not use it as a variable name. prefer lst or seq instead, for simple options (though you should usually be able to come up with something more descriptive) — Ryan Haining
– Ryan Haining, Commented Jun 5, 2015 at 1:52

Jackie Jones · Accepted Answer · 2015-06-05 01:08:48Z

1

You need a colon after your if statement/else statement.

Also, I believe in python when you use print it has to go like print(MESSAGE). Edit: Using brackets is the new way in python3.

Try this:

list= []         
list.append("item1")         
list.append("item2")         
list.append("item3")     

if 'item1' in list:
    print("yes")
else: 
    print("no")

answered Jun 5, 2015 at 1:08

Jackie Jones

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Comments

pythondetective · Accepted Answer · 2015-06-05 01:22:08Z

1

Syntax for if statement in Python is

 if *condition*:
        statements
 else:
        statements

In your code you miss a colon(:) in if and else part.

list= []         
list.append("item1")         
list.append("item2")         
list.append("item3")

print True if "item1" in list else False

answered Jun 5, 2015 at 1:22

pythondetective

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1 Comment

Ryan Haining Over a year ago

or just print 'item1' in list

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