Multiple value checks using ‘in’ operator (Python)

if 'string1' in line: ...

… works as expected but what if I need to check multiple strings like so:

if 'string1' or 'string2' or 'string3' in line: ...

… doesn’t seem to work.

Answers:

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Method 1

if any(s in line for s in ('string1', 'string2', ...)):

Method 2

If you read the expression like this

if ('string1') or ('string2') or ('string3' in line):

The problem becomes obvious. What will happen is that ‘string1’ evaluates to True so the rest of the expression is shortcircuited.

The long hand way to write it is this

if 'string1' in line or 'string2' in line or 'string3' in line:

Which is a bit repetitive, so in this case it’s better to use any() like in Ignacio’s answer

Method 3

if 'string1' in line or 'string2' in line or 'string3' in line:

Would that be fine for what you need to do?

Method 4

or does not behave that way. 'string1' or 'string2' or 'string3' in line is equivalent to ('string1') or ('string2') or ('string3' in line), which will always return true (actually, 'string1').

To get the behavior you want, you can say if any(s in line for s in ('string1', 'string2', 'string3')):.

Method 5

1. You have this confusion Because you don’t understand how the logical
   operator works with respect to string.

   Python considers empty strings as False and Non empty Strings as True.

   Proper functioning is :

   a and b returns b if a is True, else returns a.

   a or b returns a if a is True, else returns b.

   Therefore every time you put in a non empty string in place of
   string1 the condition will return True and proceed , which will
   result in an undesired Behavior . Hope it Helps :).

Method 6

Using map and lambda

a = ["a", "b", "c"]
b = ["a", "d", "e"]
c = ["1", "2", "3"]

# any element in `a` is a element of `b` ?
any(map(lambda x:x in b, a))
>>> True

# any element in `a` is a element of `c` ?
any(map(lambda x:x in c, a)) # any element in `a` is a element of `c` ?
>>> False

and high-order function

has_any = lambda b: lambda a: any(map(lambda x:x in b, a))

# using ...
f1 = has_any( [1,2,3,] )
f1( [3,4,5,] )
>>> True
f1( [6,7,8,] )
>>> False

Method 7

also as for “or”, you can make it for “and”.
and these are the functions for even more readablity:

returns true if any of the arguments are in “inside_of” variable:

def any_in(inside_of, arguments):
  return any(argument in inside_of for argument in arguments)

returns true if all of the arguments are in “inside_of” variable:

same, but just replace “any” for “all”

Method 8

You can use “set” methods

line= ["string1","string2","string3","string4"]

# check any in
any_in = not set(["string1","string2","not_in_line"]).isdisjoint(set(line))

# check all in
all_in = set(["string1","string2"]).issubset(set(line))

print(f"any_in: {any_in}, all_in:{all_in}")
# Results: 
#    any_in: True, all_in:True

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