Comparing a String to Multiple Items in Python

Comparing a string to multiple items in Python

If, OTOH, your list of strings is indeed hideously long, use a set:

accepted_strings = {'auth', 'authpriv', 'daemon'}

if facility in accepted_strings:
    do_stuff()

Testing for containment in a set is O(1) on average.

Comparing string to multiple strings in python

Assuming you are trying to check whether your rail[x].index[b] is either of ('WD' | 'POT' | 'NGI') on each iteration,

Try the below snippet if it is the case,

route=''
for b in range(10):
   route += ('-'+ rail[0].index[b] +'-') if (rail[x].index[b] in ('WD', 'POT', 'NGI'))

Quick Check:

print("WD" == ('WD' | 'POT' | 'NGI'))

TypeError: unsupported operand type(s) for |: 'str' and 'str'

print("WD" in ('WD' , 'POT' , 'NGI'))

True

Python 3: how to compare multiple strings in one line of code?

Use in for containment tests.

categoryChoice = input("what is your category choice?")
if categoryChoice not in ("category1", "category 2", "category 3"):
    print("not a valid choice")

http://ideone.com/e6n0Rr

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By the way, if you're using Python 2, you should really use raw_input instead of input.

How to use IN to compare multiple strings as exact match in Python?

Your question is asking why:

"abc" in ("abc d") 

returns True and:

"abc" in ("abc d","ab")

returns False.

Because the first expression is just wrapped in brackets it's akin to:

"abc" in "abc d"

Note that it doesn't act as a tuple because there's only 1 element.

Whereas

"abc" in ("abc d","ab")

Does use ("abc d","ab") as a tuple (because of the comma meaning multiple elements), and so it searches for a matching element.

How to test multiple variables for equality against a single value?

You misunderstand how boolean expressions work; they don't work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:

if x == 1 or y == 1 or z == 1:

x and y are otherwise evaluated on their own (False if 0, True otherwise).

You can shorten that using a containment test against a tuple:

if 1 in (x, y, z):

or better still:

if 1 in {x, y, z}:

using a set to take advantage of the constant-cost membership test (i.e. in takes a fixed amount of time whatever the left-hand operand is).

Explanation

When you use or, python sees each side of the operator as separate expressions. The expression x or y == 1 is treated as first a boolean test for x, then if that is False, the expression y == 1 is tested.

This is due to operator precedence. The or operator has a lower precedence than the == test, so the latter is evaluated first.

However, even if this were not the case, and the expression x or y or z == 1 was actually interpreted as (x or y or z) == 1 instead, this would still not do what you expect it to do.

x or y or z would evaluate to the first argument that is 'truthy', e.g. not False, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).

So for the values x = 2; y = 1; z = 0, x or y or z would resolve to 2, because that is the first true-like value in the arguments. Then 2 == 1 would be False, even though y == 1 would be True.

The same would apply to the inverse; testing multiple values against a single variable; x == 1 or 2 or 3 would fail for the same reasons. Use x == 1 or x == 2 or x == 3 or x in {1, 2, 3}.

Comparing multiple string in one line (file) to single string in other file same line number

You are comparing every line of the first file to every line of the second file. What I think you want to do is compare each file line by line.

Something like this?

lines = []
with open('actual_results.txt') as actual, open('expected_results.txt') as expected:
  try:
    while True:
      a, e = next(actual), next(expected)
      if a in e.split(','):
        lines.append((a, e, True))
      else:
        lines.append((a, e, False))
  except StopIteration:
    pass

with open('final_output_file.txt', 'w') as output:
  for actual, expected, result in lines:
    if result:
      output.write('found\n')
    else:
      output.write('not found\n')

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