Using a set would be much faster than iterating through the lists.

checklist = ['A', 'FOO']
words = ['fAr', 'near', 'A']
matches = set(checklist).intersection(set(words))
print(matches)  # {'A'}

You can use in directly on the list.

Ex:

mylist = ['cathrine', 'joey', 'bobby', 'fredrick']

if 'cat' in mylist:
    print 'Yes it is in list'
#Empty


if 'joey' in mylist:
    print 'Yes it is in list'
#Yes it is in list

It looks like you need to perform this search not only once so I would recommend to convert your list into dictionary:

>>> l = ['Hi, hello', 'hi mr 12345', 'welcome sir']
>>> d = dict()
>>> for item in l:
...     for word in item.split():
...             d.setdefault(word, list()).append(item)
...

So now you can easily do:

>>> d.get('hi')
['hi mr 12345']
>>> d.get('come')    # nothing
>>> d.get('welcome')
['welcome sir']

p.s. probably you have to improve item.split() to handle commas, point and other separators. maybe use regex and \w.

p.p.s. as cularion mentioned this won't match "welcome sir". if you want to match whole string, it is just one additional line to proposed solution. but if you have to match part of string bounded by spaces and punctuation regex should be your choice.

fruitlist is a string, not a list.
fruitlist = str(sys.argv[2:]).upper() converts the sys.argv to str then applies the upper case.
to avoid this you can do this instead:

fruitlist = [x.upper() for x in sys.argv[2:]]

full code:

import sys
fruitlist = [x.upper() for x in sys.argv[2:]]
print(sys.argv[1])
print(fruitlist)
if sys.argv[1].strip() in fruitlist:
        print(sys.argv[1], 'exact match found in list')

Try this:

def replacer():
    my_words = my_string.split(" ")
    result = []
    for key, value in dictionary.items():
        for item in value:
            if item == my_words[0]:
                result.append(key + " to me " + " ".join(my_words[1:]))
    return result

I hate to tell you this, but you're overthinking it - you can just use the regular == operator for this:

list1_element_1: ['GigabitEthernet0/6/4/1']
list2_element_1: ['GigabitEthernet0/6/4/1.100']
list3_element_1: ['GigabitEthernet0/6/4/1']

print(list1_element_1 == list2_element_1)  # False
print(list1_element_1 == list3_element_1)  # True

For strings, the == operator returns True if they match exactly. For lists, it returns True if the lengths are the same and corresponding elements match exactly (the .__eq__() method is used for this comparison, if you're trying to compare custom classes - all built-in classes have their notions of equality implemented). So if you have a list of strings, you can test whether or not it's equal to another list of strings:

print(['str1', 'str2', 'str3'] == ['str1', 'str2', 'str3'])  # True
print(['str1', 'str2', 'str3'] == ['str1', 'str3', 'str2'])  # False

A much better idea for exact matches is to store the keywords in a set

keywords = {'foo', 'bar', 'joe', 'mauer'}
listOfStrings = ['I am frustrated', 'this task is foobar', 'mauer is awesome']

[s for s in listOfStrings if any(w in keywords for w in s.split())]

This only tests each word in listOfStrings once. Your method (or using regex) looks at every word in listOfStrings for each keyword. As the number of keywords grows, that will be very inefficient.

A regex of the form r"(abc|ef|xxx)" will match with "abc", "ef", or "xxx". You can create this regex by using the string concatenation as below. Note re.search returns None if no match is found.

import re

phrases = ['hello', 'hi', 'bye']
def match(text):
  r = re.search(r'\b({})\b'.format("|".join(phrases)), text)
  return r is not None

match("hi there, how are you?"), match("hithere, how are you?")
# (True, False)