How do i check the 2 item of every string for more than one occurence

Question

[‘AC’, ‘2H’, ‘3S’, ‘4C’]

How do I check if the 1st index (eg 2nd element) of every string occurs more than once? For example, in this case, C occurs 2 times so I need to return False This must apply to other case as well such as H or S occuring more than once

Answer 1

Consider using collections.Counter to count the occurrences of interested items. And use all or any to verify the condition.

import collections

a = ['AC', '2H', '3S', '4C']
counter = collections.Counter(s[1] for s in a)
result = all(v < 2 for v in counter.values())

print(result)

Answer 2

You can use this function:

def check_amount(all_text, character):
    count = 0
    for text in all_text:
        for ch in text:
            if ch == character:
                count += 1
    return count

This returns how many times it happens, if you just want to see if it exists:

def check_amount(all_text, character):
    for text in all_text:
        for ch in text:
            if ch == character:
                return True
            else:
                return False

Those are for checking at any position this is if you need it to be at a specific position like you said:

def check_amount(all_text, character):
    count = 0
    for text in all_text:
        if text[1] == character:
            count += 1
    return count

And then you can change this if you want the boolean version using the same method of not using the count

The all_text is the list you want to pass in, and the character you want to see if is there/exists.

Answer 3

Using regular expressions, you can use re.finditer to find all (non-overlapping) occurences:

>>> import re
>>> text = 'Allowed Hello Hollow'
>>> for m in re.finditer('ll', text):
         print('ll found', m.start(), m.end())

ll found 1 3
ll found 10 12
ll found 16 18

Alternatively, if you don't want the overhead of regular expressions, you can also repeatedly use str.find to get the next index:

>>> text = 'Allowed Hello Hollow'
>>> index = 0
>>> while index < len(text):
        index = text.find('ll', index)
        if index == -1:
            break
        print('ll found at', index)
        index += 2 # +2 because len('ll') == 2

ll found at  1
ll found at  10
ll found at  16
This also works for lists and other sequences.

for an array here I'd use List Comprehension, like this:

listOfElems = ['Hello', 'Ok', 'is', 'Ok', 'test', 'this', 'is', 'a', 'test', 'Ok']

now let's find all indexes of 'ok' in the list

# Use List Comprehension Get indexes of all occurrences of 'Ok' in the list
indexPosList = [ i for i in range(len(listOfElems)) if listOfElems[i] == 'Ok' ]

print('Indexes of all occurrences of "Ok" in the list are: ', indexPosList)

output:

Indexes of all occurrences of "Ok" in the list are :  [1, 3, 9]