Python doesn't have arrays like in C++ or Java. Instead you can use a list.

my_list = ['hello','hi','hurra']

for text in my_list:
    print(text)

There are multiple ways to iterate the list, e.g.:

for i, text in enumerate(my_list):
    print(i, text)

for i in range(0, len(my_list)):
    print(my_list[i])

You could use it like "arrays", but there are many build-in functions that you would find very useful. This is a basic and essential data structure in Python. Most of books or tutorials would cover this topic in the first few chapters.

P.S.: The codes are for Python 3.

There are multiple questions here:

1. When we iterate over a string of text, why do we get individual letters not words ? The Zen of Python states "There should be one-- and preferably only one --obvious way to do it." A string is a just string of characters. It may or may not have individual words in it. The most obvious way to iterate over a string is one character at a time.

2. When we iterate using "for word in text" and modify "word", why doesn't "text" change ? The answer has two parts to it: (a) Strings are immutable in Python, so whenever you modify a string, you get a new string while the original one remains unmodified. So there is no way to modify string "text" other than by reassigning "text" to a new string. (b) Even if strings were mutable, when you iterate over an iterable, you only get copies of the elements. So modifying them has no effect on the iterable object.

3. "for elem in range(len(splitstring))" looks clumsy. What is the right way to do this in Python ?

Ok, I made up this question. But the answer is still important. The Pythonic way to do this is:

for elem in splitstring:

Or if you need the index of each element, then you do this:

for index, elem in enumerate(splitstring):

You may achieve what you need by replacing all non-letters first, then extracting pairs of letters and then applying some custom logic to extract the necessary value from the array:

>>> df['array_column'].str.replace('[^A-Z]+', '').str.findall('([A-Z]{2})').apply(lambda d: [''] if len(d) == 0 else d).apply(lambda x: 'HL' if len(x) == 1 and x[0] == 'HL' else [m for m in x if m != 'HL'][0])
0    HL
1    PG
2    PG
3    RC
Name: array_column, dtype: object
>>> 

Details

• .replace('[^A-Z]+', '') - remove all chars other the uppercase letters
• .str.findall('([A-Z]{2})') - extract pairs of letters
• .apply(lambda d: [''] if len(d) == 0 else d) will add an empty item if there is no regex match in the previous step
• .apply(lambda x: 'HL' if len(x) == 1 and x[0] == 'HL' else [m for m in x if m != 'HL'][0]) - custom logic: if the list length is 1 and it is equal to HL, keep it, else remove all HL and get the first element