Add string to each item in python list based on condition

Question

I want to add the string " Equity" to each item in a list if the third last character in the item is " ". Otherwise, I would like to add " Index". Ideally, I don't want to use loops if there is a way.

For Example, I want the list below

list = ['VOD LN','HSBA LN', 'DOKA SS','SXNE' 'KERIN FH','YORK GY','SXNP']

to look like the list below

list = ['VOD LN Equity','HSBA LN Equity', 'DOKA SS Equity','SFNE Index' 'KERIN FH Equity','YORK GY','SXKP Index']

Any idea how i can do this?

Thanks!

There is no way to do this with no loop. They question is whether it will be explicit (list comprehension) or implicit (if abusing pandas) — DeepSpace
– DeepSpace, Commented Dec 30, 2019 at 14:02
Is the length of all strings in the list at least 3? If not, solutions based on s[-3] will throw run-time errors on occasion. — John Coleman
– John Coleman, Commented Dec 30, 2019 at 14:04
DeepSpace - I pretty sure there might be a way to create a function and then apply that function to the list, but I just don't know how John Coleman - yeah they will be at least 4 — pythonlearner13
– pythonlearner13, Commented Dec 30, 2019 at 14:06
@pythonlearner13 But the loop will still have to be somewhere, in your code or in a framework/library code. That's exactly what I meant by explicit/implicit loop — DeepSpace
– DeepSpace, Commented Dec 30, 2019 at 14:10

khelwood · Accepted Answer · 2019-12-30 14:06:00Z

2

You can do this with a list comprehension containing a conditional expression.

>>> input_list = ['VOD LN','HSBA LN', 'DOKA SS','SXNE', 'KERIN FH','YORK GY','SXNP']
>>> result = [x + (' Equity' if len(x)>=3 and x[-3]==' ' else ' Index') for x in input_list]
>>> result
['VOD LN Equity', 'HSBA LN Equity', 'DOKA SS Equity', 'SXNE Index', 'KERIN FH Equity', 'YORK GY Equity', 'SXNP Index']

You could also use various string formatting operations, but since you're just concatenating two strings for each item, + works fine.

edited Dec 30, 2019 at 14:06

answered Dec 30, 2019 at 14:00

khelwood

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Comments

Andrei Berenda · Accepted Answer · 2019-12-30 14:07:02Z

2

Seems it can help you:

items = ['VOD LN','HSBA LN', 'DOKA SS','SXNE' 'KERIN FH','YORK GY','SXNP']

def update(item: str) -> str:
    if len(item) > 3 and item[-3] == ' ':
        return item + ' Equity'
    return item + ' Index'

updated_items = [
    update(item) for item in items
]

edited Dec 30, 2019 at 14:07

answered Dec 30, 2019 at 14:04

Andrei Berenda

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2 Comments

John Coleman Over a year ago

I think this is the best answer since it does a safety check on the length of the strings. Even if not needed in OP's case (they didn't specify), defensive programming is good in the long run. An explicit loop which handles edge cases is preferable to a one-liner that might fail.

DeepSpace Over a year ago

@JohnColeman A working one liner can be achieved with s[-3:-2] or the length check can be done inside the comprehension like in khelwood's answer

Guy · Accepted Answer · 2019-12-30 14:17:12Z

2

One line solution

words = ['VOD LN', 'HSBA LN', 'DOKA SS', 'SXNE' 'KERIN FH', 'YORK GY', 'SXNP']
words = [f'{word} Equity' if word[-3] == ' ' else f'{word} Index' for word in words]
print(words) # ['VOD LN Equity', 'HSBA LN Equity', 'DOKA SS Equity', 'SXNEKERIN FH Equity', 'YORK GY Equity', 'SXNP Index']

If you use Python version older then 3.6

words = [word + ' Equity' if word[-3] == ' ' else word + ' Index' for word in words]

*don't use list as variable name, it's a built in type.

edited Dec 30, 2019 at 14:17

answered Dec 30, 2019 at 13:59

Guy

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6 Comments

DeepSpace Over a year ago

... else f'{word} Index' ...

Guy Over a year ago

@DeepSpace my bad, fixed.

pythonlearner13 Over a year ago

I get the following error File "<ipython-input-48-1d20f3955ca7>", line 2 words = [f'{word} Equity' if word[-3] == ' ' else f'{word} Index' for word in words] ^ SyntaxError: invalid syntax

Guy Over a year ago

@pythonlearner13 you can't go over all the items i a list without using a loop or an equivalent of a loop.

pythonlearner13 Over a year ago

the version is 3.5.5

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