Find and replace strings in a list efficiently Python

Question

I am currently bringing a list down from an api and changing the names based on severity levels

Right now my code to do this is

priname = list((item['fields']['priority']['name']) for item in data['issues'])
priname = [w.replace('Trivial', 'Low')
            .replace('Minor', 'Low')
            .replace('Moderate', 'Medium')
            .replace('Major', 'High')
            .replace('Critical', 'High')
            .replace('Blocker', 'Emergency')for w in priname]

So I am just replacing strings in the list to keywords that generalize the content. When I do this to larger lists, it takes forever to return the altered list which i realize means this is not an efficient way of doing this at all.

Can anyone point me in a direction where i can streamline this find and replace?

EDIT:

Priname=[
'Critical',
'Moderate',
'Major',
'Moderate',
'Moderate', 
'Critical', 
'Moderate', 
'Moderate', 
'Moderate', 
'Blocker', 
'Critical', 
'Moderate', 
'Moderate', 
'Major', 
'Moderate', 
'Critical'
]

Use list-comp ([... for x in ...]) rather than list() + generator-expression. It's more Pythonic. — Joe Iddon
– Joe Iddon, Commented Nov 30, 2018 at 17:33
Can you post data or priname so we can see what we're working with? — Joe Iddon
– Joe Iddon, Commented Nov 30, 2018 at 17:34
I posted priname, i don't think there is a need to post data though? im just trying to change this list, it shouldnt matter where it comes from right? to be clear i am able to change the list with the code i have, just not efficiently — Mfreeman
– Mfreeman, Commented Nov 30, 2018 at 17:38
@Mfreeman can there be a situation where an item in Priname can have two replacements applied to it, or is it strictly one replacment operation (at most) per element? — timgeb
– timgeb, Commented Nov 30, 2018 at 17:41
I see, yeah there would be only one replacement, the whole api is pretty uniform — Mfreeman
– Mfreeman, Commented Nov 30, 2018 at 17:45

Dani Mesejo · Accepted Answer · 2018-11-30 17:43:04Z

6

You could use a dictionary:

priname = [
    'Critical',
    'Moderate',
    'Major',
    'Moderate',
    'Moderate',
    'Critical',
    'Moderate',
    'Moderate',
    'Moderate',
    'Blocker',
    'Critical',
    'Moderate',
    'Moderate',
    'Major',
    'Moderate',
    'Critical'
]

table = {'Trivial': 'Low',
         'Minor': 'Low',
         'Moderate': 'Medium',
         'Major': 'High',
         'Critical': 'High',
         'Blocker': 'Emergency'}

result = [table[e] for e in priname]
print(result)

Output

['High', 'Medium', 'High', 'Medium', 'Medium', 'High', 'Medium', 'Medium', 'Medium', 'Emergency', 'High', 'Medium', 'Medium', 'High', 'Medium', 'High']

edited Nov 30, 2018 at 17:43

answered Nov 30, 2018 at 17:41

Dani Mesejo

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3 Comments

timgeb Over a year ago

This assumes that there's at most one replacement per element in priname and that the elements match the replacements perfectly. I think we can assume that given the structure of OP's example, so +1

timgeb Over a year ago

Also, table.get(e, e) over table[e] would return the original string if there's no replacment.

Mfreeman Over a year ago

This example is definitely cleaner than the code i have, but It seems to be returning in the same amount of time though, i'm not sure anything else will be faster.

score 1 · Accepted Answer · 2018-11-30 17:45:02Z

1

One idea that comes to mind is to, rather than using a for loop, convert the list into a numpy array and use the numpy.where function:

    w = numpy.array(w)
    ind = numpy.where(w==list_elements_you_are_looking_for)
    w[ind] = new_value

edited Nov 30, 2018 at 17:45

answered Nov 30, 2018 at 17:39

user10597469

Comments

rahlf23 · Accepted Answer · 2018-11-30 17:45:29Z

1

You can use a dictionary and assign multiple values for each key:

replacements = {'Low': ['Trivial','Minor'], 'Medium': ['Moderate'], 'High': ['Major','Critical'], 'Emergency': ['Blocker']}

Priname=['Critical','Moderate','Major','Moderate','Moderate', 'Critical', 'Moderate', 'Moderate',
'Moderate', 'Blocker', 'Critical', 'Moderate', 'Moderate', 'Major', 'Moderate', 'Critical']

Priname = [k for i in Priname for k, v in replacements.items() if i in v]

Returns:

['High', 'Medium', 'High', 'Medium', 'Medium', 'High', 'Medium', 'Medium', 'Medium', 'Emergency', 'High', 'Medium', 'Medium', 'High', 'Medium', 'High']

answered Nov 30, 2018 at 17:45

rahlf23

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