Replace items in list, python

Question

I have a list of strings like this:

Item_has_was_updated_May_2010
Item_updated_Apr_2011
Item_got_updated_Sept_2011

I want to iterate through the list of string and update the last 2 parts of the string. The month and the year. The rest of the string I want to remain the same. The month and year will be taken from variables I have set earlier in my script, so let's call the month x and the year y.

My approach is to:

Iterate through the list of strings
Split each string by "_"
Replace the last 2 items
Join the items back together with the replaced items

The month and year will be taken from variables I have set earlier in my script, so let's call the month x and the year y.

If anyone can suggest a an approach, it is appreciated.

What are you asking for exactly? You've outlined a process that sounds reasonable, have you tried implementing it? — Jesse
– Jesse, Commented May 11, 2012 at 21:43
Lists in python an immutable, you can't change them.<br> However you can copy them to a new list and in doing that make changes. — Jay M
– Jay M, Commented May 11, 2012 at 21:44

Mark Byers · Accepted Answer · 2012-05-11 21:58:25Z

9

You can do it without regular expressions by using str.rsplit:

yourlist = [s.rsplit('_', 2)[0] + '_' + x + '_' + y for s in yourlist]

See it working online: ideone

If you want to use formatting instead of string concatenation, try this:

yourlist = ['{}_{}_{}'.format(s.rsplit('_', 2)[0], x, y) for s in yourlist]

See it working online: ideone

edited May 11, 2012 at 21:58

answered May 11, 2012 at 21:44

Mark Byers

844k202 gold badges1.6k silver badges1.5k bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Niklas B. Over a year ago

Could also do '_'.join(s.split('_')[:-2] + [x,y])

Mark Byers Over a year ago

@NiklasB.: +1 Yeah that works too. It's slightly more concise. It does some unnecessary splitting and joining compared to using rsplit. Probably not a huge issue considering the size of the example strings though.

Niklas B. Over a year ago

I think it might even be faster than the string concatenation (which needs to create a new string 4 times) The format approach is of course okay as well :)

Jay M · Accepted Answer · 2012-05-11 21:51:58Z

2

I don't think you really need 're'.
You could use something like this:

m="dec"
y=2012
l=["Item_has_was_updated_May_2010",
    "Item_updated_Apr_2011",
    "Item_got_updated_Sept_2011"]
r=[]
for s in l:
    t=s.split("_")
    r.append("_".join(t[:-2])+"_%s_%s"%(m,y))

answered May 11, 2012 at 21:51

Jay M

4,4831 gold badge28 silver badges39 bronze badges

1 Comment

Niklas B. Over a year ago

The r = []; for s in l: r.append(...) pattern is exactly what list comprehensions are good for :)

georg · Accepted Answer · 2012-05-12 10:54:40Z

2

lst = [
   'Item_has_was_updated_May_2010',
   'Item_updated_Apr_2011',
   'Item_got_updated_Sept_2011',
]
month='Sep';year='2012'
for s in lst:
        list=s.split('_')
        list[-2:]=(month,year)
        r=''
        for a in list:
                r=r+"%s_"%a
        r=r[:-1]
        print r

edited May 12, 2012 at 10:54

georg

216k57 gold badges324 silver badges401 bronze badges

answered May 12, 2012 at 3:47

kiranmai

171 bronze badge

Collectives™ on Stack Overflow

Replace items in list, python

3 Answers 3

3 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related