What I want to do is to remove string elements from my list that have some duplicate parts. For example, if I have given list.
ls = ['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004', '02/27', '07/21', '08/13']
I want output as
ls_out = ['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004']
That is '02/27' already existed in '02/27/1960'.
(note that I'm not sure if this question is duplicated or not)
This can also be solve with a for loop and any built-in method:
>>> ls
['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004', '02/27', '07/21', '08/13']
>>>
>>> ls_out = []
>>>
>>> for x in ls:
if not any([x in item for item in ls_out]):
ls_out.append(x)
>>> ls_out
['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004']
OR:
>>> for x in ls:
if all([x not in item for item in ls_out]):
ls_out.append(x)
>>> ls_out
['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004']
any approach seems like it would be the most efficient. Also, to me it seems like checking x not in ls_out is overkill. Better to just use: if not any(x in item for item in ls_out):I'm not sure if this is the most efficient way to do this, but it would definitely work:
ls = ['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004', '02/27', '07/21', '08/13']
ls2 = ls
for item in ls:
for dup_item in ls2:
if item == dup_item:
continue
if item.startswith(dup_item):
_ = ls.pop(ls.index(dup_item))
print ls
Basically, it creates two identical lists, loops through both and checks if they're equal - if they are, it skips. If they aren't, it checks if they start with the other one. If it does, it removes it.
cache = set()
def fun(s):
ss = s.split('/')
key = ss[0] + '/' + ss[1]
if key in cache:
return None
else:
cache.add(key)
return s
ls = ['02/27/1960', '07/21/2004', '08/13/2004', '09/12/2004', '02/27', '07/21', '08/13']
new_ls = filter(fun, ls)
print new_ls
cache to empty set and check if key is in cache, I guess.