Reputation: 47
I have one list
a = ['find old', 'find old mails', 'finding', 'find your pc', 'find your pc s ip', 'find your pc s ip address']
i want to remove SUBSET STRING from list which will give me result as
['find old mails', 'finding', 'find your pc s ip address']
Test Code
a = ['find old', 'find old mails', 'finding', 'find your pc', 'find your pc s ip', 'find your pc s ip address']
b = len(a)
def is_sub(sub, lst):
ln = len(sub)
return any(lst[i: i + ln] == sub for i in range(len(sub) - ln + 1))
# print(is_sub(a[0],a[2]))
for i in range(b-1):
for j in range(i+1,b):
if is_sub(a[i],a[j]):
index = a.index(a[i])
a.remove(a[index])
break
b = len(a)
print(a)
I am getting output as
['find old mails', 'finding', 'find your pc s ip', 'find your pc s ip address']
and expected output
['find old mails', 'finding', 'find your pc s ip address']
Upvotes: 0
Views: 495
Reputation: 1098
Why not use in
keyword?
a = ['find old', 'find old mails', 'finding', 'find your pc', 'find your pc s ip', 'find your pc s ip address']
new = []
f = 0
for i in range(0,len(a)):
f=0
for j in range(0, len(a)):
if a[i] in a[j] and not i==j:
f=1
break;
if f==0: new.append(a[i])
print(new)
This gives -
['find old mails', 'finding', 'find your pc s ip address']
Upvotes: 0
Reputation: 140
By making a few alterations to your substring checking function (specifically using in
to check for substrings) and using a list comprehension, you can achieve the desired result:
def is_sub(item, lst):
"""Check whether an item is a substring of other items in a list."""
return any([item in i for i in [i for i in lst if i != item]])
output = [i for i in a if not is_sub(i, a)]
['find old mails', 'finding', 'find your pc s ip address']
Upvotes: 1