Using two for loops
l = ['@','#','%']
out_list = []
for x in my_list:
for y in l:
if y in x:
x = x.replace(y,'')
out_list.append(x)
break
Using list comprehension
out_list = [ x.replace(y,'') for x in my_list for y in l if y in x ]
Assuming 3
in on@3
is a typo, the output will be on@3
and not one
as expected