parsing list in python
I have list in python开发者_开发知识库 which has following entries
name-1
name-2
name-3
name-4
name-1
name-2
name-3
name-4
name-1
name-2
name-3
name-4
I would like remove name-1 from list except its first appearance -- resultant list should look like
name-1
name-2
name-3
name-4
name-2
name-3
name-4
name-2
name-3
name-4
How to achieve this ?
def remove_but_first( lst, it):
first = lst.index( it )
# everything up to the first occurance of it, then the rest of the list without all it
return lst[:first+1] + [ x for x in lst[first:] if x != it ]
s = [1,2,3,4,1,5,6]
print remove_but_first( s, 1)
Assuming name-1 denotes "the first element":
[names[0]] + [n for n in names[1:] if n != names[0]]
EDIT: If the overall goal is to de-duplicate the entire list, just use set(names)
.
Based on Marcelo's solution:
[name for cnt,name in enumerate(names) if (name != names[0] or cnt > 0)]
Find the index of the first element you wish to remove, then filter the rest of the list. The following works in Python 2.5:
def removeAllButFirst(elem, myList):
idx = myList.index(elem)
return myList[0:idx+1] + filter(lambda x: x != elem, myList[idx+1:])
mylist = ['name-1', 'name-2', 'name-3', 'name-4', 'name-1', 'name-2', 'name-3', 'name-4', 'name-1', 'name-2', 'name-3', 'name-4']
newlist = filter(lambda x: x != 'name-1', mylist)
newlist.insert(mylist.index('name-1'), 'name-1')
print newlist
['name-1', 'name-2', 'name-3', 'name-4', 'name-2', 'name-3', 'name-4', 'name-2', 'name-3', 'name-4']
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