remove an item from string list

[python101]

I have a string list

items = ['B','D','B','A' ,'E']

Assume that I don't know the order (index) of these items. I would like to remove the second 'B' out of the list without sorting [using items.sort() ] or changing the the order of other items in the list. How can I do that?

The remove() and pop(), only take the first 'B' or I have to know the index of the second 'B' to remove second 'B' out.

[bartonc]

I have a string list

items = ['B','D','B','A' ,'E']

Assume that I don't know the order (index) of these items. I would like to remove the second 'B' out of the list without sorting [using items.sort() ] or changing the the order of other items in the list. How can I do that?

The remove() and pop(), only take the first 'B' or I have to know the index of the second 'B' to remove second 'B' out.

[CODE=python]
>>> items = ['B','D','B','A' ,'E']
>>> list(set(items) )
['A', 'B', 'E', 'D']
>>> [/CODE]Hmmm... That shouldn't have reordered the list. I'll keep at it..

[bartonc]

[CODE=python]
>>> items = ['B','D','B','A' ,'E']
>>> list(set(items) )
['A', 'B', 'E', 'D']
>>> [/CODE]Hmmm... That shouldn't have reordered the list. I'll keep at it..

[CODE=python]
>>> for i, item in enumerate(items[:]): #loop through a copy of the list
... try:
... j = items.index(ite m, i + 1)
... items.pop(j)
... except ValueError:
... pass
...
>>> items
['B', 'D', 'A', 'E']
>>> [/CODE]

[python101]

[CODE=python]
>>> items = ['B','D','B','A' ,'E']
>>> list(set(items) )
['A', 'B', 'E', 'D']
>>> [/CODE]Hmmm... That shouldn't have reordered the list. I'll keep at it..

I mentioned that I did not know the order of the list in items, I only knew that there was two 'B' in the list. In your case, we would have to know the order of all items.

[python101]

[CODE=python]
>>> for i, item in enumerate(items[:]): #loop through a copy of the list
... try:
... j = items.index(ite m, i + 1)
... items.pop(j)
... except ValueError:
... pass
...
>>> items
['B', 'D', 'A', 'E']
>>> [/CODE]

Thank you very much. However, is there a simpler way to pop second 'B' out without using loop?

[bartonc]

Thank you very much. However, is there a simpler way to pop second 'B' out without using loop?

Two tricks here:
1) Start searching for the index one past the location of the first occurrence.
2) Wrap the index() method in a try block in case it fails.[CODE=python]
>>> items = ['B','D','B','A' ,'E']
>>> try:
... j = items.index('B' , 1) # one past the location of the first occurrence
... items.pop(j)
... except ValueError:
... pass
...
>>> items
['B', 'D', 'A', 'E']
>>> [/CODE]

[python101]

Two tricks here:
1) Start searching for the index one past the location of the first occurrence.
2) Wrap the index() method in a try block in case it fails.[CODE=python]
>>> items = ['B','D','B','A' ,'E']
>>> try:
... j = items.index('B' , 1) # one past the location of the first occurrence
... items.pop(j)
... except ValueError:
... pass
...
>>> items
['B', 'D', 'A', 'E']
>>> [/CODE]

Thank you. I got the trick

[python101]

I just thought another way to approach the problem. Could I determine all indexs of 'B' in the list, so when there were numberous 'B's in the list I would be able to pop B at any certain index?

[bvdet]

I just thought another way to approach the problem. Could I determine all indexs of 'B' in the list, so when there were numberous 'B's in the list I would be able to pop B at any certain index?

I have used this function many times:[code=Python]"""
Return an index list of all occurrances of 'item' in string/list 's'.
Optional start search position 'i'
"""
def indexList(s, item, i=0):
i_list = []
while True:
try:
i = s.index(item, i)
i_list.append(i )
i += 1
except:
break
return i_list[/code]

>>> items = ['B','D','B','A' ,'E']
>>> indexList(items , 'B')
[0, 2]
>>>

[ilikepython]

I have used this function many times:[code=Python]"""
Return an index list of all occurrances of 'item' in string/list 's'.
Optional start search position 'i'
"""
def indexList(s, item, i=0):
i_list = []
while True:
try:
i = s.index(item, i)
i_list.append(i )
i += 1
except:
break
return i_list[/code]

>>> items = ['B','D','B','A' ,'E']
>>> indexList(items , 'B')
[0, 2]
>>>

I don't know about you, but I think this is clearer:
[code=python]
def indexList(s, item, start = 0):
i_list = []
for (i, obj) in enumerate(s[start:]):
if obj == item:
i_list.append(i + start)
return i_list
[/code]
or with map:
[code=python]
def indexList(s, item, start = 0):
return filter(lambda x: x, map(lambda (i, obj): i + start if obj == item else None, enumerate(s[start:]))) # i think i over did it
[/code]I think yours is faster for larger lists though.

EDIT:
List comprhension is definately clearer:
[code=python]
def indexList(s, item, start = 0):
return [i + start for (i, obj) in enumerate(s[start:]) if obj == item]
[/code]

[bvdet]

I don't know about you, but I think this is clearer:
[code=python]
def indexList(s, item, start = 0):
i_list = []
for (i, obj) in enumerate(s[start:]):
if obj == item:
i_list.append(i + start)
return i_list
[/code]
or with map:
[code=python]
def indexList(s, item, start = 0):
return filter(lambda x: x, map(lambda (i, obj): i + start if obj == item else None, enumerate(s[start:]))) # i think i over did it
[/code]I think yours is faster for larger lists though.

EDIT:
List comprhension is definately clearer:
[code=python]
def indexList(s, item, start = 0):
return [i + start for (i, obj) in enumerate(s[start:]) if obj == item]
[/code]

Thanks ilikepython. I like your code. I have not done any tests for efficiency. It's good to see alternate methods that accomplish the same goal.

[ghostdog74]

I have a string list

items = ['B','D','B','A' ,'E']

Assume that I don't know the order (index) of these items. I would like to remove the second 'B' out of the list without sorting [using items.sort() ] or changing the the order of other items in the list. How can I do that?

The remove() and pop(), only take the first 'B' or I have to know the index of the second 'B' to remove second 'B' out.

go with the basics. The basics are very important

>>> items = ['B','D','B','A','E']
>>> start=items.index('B') #get the first "B" position
>>> print start
0
>>> second = items[start+1:].index("B") # get the second "B"
>>> print second 
1
>>> index_to_remove = second + 1 #add 1 to get exact position of second "B" since we started search 1 position after the first "B"
>>> items.pop(index_to_remove)
'B'
>>> items
['B', 'D', 'A', 'E']
>>>

[ilikepython]

Thanks ilikepython. I like your code. I have not done any tests for efficiency. It's good to see alternate methods that accomplish the same goal.

The only problem is it doesn't work for patterns. Like, to search for "ACC" in "ACACCAAACC ".