I have the following list:
l = [["a", "done"], ["c", "not done"]]
If the second element of each sub list is "done" I want to remove that the sub list. So the output should be:
l = [["c", "not done"]]
Obviously the below doesn't work:
for i in range(len(l)):
if l[i][1] == "done":
l.pop(0)
Use list_comprehension. It just builts a new list by iterating over the sublists where the second element in each sublist won't contain the string done
>>> l = [["a", "done"], ["c", "not done"]]
>>> [subl for subl in l if subl[1] != 'done']
[['c', 'not done']]
>>>
l = [["a", "done"], ["c", "not done"]]
print [i for i in l if i[1]!="done"]
or use filter
l = [["a", "done"], ["c", "not done"]]
print filter(lambda x:x[1]!="done",l)
Apply a filter for your criteria:
l = [["a", "done"], ["c", "not done"]]
l = filter(lambda x: len(x)>=2 and x[1]!='done', l)
status index is 1, you checked index 0
for i in range(len(l)):
if(l[i][1] == "done"):
l.pop(i)
Use list comprehension:
l = [ item for item in l if item[-1] != 'done']
