Use the insert method of a list:
l = list(...)
l.insert(index, item)
Alternatively, you can use a slice notation:
l[index:index] = [item]
If you want to move an item that's already in the list to the specified position, you would have to delete it and insert it at the new position:
l.insert(newindex, l.pop(oldindex))
If you don't know the position of the item, you may need to find the index first:
old_index = list1.index(item)
then move it:
list1.insert(new_index, list1.pop(old_index))
or IMHO a cleaner way:
try:
list1.remove(item)
list1.insert(new_index, item)
except ValueError:
pass
A slightly shorter solution, that only moves the item to the end, not anywhere is this:
l += [l.pop(0)]
For example:
>>> l = [1,2,3,4,5]
>>> l += [l.pop(0)]
>>> l
[2, 3, 4, 5, 1]
I profiled a few methods to move an item within the same list with timeit. Here are the ones to use if j>i:
+---------------------------------+ ¦ 14.4usec ¦ x[i:i]=x.pop(j), ¦ ¦ 14.5usec ¦ x[i:i]=[x.pop(j)] ¦ ¦ 15.2usec ¦ x.insert(i,x.pop(j)) ¦ +---------------------------------+
and here the ones to use if j<=i:
+--------------------------------------+ ¦ 14.4usec ¦ x[i:i]=x[j],;del x[j] ¦ ¦ 14.4usec ¦ x[i:i]=[x[j]];del x[j] ¦ ¦ 15.4usec ¦ x.insert(i,x[j]);del x[j] ¦ +--------------------------------------+
Not a huge difference if you only use it a few times, but if you do heavy stuff like manual sorting, it's important to take the fastest one. Otherwise, I'd recommend just taking the one that you think is most readable.
A solution very simple, but you have to know the index of the original position and the index of the new position:
list1[index1],list1[index2]=list1[index2],list1[index1]
l = list(...)
if item in l:
l.remove(item) # checks if the item to be moved is present in the list
l.insert(new_index,item)
Source: Stackoverflow.com