How to switch position of two items in a Python list?

231
votes

I haven’t been able to find a good solution for this problem on the net (probably because switch, position, list and Python are all such overloaded words).

It’s rather simple – I have this list:

['title', 'email', 'password2', 'password1', 'first_name', 'last_name', 'next', 'newsletter']

I’d like to switch position of 'password2' and 'password1' – not knowing their exact position, only that they’re right next to one another and password2 is first.

I’ve accomplished this with some rather long-winded list-subscripting, but I wondered its possible to come up with something a bit more elegant?

7
pythonlist
Is your problem the efficiency of trying to find 'password2' in the list? Can 'password1' come before 'password2'? Is there some complexity here that doesn't come across in your original question? Otherwise I agree with @unwind. - Brent Writes Code
You should post what you've got - I'm curious about what you mean by a "rather long-winded list-subscripting." - samtregar
Something along the lines of index1 = index('password1'); index2 = index('password2'); order = order[:index2].append(order[index1]).append(order[index2]).append(order[index1 + 1:]); spread on a few more lines. Pretty, no. - mikl

7 Answers

430
votes
i = ['title', 'email', 'password2', 'password1', 'first_name', 
     'last_name', 'next', 'newsletter']
a, b = i.index('password2'), i.index('password1')
i[b], i[a] = i[a], i[b]
184
votes

The simple Python swap looks like this:

foo[i], foo[j] = foo[j], foo[i]

Now all you need to do is figure what i is, and that can easily be done with index:

i = foo.index("password2")
17
votes

Given your specs, I'd use slice-assignment:

>>> L = ['title', 'email', 'password2', 'password1', 'first_name', 'last_name', 'next', 'newsletter']
>>> i = L.index('password2')
>>> L[i:i+2] = L[i+1:i-1:-1]
>>> L
['title', 'email', 'password1', 'password2', 'first_name', 'last_name', 'next', 'newsletter']

The right-hand side of the slice assignment is a "reversed slice" and could also be spelled:

L[i:i+2] = reversed(L[i:i+2])

if you find that more readable, as many would.

7
votes

How can it ever be longer than

tmp = my_list[indexOfPwd2]
my_list[indexOfPwd2] = my_list[indexOfPwd2 + 1]
my_list[indexOfPwd2 + 1] = tmp

That's just a plain swap using temporary storage.

1
votes
for i in range(len(arr)):
    if l[-1] > l[i]:
        l[-1], l[i] = l[i], l[-1]
        break

as a result of this if last element is greater than element at position i then they both get swapped .

-1
votes

you can use for example:

>>> test_list = ['title', 'email', 'password2', 'password1', 'first_name',
                 'last_name', 'next', 'newsletter']
>>> reorder_func = lambda x: x.insert(x.index('password2'),  x.pop(x.index('password2')+1))
>>> reorder_func(test_list)
>>> test_list
... ['title', 'email', 'password1', 'password2', 'first_name', 'last_name', 'next', 'newsletter']
-1
votes

I am not an expert in python but you could try: say 

i = (1,2)

res = lambda i: (i[1],i[0])
print 'res(1, 2) = {0}'.format(res(1, 2))

above would give o/p as:

res(1, 2) = (2,1)