Mr_and_Mrs_D - 1 year ago 51

Python Question

So having a list say

`b = [b1, b2, b3]`

`a`

`bi`

`a`

`b`

`a`

`a = [ b1, x, b3, y, b2] -> [ b1, x, b2, y, b3]`

a = [ b1, x, b2, y, b3] -> no change

a = [ b1, x, y, b2] -> no change

a = [ b3, x, b1, y, b2] -> [ b1, x, b2, y, b3]

b of course may be a tuple or any other ordered structure. What I came up with

`bslots = dict((x, a.index(x)) for x in a if x in b)`

bslotsSorted = sorted(bslots.keys(), key=lambda y: b.index(y))

indexes = sorted(bslots.values())

for x,y in zip(bslotsSorted, indexes):

a[y] = x

is clumsy and O(n^2)

Answer

First create a dictionary using items from

`b`

where the key is the item and value is its index, we will use this to sort the matched items in`a`

later on.Now filter out item from

`a`

that are present in that dict, dict provides O(1) lookup.Now sort this list of filtered items and convert it to an iterator.

Now loop over

`a`

again and for each item check if is present in dict then fetch its value from iterator otherwise use it as is.

```
def solve(a, b):
dct = {x: i for i, x in enumerate(b)}
items_in_a = [x for x in a if x in dct]
items_in_a.sort(key=dct.get)
it = iter(items_in_a)
return [next(it) if x in dct else x for x in a]
...
>>> b = ['b1', 'b2', 'b3']
>>> a = [ 'b1', 'x', 'b3', 'y', 'b2']
>>> solve(a, b)
['b1', 'x', 'b2', 'y', 'b3']
>>> a = [ 'b1', 'x', 'b2', 'y', 'b3']
>>> solve(a, b)
['b1', 'x', 'b2', 'y', 'b3']
>>> a = [ 'b1', 'x', 'y', 'b2']
>>> solve(a, b)
['b1', 'x', 'y', 'b2']
>>> a = [ 'b3', 'x', 'b1', 'y', 'b2']
>>> solve(a, b)
['b1', 'x', 'b2', 'y', 'b3']
```

Overall time complexity is going to be max of `(O(len(a)), O(len(b)), O(items_in_a_length log items_in_a_length)`

.

Source (Stackoverflow)