Sorting a list of dictionaries based on the order of values of another list

Question

I'm using python 2.7.3, and I'm trying to sort a list of dictionaries based on the order of values of another list. IE: ``` listOne = ['hazel', 'blue', 'green', 'brown'] listTwo = [{'name': 'Steve', 'eyecolor': 'hazel', 'height': '5 ft. 11 inches'}, {'name': 'Mark', 'eyecolor': 'brown', 'height': '6 ft. 2 inches'}, {'name': 'Mike', 'eyecolor': 'blue', 'height': '6 ft. 0 inches'}, {'name': 'Ryan', 'eyecolor': 'brown', 'height': '6 ft, 0 inches'}, {'name': 'Amy', 'eyecolor': 'green', 'height': '5 ft, 6 inches'}] ``` Sorting listTwo based off of the order of values in listOne, we would end up with the following: ``` print listTwo [{'name': 'Steve', 'eyecolor': 'hazel', 'height': '5 ft. 11 inches'}, {'name': 'Mike', 'eyecolor': 'blue', 'height': '6 ft. 0 inches'}, {'name': 'Amy', 'eyecolor': 'green', 'height': '5 ft, 6 inches'}, {'name': 'Mark', 'eyecolor': 'brown', 'height': '6 ft. 2 inches'}, {'name': 'Ryan', 'eyecolor': 'brown', 'height': '6 ft, 0 inches'}] ``` I eventually need to out…

Accepted Answer

The simplest way would be to use `list.index` to generate a sort value for your list of dictionaries:

```
listTwo.sort(key=lambda x: listOne.index(x["eyecolor"]))
```

This is a little bit inefficient though, since `list.index` does a linear search through the eye-color list. If you had many eye colors to check against, it would be slow. A somewhat better approach would build an index dictionary instead:

```
order_dict = {color: index for index, color in enumerate(listOne)}
listTwo.sort(key=lambda x: order_dict[x["eyecolor"]])
```

If you don't want to modify `listTwo`, you can use the built-in `sorted` function instead of the `list.sort` method. It returns a sorted copy of the list, rather than sorting in-place.

Sorting a list of dictionaries based on the order of values of another list

Solution

Problem