How to sort a dictionary based on a list in python
Question:
I have a dictionary
a = {'ground': obj1, 'floor 1': obj2, 'basement': obj3}
I have a list.
a_list = ['floor 1', 'ground', 'basement']
I want to sort dictionary a using its keys based on the list. Is it possible to do that?
i.e.:
sort(a).based_on(a_list) #this is wrong. But I want something like this.
The output doesn’t have to be another dictionary, I don’t mind converting the dictionary to tuples and then sort those.
Answers:
The naive way, sorting the list of (key, value) tuples, using the sorted() function and a custom sort key (called for each (key, value) pair produced by dict.items())):
sorted(a.items(), key=lambda pair: a_list.index(pair[0]))
The faster way, creating an index map first:
index_map = {v: i for i, v in enumerate(a_list)}
sorted(a.items(), key=lambda pair: index_map[pair[0]])
This is faster because the dictionary lookup in index_map takes O(1) constant time, while the a_list.index() call has to scan through the list each time, so taking O(N) linear time. Since that scan is called for each key-value pair in the dictionary, the naive sorting option takes O(N^2) quadratic time, while using a map keeps the sort efficient (O(N log N), linearithmic time).
Both assume that a_list contains all keys found in a. However, if that’s the case, then you may as well invert the lookup and just retrieve the keys in order:
[(key, a[key]) for key in a_list if key in a]
which takes O(N) linear time, and allows for extra keys in a_list that don’t exist in a.
To be explicit: O(N) > O(N log N) > O(N^2), see this cheat sheet for reference.
Demo:
>>> a = {'ground': 'obj1', 'floor 1': 'obj2', 'basement': 'obj3'}
>>> a_list = ('floor 1', 'ground', 'basement')
>>> sorted(a.items(), key=lambda pair: a_list.index(pair[0]))
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
>>> index_map = {v: i for i, v in enumerate(a_list)}
>>> sorted(a.items(), key=lambda pair: index_map[pair[0]])
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
>>> [(key, a[key]) for key in a_list if key in a]
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
You could just retrieve the values in order of the keys provided by the list and make a new list out of the key-value pairs.
Example:
d = a # dictionary containing key-value pairs that are to be ordered
l = a_list # list of keys that represent the order for the dictionary
# retrieve the values in order and build a list of ordered key-value pairs
ordered_dict_items = [(k,d[k]) for k in l]
This works for me:
a = {'ground': 'obj1', 'floor 1': 'obj2', 'basement': 'obj3'}
a_list = ['floor 1', 'ground', 'basement']
ordered_dict = {}
for item in a_list:
ordered_dict[item] = a[item]
print(ordered_dict)
{'floor 1': 'obj2', 'ground': 'obj1', 'basement': 'obj3'}
I have a dictionary
a = {'ground': obj1, 'floor 1': obj2, 'basement': obj3}
I have a list.
a_list = ['floor 1', 'ground', 'basement']
I want to sort dictionary a using its keys based on the list. Is it possible to do that?
i.e.:
sort(a).based_on(a_list) #this is wrong. But I want something like this.
The output doesn’t have to be another dictionary, I don’t mind converting the dictionary to tuples and then sort those.
The naive way, sorting the list of (key, value) tuples, using the sorted() function and a custom sort key (called for each (key, value) pair produced by dict.items())):
sorted(a.items(), key=lambda pair: a_list.index(pair[0]))
The faster way, creating an index map first:
index_map = {v: i for i, v in enumerate(a_list)}
sorted(a.items(), key=lambda pair: index_map[pair[0]])
This is faster because the dictionary lookup in index_map takes O(1) constant time, while the a_list.index() call has to scan through the list each time, so taking O(N) linear time. Since that scan is called for each key-value pair in the dictionary, the naive sorting option takes O(N^2) quadratic time, while using a map keeps the sort efficient (O(N log N), linearithmic time).
Both assume that a_list contains all keys found in a. However, if that’s the case, then you may as well invert the lookup and just retrieve the keys in order:
[(key, a[key]) for key in a_list if key in a]
which takes O(N) linear time, and allows for extra keys in a_list that don’t exist in a.
To be explicit: O(N) > O(N log N) > O(N^2), see this cheat sheet for reference.
Demo:
>>> a = {'ground': 'obj1', 'floor 1': 'obj2', 'basement': 'obj3'}
>>> a_list = ('floor 1', 'ground', 'basement')
>>> sorted(a.items(), key=lambda pair: a_list.index(pair[0]))
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
>>> index_map = {v: i for i, v in enumerate(a_list)}
>>> sorted(a.items(), key=lambda pair: index_map[pair[0]])
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
>>> [(key, a[key]) for key in a_list if key in a]
[('floor 1', 'obj2'), ('ground', 'obj1'), ('basement', 'obj3')]
You could just retrieve the values in order of the keys provided by the list and make a new list out of the key-value pairs.
Example:
d = a # dictionary containing key-value pairs that are to be ordered
l = a_list # list of keys that represent the order for the dictionary
# retrieve the values in order and build a list of ordered key-value pairs
ordered_dict_items = [(k,d[k]) for k in l]
This works for me:
a = {'ground': 'obj1', 'floor 1': 'obj2', 'basement': 'obj3'}
a_list = ['floor 1', 'ground', 'basement']
ordered_dict = {}
for item in a_list:
ordered_dict[item] = a[item]
print(ordered_dict)
{'floor 1': 'obj2', 'ground': 'obj1', 'basement': 'obj3'}