Sorting and filtering one list based on another-CodePudding

Problem: I would like to sort one list based on another list. Example: below I would like to sort list_sec based on 'key' in this list and the order would be from list_main.

list_main = [3, 33, 2]

list_sec = [{'key': 2, 'rocket': 'mark11'}, {'key': 332, 'rocket': 'mark23'}, {'key': 3, 'rocket': 'mark1'} ]

Output would be as below. (exp: first entry in list_main is 3, so 'key' : 3 should come to index = 0, second value is 33 but this key is missing in list_sec so will discard this. third key is 2, so this will come next.

output = [{'key': 3, 'rocket': 'mark1'}, {'key': 2, 'rocket': 'mark11'}]

I have read couple of answers on similar lines: Python, sort a list by another list , Python 3.x -- sort one list based on another list and then return both lists sorted

But stuck without attempt. Any way this can be done.

CodePudding user response：

Transform your list of dicts into a dict, then you'll be able to retrieve items by value of key.

In [2]: rocket_dict = {item['key']: item for item in list_sec}

In [3]: rocket_dict
Out[3]: 
{2: {'key': 2, 'rocket': 'mark11'},
 332: {'key': 332, 'rocket': 'mark23'},
 3: {'key': 3, 'rocket': 'mark1'}}

If you need a list in that particular order, you can do:

In [4]: output = []

In [5]: for key in list_main:
   ...:     if key in rocket_dict:
   ...:         output.append(rocket_dict[key])
   ...: 

In [6]: output
Out[6]: [{'key': 3, 'rocket': 'mark1'}, {'key': 2, 'rocket': 'mark11'}]

CodePudding user response：

Try this simple method it will work in your case

list_main = [3, 33, 2]

list_sec = [{'key': 2, 'rocket': 'mark11'}, {'key': 332, 'rocket': 'mark23'}, {'key': 3, 'rocket': 'mark1'} ]

sorted_list = [] # The final sorted list
for key in list_main:  # iterate on each key
    for item in list_sec:
        if item['key'] == key: # find the item based on your key
            sorted_list.append(item) # append in sequence
print(sorted_list) # your sorted list

CodePudding user response：

sorted(list(filter(lambda data: data["key"] in list_main, list_sec)),
       key=lambda data: list_main.index(data["key"]))

output:

[{'key': 3, 'rocket': 'mark1'}, {'key': 2, 'rocket': 'mark11'}]

CodePudding user response：

Essentially @LevLevitsky's answer, but with list comprehension

wrap_dict = {item['key']: item for item in list_sec}
output = [wrap_dict[key] for key in list_main if key in wrap_dict]

Or we can use the coolest walrus operator introduced in Python 3.8

output = [rkt_dict for key in list_main if (rkt_dict := wrap_dict.get(key))]