Python, Grouping a list by specific order-CodePudding

I want to group items in a list with a specific order. The order I want is in groups and the items must be grouped by that order. The solution I've tried is to pop the item[index] in items when I append it to temp, but it's not a good idea to manipulate a list while iterating it.

Is there a better approach or any other way to optimize this sample code?

groups = ["d", "c", "a", "b"]
items = ["a", "b", "b", "c", "d", "a"]
result = []
for group in groups:
    temp = []
    for item in items:
        if item == group:
            temp.append(item)
    result.extend(temp)
    
print(result)

CodePudding user response：

You could get the frequency of all the items, then flatten in order of groups... eg:

from collections import Counter

counts = Counter(items)
result = [k for k in groups for n in range(counts[k])]
# ['d', 'c', 'a', 'a', 'b', 'b']

CodePudding user response：

It looks like you're sorting more than grouping. There are a couple of ways to approach this.

One way would be a sort with a key based on groups. To avoid the O(n) lookup every time you call groups.index, convert it to dict and use get or __getitem__ instead:

groupdict = {g: i for i, g in enumerate(groups)}
result = sorted(items, key=groupdict.__getitem__)

CodePudding user response：

You don't really need to sort the list; you can just count how many instances of an item there are, then add that many instances to your result list. Iterate through your groups list. You begin with "d". There is one instance of "d" in items. Therefore, you append "d" to result one time.

groups = ["d", "c", "a", "b"]
items = ["a", "b", "b", "c", "d", "a"]
result = []

for group in groups:
    for i in range(items.count(group)):
        result.append(group)

print(result)
# ['d', 'c', 'a', 'a', 'b', 'b']