当一个列表的所有元素都在另一个列表中时，如何分组和求和

Question

I have a data frame df1. "transactions" column has an array of int.

id     transactions
1      [1,2,3]
2      [2,3]

data frame df2. "items" column has an array of int.

items  cost
[1,2]  2.0
[2]    1.0
[2,4]  4.0

I need to check whether all elements of items are in each transaction if so sum up the costs.

Expected Result

id    transaction score
 1      [1,2,3]     3.0
 2      [2,3]       1.0

I did the following

#cross join
-----------
def cartesian_product_simplified(left, right):
   la, lb = len(left), len(right)
   ia2, ib2 = np.broadcast_arrays(*np.ogrid[:la,:lb])

    return pd.DataFrame(
    np.column_stack([left.values[ia2.ravel()], 
     right.values[ib2.ravel()]]))

out=cartesian_product_simplified(df1,df2) 

#column names assigning        
out.columns=['id', 'transactions', 'cost', 'items']

#converting panda series to list
t=out["transactions"].tolist()
item=out["items"].tolist()


#check list present in another list
-------------------------------------
def check(trans,itm):
out_list=list() 
for row in trans:
   ret =np.all(np.in1d(itm, row))
   out_list.append(ret)
return out_list

if true: group and sum
-----------------------
a=check(t,item)
for i in a:
  if(i):
   print(out.groupby(['id','transactions']))['cost'].sum()      
  else:
   print("no")

Throws TypeError: 'NoneType' object is not subscriptable.

I am new to python and don't know how to put all these together. How to group by and sum the cost when all items of one list in another list?

Answer 1

The simplies way is just to check all items for all transactions:

# df1 and df2 are initialized

def sum_score(transaction):
    score = 0
    for _, row in df2.iterrows():
        if all(item in transaction for item in row["items"]):
            score += row["cost"]
    return score

df1["score"] = df1["transactions"].map(sum_score)

It will be extremely slow on big scale. If this is a problem, we need to iterate not over every item, but preselect only possible. If you have enough memory, it can be done like that. For each item we remember all the row numbers in df2 , where it appeared. So for each transaction we get the items, get all the possible lines and check only them.

import collections

# df1 and df2 are initialized

def get_sum_score_precalculated_func(items_cost_df):

    # create a dict of possible indexes to search for an item
    items_search_dict = collections.default_dict(set)
    for i, (_, row) in enumerate(items_cost_df.iterrow()):
        for item in row["items"]:
            items_search_dict[item].add(i)

    def sum_score(transaction):
        possible_indexes = set()
        for i in transaction:
            possible_indexes += items_search_dict[i]

        score = 0
        for i in possible_indexes:
            row = items_cost_df.iloc[i]
            if all(item in transaction for item in row["items"]):
                score += row["cost"]
        return score

    return sum_score

df1["score"] = df1["transactions"].map(get_sum_score_precalculated_func(df2))

Here I use set which is an unordered storage of unique values (it helps to join possible line numbers and avoid double count). collections.defaultdict which is a usual dict , but if you are trying to access uninitialized values it fill it with the given data (blank set in my case). It help to avoid if x not in my_dict: my_dict[x] = set() . I also use so called "closure", which means sum_score function will have access to items_cost_df and items_search_dict which were accessible at the level the sum_score function was declared even after it was returned and get_sum_score_precalculated_func

That should be much faster in case the items are quite unique and can be found only in a few lines of df2 .

If you have quite a few unique items and so many identical transactions, you'd better calculate score for each unique transaction first. And then just join the result.

transactions_score = []
for transaction in df1["transactions"].unique():
    score = sum_score(transaction)
    transaction_score.append([transaction, score])
transaction_score = pd.DataFrame(
    transaction_score,
    columns=["transactions", "score"])
df1 = df1.merge(transaction_score, on="transactions", how="left")

Here I use sum_score from first example of code

PS With the python error message there should be a line number which helps a lot to understand the problem.

Answer 2

# convert df_1 to dictionary for iteration
df_1_dict = dict(zip(df_1["id"], df_1["transactions"]))
# convert df_2 to list for iteration as there is no unique column
df_2_list = df_2.values.tolist()

# iterate through each combination to find a valid one
new_data = []
for rows in df_2_list:
    items = rows[0]
    costs = rows[1]
    for key, value in df_1_dict.items():
        # find common items in both
        common = set(value).intersection(set(items))
        # execute of common item exist in second dataframe 
        if len(common) == len(items):
            new_row = {"id": key, "transactions": value, "costs": costs}
            new_data.append(new_row)

merged_df = pd.DataFrame(new_data)
merged_df = merged_df[["id", "transactions", "costs"]]

# group the data by id to get total cost for each id
merged_df = (
    merged_df
    .groupby(["id"])
    .agg({"costs": "sum"})
    .reset_index()
)