Finding number of occurrences of items in a dictionary of lists

Question

I have two supplied bits of information: A dictionary of transactions and a list of the unique items in the transactions.

transactions = {
   "T1": ["A", "B", "C", "E"],
    "T2": ["A", "D", "E"],
    "T3": ["B", "C", "E"],
    "T4": ["B", "C", "D", "E"],
    "T5": ["B", "D", "E"]
}
items = ["A", "B", "C", "D", "E"]

What I need to do is find the number of occurrences of these items in the transactions. I created a dictionary that has keys representing the unique items and the value for each key initialized to 0, but I am unsure of how to update these values to represent the number of occurrences in the transactions.

occurr = dict()
for x in items:
    occurr[x] = 0

This is my occurrences dictionary which yields the output:

{'A': 0, 'B': 0, 'C': 0, 'D': 0, 'E': 0}

The final dictionary should look like:

{'A': 2, 'B':4, 'C': 3, 'D': 3, 'E': 5}

as ‘A’ occurs 2 times in the transactions, ‘B’ occurs 4 times, etc.

Asked By: LeGOATJames23

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Source

Answer 1

You can use Counter, for example:

from collections import Counter

transactions = {
    "T1": ["A", "B", "C", "E"],
    "T2": ["A", "D", "E"],
    "T3": ["B", "C", "E"],
    "T4": ["B", "C", "D", "E"],
    "T5": ["B", "D", "E"]
}

c = Counter()

for dd in transactions.values():
    c.update(dd)

print(c) # or c.items(), c.keys() or c.values()

# Result: Counter({'E': 5, 'B': 4, 'C': 3, 'D': 3, 'A': 2})
# Note that the result is a subclass of dict

This will count the frequency of all values in transactions. If you need to restrict to those keys present in items then filter for that.

Alternatively, flatten the transaction list values into a single list, and count that in one call. For example:

flatlist = [item for sublist in transactions.values() for item in sublist]
print(Counter(flatlist))

Answered By: jarmod

Answer 2

Well, you are in the right direction. You need to iterate over the values of dictionary.

occurr = dict()
for x in items:
    occurr[x] = 0

for transaction in transactions.values():
    for item in transaction:
        occurr[item] += 1

Alternatively, you can concatenate all lists to single list and call collections.Counter:

import collections
items = [item for transaction in transactions.values() for item in transaction]
print(collections.Counter(items))

Answered By: Piyush Singh

Answer 3

Try:

transactions = {
    "T1": ["A", "B", "C", "E"],
    "T2": ["A", "D", "E"],
    "T3": ["B", "C", "E"],
    "T4": ["B", "C", "D", "E"],
    "T5": ["B", "D", "E"],
}
items = ["A", "B", "C", "D", "E"]

out = {}
for l in transactions.values():
    for v in l:
        out[v] = out.get(v, 0) + 1

out= {k: out.get(k) for k in items}
print(out)

Prints:

{'A': 2, 'B': 4, 'C': 3, 'D': 3, 'E': 5}

Answered By: Andrej Kesely

Finding number of occurrences of items in a dictionary of lists

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