Grouping Transactions by Items' Names-CodePudding

Using Python3

For a given array of transactions, group all of the transactions by item name. Return an array of string where each string contains the item name followed by a space and the number of associated transactions.

Note: Sort the array descending by transaction count, then ascending alphabetically by item name for items with matching transaction counts.

Example

transactions = ['notebook', 'notebook', 'mouse', 'keyboard', 'mouse']

There are two items with 2 transactions each: 'notebook' and 'mouse'. In alphabetical order, they are 'mouse', 'notebook'. There is one item with 1 transaction: 'keyboard'. The return array sorted as required is ['mouse 2', 'notebook 2', 'keyboard 1']

What I've got

def get_trans(lst):
    ans = {}

   for i in lst:
      ans[i] = ans.get(i, 0)   1
   return ans

dictionary = get_trans(transactions)

print(dictionary)

{'notebook' : 2, 'mouse' : 2, 'keyboard' : 1}

data = list(dictionary.items())
an_array = np.array(data)
print(an_array)

Current output: 'notebook' comes before 'mouse', separate strings for word/count

[['notebook' '2']
 ['mouse' '2']
 ['keyboard' '1']]

Desired output: 'notebook' alphabetically ordered after 'mouse', word/count is one string

['mouse 2', 'notebook 2', 'keyboard 1']

CodePudding user response：

Your function is almost correct, only you need to pass the key first and a default value second to the get method. Then you can use the sorted function to sort the tuples of key-value pairs in dictionary with counts in descending order and item names in ascending order:

def get_trans(lst):
    ans = {}
    for i in lst:
        ans[i] = ans.get(i, 0)   1
    return ans

out = [' '.join([t[0],str(t[1])]) for t in sorted(get_trans(transactions).items(), key=lambda x:(-x[1],x[0]))]

Output:

['mouse 2', 'notebook 2', 'keyboard 1']

CodePudding user response：

Manlai A provided the correct answer. I've converted it to a function:

def get_trans(lst):
    ans = {}
    for i in lst:
        ans[i] = ans.get(i, 0)   1
    return ans

# A function using get_trans() function with Manlai A's answer
def grouping(lst):
    out = sorted(get_trans(lst).items(), key=lambda x:(-x[1],x[0]))
    out = [' '.join([t[0],str(t[1])]) for t in out]
    return out

# Passing transactions list into the new function

grouping(transactions)

Output:

['mouse 2', 'notebook 2', 'keyboard 1']