return items from dictionary not as tuple

Question

I have an excel file and I accumulate thre values for each fruit sort with each other.

So I do it like this:

def calulate_total_fruit_NorthMidSouth():

    import openpyxl
    import tabula

    excelWorkbook = openpyxl.load_workbook(path, data_only=True)

    sheet_factuur = excelWorkbook['Facturen ']
    new_list =[]

    fruit_sums = {
        'ananas': 0,
        'apple': 0,
        'waspeen': 0,
    }

    fruit_name_rows = {
        'ananas': [6, 7, 8],
        'apple': [9, 10, 11],
        'waspeen': [12, 13, 14],
    }
    array = [row for row in sheet_factuur.values]  # type: ignore

    # excel does not have a row 0
    for row_num, row_values in enumerate(array, 1):
        for fruit in ['ananas', 'apple', 'waspeen']:  # loop through specific fruits
            if row_num in fruit_name_rows[fruit]:
                # index 4 is column 5 in excel
                fruit_sums[fruit] += row_values[4]  # type: ignore
    return list(fruit_sums.items())

But the output is this:

[('ananas', 3962), ('apple', 3304.08), ('waspeen', 3767.3999999999996)]

But the output has to look like this:

ananas 3962
apple 3304.08
waspeen 3767.39

How to archive this with return statement?

Asked By: mightycode Newton

||

Source

Answer 1

Not sure if this is the best way to solve this problem but you could add this to your code before printing:

mylist = list(fruit_sums.items())
for i in mylist:
  newlist = list(i)
  for x in range(len(newlist)):
    newlist[x] = str(newlist[x])
print(" ".join(newlist))

Answered By: Cole Bechtel

Answer 2

Try something like this:

def f():
    x = {'a': 1, 'b': 2, 'c': 3}
    return 'n'.join(f'{a} {b}' for a, b in x.items())

print(f())
# a 1
# b 2
# c 3

Answered By: Adrian Kurzeja

return items from dictionary not as tuple

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