CODEWITHSUNDEEP
pythonsortingdictionary
Sapoora Alavi
Sapoora Alavi

Reputation: 1

sorting a OrderedDict dictionary with duplicated float values

I have a dictionary named dict that stores names and float averages, and I want to store the sorted averages with names in a file, but when I use a loop to find the key (students' names) it prints duplicated name.

print(dict)
sorted_list = sorted(dict.values() , key = float)

sorted_dict = list()
for i in sorted_list:
    for j in dict:
        if dict[j] == i:
            sorted_dict.append([j,dict[j]])
            break
print(sorted_dict)

Take a look at the output:

OrderedDict([('mandana', '5'), ('hamid', '6.066666666666666'), ('sina', '11.285714285714286'), ('sara', '9.75'), ('soheila', '7.833333333333333'), ('ali', '5'), ('sarvin', '11.375')])

[['mandana', '5'], ['mandana', '5'], ['hamid', '6.066666666666666'], ['soheila', '7.833333333333333'], ['sara', '9.75'], ['sina', '11.285714285714286'], ['sarvin', '11.375']]

Upvotes: 0

Views: 67

Answers (1)

mkrieger1
mkrieger1

Reputation: 23192

Your algorithm looks up the name corresponding to a given number. This leads to wrong results if the same number occurs for more than one name. Here, for 5, which actually belongs to 'ali', it finds 'mandana' first and includes it instead.

You can directly sort the name, value pairs by passing dict.items instead of dict.values to sorted. The key function would then have to take the second item of each pair and convert it to a float.

sorted_dict = sorted(dict.items(), key=lambda pair: float(pair[1]))

Upvotes: 2

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