对包含浮点数和字符串的列表值中的浮点数执行操作

Question

I have a dictionary in Python 3 in which all key-value pairs have the following structure:

my_dict = {'Jerry': [0.4, 'Queens', 0.6, 'Knicks']}

This means that the value is always a list with float, string, float, string.

I need to modify the values so that the two floats are averaged, and that the new dictionary looks as follows:

new_dict = {'Jerry': [0.5, 'Queens', 'Knicks]}

I have tried a for loop that selects the floats from the value list based on their indexes, but Python doesn't let me index through it.

for k, v in my_dict.items():
    average = sum(v[0], v[1])/2
    my_dict[k] = [average, v[1], v[3]]

This returns

TypeError: 'float' object is not iterable

How could I select the floats to work out the average?

Answer 1

You need to pass an iterable to the sum() function, so make the values v[0] and v[1] to a list by adding a [ and ]

With minimal modification to your code

my_dict = {'Jerry': [0.4, 'Queens', 0.6, 'Knicks']}

for k, v in my_dict.items():
    average = sum([v[0], v[2]])/2
    my_dict[k] = [average, v[1], v[3]]

print(my_dict)

Output:

{'Jerry': [0.5, 'Queens', 'Knicks']}

Or a shorter solution

my_dict = {'Jerry': [0.4, 'Queens', 0.6, 'Knicks'], 'Berry': [0.5, 'Queens', 0.1, 'Knicks']}

my_dict = {k:[(v[0]+v[2])/2, v[1], v[3]] for k,v in my_dict.items()}

print(my_dict)

Output:

{'Jerry': [0.5, 'Queens', 'Knicks'], 'Berry': [0.3, 'Queens', 'Knicks']}

Answer 2

Iterate through the key , value pairs, and create a temporary list to store numeric and string data, and iterate each value list and append to numbers / strings lists based on type , finally find the average and add it to the resulting dictionary.

my_dict = {'Jerry': [0.4, 'Queens', 0.6, 'Knicks']}
result = {}

for key,value in my_dict.items():
    numbers = []
    strings =[]
    for item in value:
        if type(item) is float:
            numbers.append(item)
        else:
            strings.append(item)
    result[key] = [sum(numbers)/len(numbers)] + strings

OUTPUT :

{'Jerry': [0.5, 'Queens', 'Knicks']}

PS : This solution will work even if you have more or less number of items in the value list of the given dictionary.

Answer 3

The first argument for sum() is supposed to take an iterable (eg, a list).

You can either use average = sum([v[0], v[2]])/2 as suggested by @abhi-j or just average = (v[0]+v[2])/2 .