writing a loop with dictionary in python

Question

I have a dictionary of length 50 which has numbers(from 0 to 50) as keys and BMI(float from 0.1 to 0.98) as the values.The code that i have tried only extract first and 5th value from the keys..but i need to extract every first and 5th value.

I need to create a new dict such that the output has

key:(0,5), value: average of bmi(first 5 bmi) 
key:(6,10), value: average of bmi(next 5 bmi) and so on

key_new=[]
val_new=[]

num=[0,5]
for i in num:
    if(i==0):
        number=list(dictionary.keys())[i]
        key_new.append(number)

    elif(i==5):
        number=list(dictionary.keys())[i]
        key_new.append(number)

print(key_new)

If your dict keys are numbers 0-50, you should just treat it as a list. — mattmess1221
– mattmess1221, Commented Oct 29, 2019 at 17:52

Rithin Chalumuri · Accepted Answer · 2019-10-29 18:38:05Z

1

As mentioned in the comments, if your dict keys are numbers 0-50, you should just treat it as a list -> allows you to use list slicing when calculating average.

You can set step=5 in for loop range method.

Example:


bmi_vals = [0.23, 0.24, 0.3, 0.32, 0.16 ...] #Example format for bmi values list

bmi_vals = [your_dictionary[key] for key in range(len(your_dictionary))] #Constructing bmi values list from your dictionary

result = {}

total_size = len(bmi_vals)
step_size = 5

for i in range(0, total_size, step_size):
    start_index = i
    end_index = i+step_size

    result[(start_index , end_index)] = sum(bmi_vals[start_index:end_index])/step_size #Eg Key: Tuple (0, 5), Val: average of first 5

print(result)

You can replace step_size to any other number also -> if you want average of every 10 values then set step_size=10.

edited Oct 29, 2019 at 18:38

answered Oct 29, 2019 at 18:27

Rithin Chalumuri

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