How to Calculate Average a Dictionary from List of Dictionary Data

Get the average of dictionary values? [closed]

This answer was intended for Python2, which is now dead

Okay, so let's iterate over all dictionary keys and average the items:

avgDict = {}
for k,v in StudentGrades.iteritems():
    # v is the list of grades for student k
    avgDict[k] = sum(v)/ float(len(v))

In Python3, the iteritems() method is no longer necessary, can use items() directly.

now you can just see :

avgDict
Out[5]: 
{'Ivan': 3.106666666666667,
 'Martin': 4.816666666666666,
 'Stoyan': 3.89,
 'Vladimir': 5.433333333333334}

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From your question I think you're queasy about iteration over dicts, so here is the same with output as a list :

avgList = []
for k,v in StudentGrades.iteritems():
    # v is the list of grades for student k
    avgDict.append(sum(v)/ float(len(v)))

Be careful though : the order of items in a dictionary is NOT guaranteed; this is, the order of key/values when printing or iterating on the dictionary is not guaranteed (as dicts are "unsorted").
Looping over the same identical dictionary object(with no additions/removals) twice is guaranteed to behave identically though.

How to calculate an average from average values in a dictionary

You can take the values of your newly created dictionary using .values and then apply the same logic you applied earlier to get an average of averages.

def avg(classes):
    average = {}
    for classnames, grades in classes.items():
        average[classnames] = sum(grades) / len(grades)
    return average

classes = {"Spanish II": [100, 99, 100, 98], "US History I": [95, 96, 97, 94]}
averages = avg(classes)
#{'Spanish II': 99.25, 'US History I': 95.5}

average_of_averages = sum(averages.values())/len(averages)
#97.375

Get average value from list of dictionary

Just divide the sum of values by the length of the list:

print sum(d['value'] for d in total) / len(total)

Note that division of integers returns the integer value. This means that average of the [5, 5, 0, 0] will be 2 instead of 2.5. If you need more precise result then you can use the float() value:

print float(sum(d['value'] for d in total)) / len(total)

How to calculate the mean of elements of a lists inside a dictionary in python?

First: your example is not correct dictionary. You missed {} in some places.

You should have

{
    'NN3-001': {'diffe_1':[1,2,3,4],'mas_1':[10,20,30,40],'diffe_2':[5,6,7,8],'mas_2':[50,60,70,80]},
    'NN3-002': {'diffe_1':[14,15,16,17],'mas_1':[100,200,300,400],'diffe_2':[18,19,20,21],'mas_2':[500,600,700,800]}
}

---

To get single list you can use

values = data['NN3-001']['diffe_1'] 

and you can calculate mean

mean = sum(values)/len(values)

---

For all list you have to use for-loops with dict.items()

dictionary = {
    'NN3-001': {'diffe_1':[1,2,3,4],'mas_1':[10,20,30,40],'diffe_2':[5,6,7,8],'mas_2':[50,60,70,80]},
    'NN3-002': {'diffe_1':[14,15,16,17],'mas_1':[100,200,300,400],'diffe_2':[18,19,20,21],'mas_2':[500,600,700,800]}
}

for name, values in dictionary.items():
    print('=== time serie:', name, '===')
    for key, data in values.items():
        print('  key:', key)
        print(' data:', data)
        print(' mean:', sum(data)/len(data))
        print('---')

Result:

=== time serie: NN3-001 ===
  key: diffe_1
 data: [1, 2, 3, 4]
 mean: 2.5
---
  key: mas_1
 data: [10, 20, 30, 40]
 mean: 25.0
---
  key: diffe_2
 data: [5, 6, 7, 8]
 mean: 6.5
---
  key: mas_2
 data: [50, 60, 70, 80]
 mean: 65.0
---
=== time serie: NN3-002 ===
  key: diffe_1
 data: [14, 15, 16, 17]
 mean: 15.5
---
  key: mas_1
 data: [100, 200, 300, 400]
 mean: 250.0
---
  key: diffe_2
 data: [18, 19, 20, 21]
 mean: 19.5
---
  key: mas_2
 data: [500, 600, 700, 800]
 mean: 650.0

---

EDIT:

After changes in question I see that you need zip(diffe_1, diffe_2) to create pairs.

dictionary = {
    'NN3-001': {'diffe_1':[1,2,3,4],'mas_1':[10,20,30,40],'diffe_2':[5,6,7,8],'mas_2':[50,60,70,80]},
    'NN3-002': {'diffe_1':[14,15,16,17],'mas_1':[100,200,300,400],'diffe_2':[18,19,20,21],'mas_2':[500,600,700,800]}
}

result = {}

for name, values in dictionary.items():
    print('=== time serie:', name, '===')
    
    result[name] = {'diff':[], 'mas':[]}
    
    print('--- diffe_1, diffe_2 ---')
    for a, b in zip(values['diffe_1'],values['diffe_2']):
        mean = int( (a+b)/2 )
        print(a, '&', b, '=>', mean)
        result[name]['diff'].append(mean)
        
    print('--- mas_1, mas_2 ---')
    for a, b in zip(values['mas_1'],values['mas_2']):
        mean = int( (a+b)/2 )
        print(a, '&', b, '=>', mean)
        result[name]['mas'].append(mean)

print(result)      

gives

=== time serie: NN3-001 ===
--- diffe_1, diffe_2 ---
1 & 5 => 3.0
2 & 6 => 4.0
3 & 7 => 5.0
4 & 8 => 6.0
--- mas_1, mas_2 ---
10 & 50 => 30.0
20 & 60 => 40.0
30 & 70 => 50.0
40 & 80 => 60.0
=== time serie: NN3-002 ===
--- diffe_1, diffe_2 ---
14 & 18 => 16.0
15 & 19 => 17.0
16 & 20 => 18.0
17 & 21 => 19.0
--- mas_1, mas_2 ---
100 & 500 => 300.0
200 & 600 => 400.0
300 & 700 => 500.0
400 & 800 => 600.0


{
'NN3-001': {'diff': [3, 4, 5, 6], 'mas': [30, 40, 50, 60]},  
'NN3-002': {'diff': [16, 17, 18, 19], 'mas': [300, 400, 500, 600]}
}

You may also use loop for prefix in ['diffe', 'mas']: to reduce code.

dictionary = {
    'NN3-001': {'diffe_1':[1,2,3,4],'mas_1':[10,20,30,40],'diffe_2':[5,6,7,8],'mas_2':[50,60,70,80]},
    'NN3-002': {'diffe_1':[14,15,16,17],'mas_1':[100,200,300,400],'diffe_2':[18,19,20,21],'mas_2':[500,600,700,800]}
}

result = {}

for name, values in dictionary.items():
    print('=== time serie:', name, '===')
    
    
    result[name] = {}
    
    for prefix in ['diffe', 'mas']:

        print('--- prefix:', prefix, '---')
        
        result[name][prefix] = []

        for a, b in zip(values[prefix+'_1'],values[prefix+'_2']):
            mean = int( (a+b)/2 )
            print(a, '&', b, '=>', mean)
            result[name][prefix].append(mean)
        
print(result)

Average of key values in a list of dictionaries

You can use zip and numpy functions mean and round for this task:

In [8]: import numpy as np

In [9]:  [dict(zip(d.keys(), [int(np.round(np.mean(d.values())))])) for d in L]

#Out[9]: [{'Eva': 5}, {'Ana': 53}, {'Ada': 12}]

Version with "less" parenthesis:

[dict(zip(d.keys(), [np.array(d.values()).mean().round().astype(int)])) for d in L]

average list of dictionaries in python

You can use the following

sum([item['Weight'] for item in data])/len(data)

and use

float(len(data)) 

if you want a more exact value.

Python - Find average in dict elements

You could create an intermediate dictionary that collects all encountered values as lists:

dct = [{'a':2, 'b':3}, {'b':4}, {'a':1, 'c':5}]
from collections import defaultdict
intermediate = defaultdict(list)

for subdict in dct:
    for key, value in subdict.items():
        intermediate[key].append(value)

# intermediate is now: defaultdict(list, {'a': [2, 1], 'b': [3, 4], 'c': [5]})

And finally calculate the average by dividing the sum of each list by the length of each list:

for key, value in intermediate.items():
    print(key, sum(value)/len(value))

which prints:

b 3.5
c 5.0
a 1.5

Average each dictionary list value

Assuming you meant for each value in the dictionary to be a list of numbers, here's a way to do what you're asking:

d = { "numbers1" : [4,5,6,3,2] , "numbers2" : [40,5,63,2] , "numbers3" : [10,25,96,3]}
avg = {k: sum(v) / len(v) for k, v in d.items()}
print(avg)

sortedAvg = sorted(d.items(), key=lambda x: sum(x[1]) / len(x[1]))
print(sortedAvg)

Output:

{'numbers1': 4.0, 'numbers2': 27.5, 'numbers3': 33.5}
[('numbers1', [4, 5, 6, 3, 2]), ('numbers2', [40, 5, 63, 2]), ('numbers3', [10, 25, 96, 3])]

Explanation:

• avg uses a dict comprehension to create a dictionary with the same keys as the input containing the average of each key's list
• sortedAvg uses a list comprehension to create a list of tuples of key/value pairs in the input dict sorted in the order of ascending average.

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