How can I aggregate the same key values and take an average of that value?

[['Absard', 140000.0], ['Absard', 150000.0], ['Absard', 150000.0], ['Absard', 173333.3], ['Abuzar', 28333.3], ['Abuzar', 34000.0], ['Abuzar', 90500.0], ['Afsarieh', 37333.3], ['Afsarieh', 44333.3], ['Afsarieh', 51666.6], ['Afsarieh', 55000.0], ['Afsarieh', 80000.0], ['Afsarieh', 105000.0], ['Ahang', 26666.6], ['Ahang', 46666.6], ['Air force', 55000.0], ['Air force', 56333.3]]

I want to print out this list like this,

NAME        AVG.
----------------------
Abazar      3033333.33
Ahang       2666666.67
Air force   2333333.33
Afsarieh    1916666.67

These 5 are expected to store in a dictionary (Average values are randomly written)

Wrote something like this but this aggregates not properly

 def takeAvg(addressList):
    resultDict = {}
    tot = 0
    startIndex = 0
    inner_i = 1
    for inner_i in range(len(addressList)):
        if(addressList[startIndex][0] == addressList[inner_i][0]):
            tot  = float(addressList[inner_i][1])
        else:
            tot  = float(addressList[startIndex][1])
            resultDict.update({addressList[startIndex][0]: format(float(tot / (inner_i-startIndex)), ".2f")})
            tot = 0
            startIndex = inner_i
    return resultDict

CodePudding user response：

You can iterate your data, sorting into a defaultdict of lists, and then average out the results for each key:

from collections import defaultdict

data = [['Absard', 140000.0], ['Absard', 150000.0], ['Absard', 150000.0], ['Absard', 173333.3], ['Abuzar', 28333.3], ['Abuzar', 34000.0], ['Abuzar', 90500.0], ['Afsarieh', 37333.3], ['Afsarieh', 44333.3], ['Afsarieh', 51666.6], ['Afsarieh', 55000.0], ['Afsarieh', 80000.0], ['Afsarieh', 105000.0], ['Ahang', 26666.6], ['Ahang', 46666.6], ['Air force', 55000.0], ['Air force', 56333.3]]

acc = defaultdict(list)
for name, value in data:
    acc[name].append(value)

result = { k : sum(v)/len(v) for k, v in acc.items() }

Output:

{
 'Absard': 153333.325,
 'Abuzar': 50944.43333333333,
 'Afsarieh': 62222.200000000004,
 'Ahang': 36666.6,
 'Air force': 55666.65
}

For display purposes you can print the values with using an f-string to format to 2 decimal places. e.g.

print(*[f'{k:16}\t{v:.2f}\n' for k, v in result.items()], end='')

Output:

 Absard                 153333.33
 Abuzar                 50944.43
 Afsarieh               62222.20
 Ahang                  36666.60
 Air force              55666.65

CodePudding user response：

If you convert it to a dict with lists as values, and then iterate the dict to get the average, you can do it like this:

data = [['Absard', 140000.0], ['Absard', 150000.0], ['Absard', 150000.0], ['Absard', 173333.3], ['Abuzar', 28333.3], ['Abuzar', 34000.0], ['Abuzar', 90500.0], ['Afsarieh', 37333.3], ['Afsarieh', 44333.3], ['Afsarieh', 51666.6], ['Afsarieh', 55000.0], ['Afsarieh', 80000.0], ['Afsarieh', 105000.0], ['Ahang', 26666.6], ['Ahang', 46666.6], ['Air force', 55000.0], ['Air force', 56333.3]]

res = {_lst[0]: [] for _lst in data}

for _lst in data:
    name, num = _lst
    res[name].append(num)

res = {k: round(sum(v) / len(v), 2) for k, v in res.items()}
print(res)

{'Absard': 153333.33, 'Abuzar': 50944.43, 'Afsarieh': 62222.2, 'Ahang': 36666.6, 'Air force': 55666.65}

Not the most efficient solution for all the iteration, but I hope it's easy to follow.

CodePudding user response：

You can use itertools.groupby

l = [['Absard', 140000.0], ['Absard', 150000.0], ['Absard', 150000.0], ['Absard', 173333.3], ['Abuzar', 28333.3], ['Abuzar', 34000.0], ['Abuzar', 90500.0], ['Afsarieh', 37333.3], ['Afsarieh', 44333.3], ['Afsarieh', 51666.6], ['Afsarieh', 55000.0], ['Afsarieh', 80000.0], ['Afsarieh', 105000.0], ['Ahang', 26666.6], ['Ahang', 46666.6], ['Air force', 55000.0], ['Air force', 56333.3]]

for k, g in groupby(l, key=lambda x: x[0]):
    values = [_[1] for _ in g]
    print(k, sum(values) / len(values), sep='\t')

Output

Absard  153333.325
Abuzar  50944.43333333333
Afsarieh        62222.200000000004
Ahang   36666.6
Air force       55666.65

This assumes that all the "keys" occur together - if, for example, "Absard" was to occur again at the end of the list, you would get 2 means for "Absard". You can ensure that by sorting the list before passing it into itertools.groupby -

l = sorted(l, key=lambda x: x[0])