平均Python中两个配对列表的重复值

Question

in my code I obtain two different lists from different sources, but I know they are in the same order. The first list ("names") contains a list of keys strings, while the second ("result_values") is a series of floats. I need to make the pair unique, but I can't use a dictionary as only the last value inserted would be kept: instead, I need to make an average (arithmetic mean) of the values that have a duplicate key.

Example of the wanted results:

names = ["pears", "apples", "pears", "bananas", "pears"]
result_values = [2, 1, 4, 8, 6] # ints here but it's the same conceptually

combined_result = average_duplicates(names, result_values)

print combined_result

{"pears": 4, "apples": 1, "bananas": 8}

My only ideas involve multiple iterations and so far have been ugly... is there an elegant solution to this problem?

Answer 1

from collections import defaultdict
def averages(names, values):
    # Group the items by name.
    value_lists = defaultdict(list)
    for name, value in zip(names, values):
        value_lists[name].append(value)

    # Take the average of each list.
    result = {}
    for name, values in value_lists.iteritems():
        result[name] = sum(values) / float(len(values))
    return result

names = ["pears", "apples", "pears", "bananas", "pears"]
result_values = [2, 1, 4, 8, 6]
print averages(names, result_values)

Answer 2

I would use a dictionary anyways

averages = {}
counts = {}
for name, value in zip(names, result_values):
    if name in averages:
        averages[name] += value
        counts[name] += 1
    else:
        averages[name] = value
        counts[name] = 1
for name in averages:
    averages[name] = averages[name]/float(counts[name]) 

If you're concerned with large lists, then I would replace zip with izip from itertools.

Answer 3

You could calculate the mean using a Cumulative moving average to only iterate through the lists once:

from collections import defaultdict
averages = defaultdict(float)
count = defaultdict(int)

for name,result in zip(names,result_values):
    count[name] += 1
    averages[name] += (result - averages[name]) / count[name]

Answer 4

I think what you're looking for is itertools.groupby :

import itertools

def average_duplicates(names, values):
  pairs = sorted(zip(names, values))
  result = {}
  for key, group in itertools.groupby(pairs, key=lambda p: p[0]):
    group_values = [value for (_, value) in group]
    result[key] = sum(group_values) / len(group_values)
  return result

See also zip and sorted .

Answer 5

>>> def avg_list(keys, values):
...     def avg(series):
...             return sum(series) / len(series)
...     from collections import defaultdict
...     d = defaultdict(list)
...     for k, v in zip(keys, values):
...             d[k].append(v)
...     return dict((k, avg(v)) for k, v in d.iteritems())
... 
>>> if __name__ == '__main__':
...     names = ["pears", "apples", "pears", "bananas", "pears"]
...     result_values = [2, 1, 4, 8, 6]
...     print avg_list(names, result_values)
... 
{'apples': 1, 'pears': 4, 'bananas': 8}

You can have avg() return float(len(series)) if you want a floating point average.