对Python中两个配对列表中的重复值求平均值 [英] Average the duplicated values from two paired lists in Python

问题描述

在我的代码中，我从不同的来源获得两个不同的列表，但是我知道它们的顺序相同.第一个列表(名称")包含一个键字符串列表，而第二个列表("result_values")是一系列浮点数.我需要使该对唯一，但是我不能使用字典，因为只会保留最后插入的值:相反，我需要对具有重复键的值进行平均(算术平均值).

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.

所需结果示例:

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}

我唯一的想法涉及多次迭代，到目前为止，它是丑陋的...是否有解决此问题的好方法?

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

推荐答案

我还是会使用字典

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]) 

如果您担心大型列表，那么我将使用itertools中的izip替换zip.

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

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