Question

我在Python中有以下词典列表。

[{"country": "IE", "values": ["Server1-17.6650", "Server3-78.6064", "Server2-3.7286"]}, {"country": "CA", "values": ["Server1-100.0000"]}, {"country": "DE", "values": ["Server2-100.0000"]}, {"country": "JP", "values": ["Server2-100.0000"]}, {"country": "IT", "values": ["Server1-100.0000"]}, {"country": "US", "values": ["Server1-6.3158", "Server3-15.7895", "Server2-77.8947", "Server1-5.5556", "Server3-2.7778", "Server2-91.6667", "Server1-12.6145", "Server3-86.8043", "Server2-0.5811"]}, {"country": "CZ", "values": ["Server1-100.0000"]}, {"country": None, "values": ["Server1-100.0000", "Server2-100.0000", "Server2-100.0000", "Server1-100.0000"]}, {"country": "A", "values": ["Server2-100.0000"]}, {"country": "IL", "values": ["Server1-100.0000"]}, {"country": "BR", "values": ["Server2-100.0000"]}, {"country": "KP", "values": ["Server1-100.0000"]}, {"country": "SG", "values": ["Server1-79.2000", "Server2-20.8000"]}, {"country": "ES", "values": ["Server1-100.0000"]}]

现在对于每个values，如果列表中重复了服务器名称，我必须在服务器的-之后平均它的值。基本上对于上面的列表，最终输出变为。

[{"country": "IE", "values": ["Server1-17.6650", "Server3-78.6064", "Server2-3.7286"]}, {"country": "CA", "values": ["Server1-100.0000"]}, {"country": "DE", "values": ["Server2-100.0000"]}, {"country": "JP", "values": ["Server2-100.0000"]}, {"country": "IT", "values": ["Server1-100.0000"]}, {"country": "US", "values": ["Server1-8.1619", "Server3-35.1238", "Server2-56.7141"]}, {"country": "CZ", "values": ["Server1-100.0000"]}, {"country": None, "values": ["Server1-100.0000", "Server2-100.0000", "Server2-100.0000", "Server1-100.0000"]}, {"country": "AU", "values": ["Server2-100.0000"]}, {"country": "IL", "values": ["Server1-100.0000"]}, {"country": "BR", "values": ["Server2-100.0000"]}, {"country": "KP", "values": ["Server1-100.0000"]}, {"country": "SG", "values": ["Server1-79.2000", "Server2-20.8000"]}, {"country": "ES", "values": ["Server1-100.0000"]}]

我在Python中尝试了以下代码

for key_dict in resp:
    for i, value in enumerate(key_dict['values']):
        for j, new_value in enumerate(key_dict['values']):
            if value[:value.index('-')] == new_value[:new_value.index('-')]:
                key_dict['values'][i] = value[:value.index('-')] + str(float(value[value.index('-'):]) + float(new_value[new_value.index('-'):]))
                del key_dict['values'][j]

但这不会产生我需要的结果。有人可以在python中指出如何做到这一点。

Answer 1

这是对正确的数据结构微不足道的问题之一，没有痛苦的问题。如果values是一个字典映射服务器名称到数字列表，而不是一个大的字符串列表，这将很容易：

如果您可以首先控制值的到达方式，则应该这样做。如果你不能，你可能想手动转换它们。像这样：

for key_dict in resp:
    new_values = {}
    for value in key_dict['values']:
        name, number = value.split('-', 1)
        new_values.setdefault(name, []).append(float(number))
    key_dict['values'] = new_values

现在，平均它们是微不足道的：

for key_dict in resp:
    averages = {}
    for name, numbers in key_dict['values'].items():
        averages[name] = sum(numbers) / len(numbers)
    key_dict['values'] = averages

如果你真的需要在最后把它变成一个字符串，你可以：

for key_dict in resp:
    key_dict['values'] = ['{}-{}'.format(name, value) 
                          for name, value in key_dict['values'].items()]

当然，如果您真的想要，可以将这些内容整合在一起：

for key_dict in resp:
    values = {}
    for value in key_dict['values']:
        name, number = value.split('-', 1)
        values.setdefault(name, []).append(float(number))
    values = ['{}-{}'.format(name, sum(numbers)/len(numbers))
              for name, numbers in values.items()]
    key_dict['values'] = values

Answer 2

您可以在此处使用groupby：

import numpy as np
from itertools import groupby

def average_servers(server_list):

    post_split = [x.split('-') for x in server_list]
    averages = []

    for server, data in groupby(sorted(post_split), lambda x: x[0]):

         cur_average = np.mean([float(x[1]) for x in list(data)])
         averages.append('{}-{}'.format(server, cur_average))

    return averages

然后应用该函数生成一个新的字符串列表，作为values键的值：

for entry in your_data_structure:
    entry['values'] = average_servers(entry['values'])

如何在Python dict中平均输出相同键的值

2 个答案: