如何使用python从列表中获取最后一个值？

Question

data = ['reply': '{"osc":{"version":"1.0"}}']
data1 = ['reply':'{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}']

我需要使用python 3.6从data和data1中获取“1.0”值。

我怎样才能做到这一点？

Answer 1

您的数据和数据1在python中是无效的数据类型，因此我将它们转换为有效的字典。

from operator import getitem
data = {'reply': {"osc":{"version":"1.0"}}}
data1 = {'reply':{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}}
def get_item(keys, dict_):
    return reduce(getitem, keys, dict_)
print(get_item(['reply','osc','version'], data))
print(get_item(['reply','device','network', 'ipv4_dante',"auto"],data1))
>>>1.0
   1.0

另一种方法，数据和data1为字符串：

class GetValue:

      def __init__(self, string):
            self.string = string
            self.new_keys  = {}


     def clean_data(self, data):
        if data[2] == '{':
            get_json_data = len(data) - 3
        else:
            get_json_data = len(data) - 1
        modified_data = [val for val in list(data[data.find(':')+1:get_json_data])
                        if val is not "'" ]
        return json.loads(''.join(modified_data))

      def get_recurvise_key(self, data, dict_):
            for key, val in dict_.items():
                self.new_keys.setdefault(data,[]).append(key)
                if isinstance(val, dict):
                    self.get_recurvise_key(data, val)
            return self.new_keys.get(data)

      def get_value(self):
            get_data = self.clean_data(self.string)
            get_keys = self.get_recurvise_key(self.string,get_data)
            value = reduce(getitem, get_keys, get_data)
            return value

data = """['{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}']"""
data1 = """['reply':'{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}']"""

obj_data = GetValue(data)

obj_data1 = GetValue(data1)

print(obj_data.get_value(), obj_data1.get_value())
>>> 1.0 1.0

Answer 2

您可以递归地获得该结果，如：

代码：

def bottom_value(in_data):
    def recurse_to_bottom(a_dict):
        if isinstance(a_dict, dict):
            a_key = list(a_dict)[0]
            return recurse_to_bottom(a_dict[a_key])
        return a_dict

    return recurse_to_bottom(json.loads(in_data['reply']))

测试代码：

import json

data = {'reply': '{"osc":{"version":"1.0"}}'}
data1 = {'reply':'{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}'}

print(bottom_value(data))
print(bottom_value(data1))

结果：

1.0
1.0

Answer 3

修复数据后：

data1 = {'reply':{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}}

只要它仍然是字典，

继续从字典中获取值：

d = data1
while isinstance(d, dict): 
    d = list(d.values())[0]
print(d)
#1.0

Answer 4

您可以尝试正则表达式：

import re
pattern=r'(?<=")[0-9.]+'

data1="""['reply': '{"osc":{"version":"1.0"}}']"""
data2="""['reply':'{"device":{"network":{"ipv4_dante":{"auto":"1.0"}}}}']"""

def find_value(data):
    return re.findall(pattern,data)[0]

输出：

print(find_value(data1))

输出：

1.0

第二

print(find_value(data2))

输出

1.0