返回 JSON 列表和 PYTHON 请求库答案

【问题标题】：return JSON list and PYTHON requests library返回 JSON 列表和 PYTHON 请求库
【发布时间】：2018-12-29 17:00:04
【问题描述】：

我正在使用 python 请求模块从 3 个不同的服务器获取 JSON 响应。 2 JSON 响应如下所示：

JSON 响应 1：

{"MaleName1":"John","MaleAge1":"1.40531900","FemaleName1":"Anna","FemaleAge1":"14"}

JSON 响应 2：

{"male":[{"name":"John","age":"12"}],"female":[{"name":"Anna","age":"14"}]}

JSON 响应 3：

{"male":[["John","12",[]],["Alex","13",[]],["Glenn","12",[]],["Patrick","14",[]],["Gerard","14",[]]],"female":[["Anna","14",[]],["Lena","12",[]],["Martha","13",[]],["Penelope","13",[]],["Brenda","13",[]]]}

我的问题是解析第二个和第三个 JSON 响应的正确方法是什么，以便我可以打印以下所需值：

1st Male Name: John
1st Male Age: 12
1st Female Name: Anna
1st Female Age: 14

对于第一个 JSON 响应，使用下面的 float() 参数获得所需的响应没有问题：

import json, requests

def 1stMaleName():
    1stMaleNameData = requests.get('url')
    return 1stMaleNameData.json()['MaleName1']
    1stMaleNameValue = float(1stMaleName())
    Print ("!st Male Name: ", 1stMaleNameValue)

【问题讨论】：

既然您需要的所有信息都在第二个响应中，为什么您需要解析第三个响应？
@AChampion 因为第二和第三响应将来自不同的来源，并且值可能彼此不同，所以我需要从两个来源获取数据以进行一些验证。

标签： python arrays json python-3.x

【解决方案1】：

对于您的第二个 JSON 查询： 1) 遍历字典

2) 遍历列表

3) 遍历列表中的字典

这将使您更接近您正在搜索的内容。

请查看example。

【讨论】：

【解决方案2】：

第二个响应可以解析为：

In []:
ordinal = lambda n: {1:'st', 2:'nd', 3:'rd'}.get(n%10, 'th')

d = {"male":[{"name":"John","age":"12"}],"female":[{"name":"Anna","age":"14"}]}
for k, v in d.items():
    for i, s in enumerate(v, 1):
        print(f"{i}{ordinal(i)} {k.capitalize()} Name: {s['name']}")
        print(f"{i}{ordinal(i)} {k.capitalize()} Age: {s['age']}")

Out[]:
1st Male Name: John
1st Male Age: 12
1st Female Name: Anna
1st Female Age: 14

第三个回应是：

In []:
d = {"male":[["John","12",[]],["Alex","13",[]],["Glenn","12",[]],["Patrick","14",[]],["Gerard","14",[]]],"female":[["Anna","14",[]],["Lena","12",[]],["Martha","13",[]],["Penelope","13",[]],["Brenda","13",[]]]}
for k, v in d.items():
    for i, s in enumerate(v, 1):
        print(f"{i}{ordinal(i)} {k.capitalize()} Name: {s[0]}")
        print(f"{i}{ordinal(i)} {k.capitalize()} Age: {s[1]}")

Out[]:
1st Male Name: John
1st Male Age: 12
2nd Male Name: Alex
2nd Male Age: 13
3rd Male Name: Glenn
3rd Male Age: 12
4th Male Name: Patrick
4th Male Age: 14
5th Male Name: Gerard
5th Male Age: 14
1st Female Name: Anna
1st Female Age: 14
2nd Female Name: Lena
2nd Female Age: 12
3rd Female Name: Martha
3rd Female Age: 13
4th Female Name: Penelope
4th Female Age: 13
5th Female Name: Brenda
5th Female Age: 13

注意：这个f-strings 在 Py3.6+ 中工作。对于 .format()，例如：

f"{i}{ordinal(i)} {k.capitalize()} Name: {s['name']}"

相当于：

"{}{} {} Name: {}".format(i, ordinal(i), k.capitalize(), s['name'])

【讨论】：