python按值列出的json排序列表

时间:2014-11-14 07:14:32

标签: python json lambda sorted

我有一个由JSON组成的文件,每行一行,并希望通过update_time对文件进行排序。

示例JSON文件:

{ "page": { "url": "url1", "update_time": "1415387875"}, "other_key": {} }
{ "page": { "url": "url2", "update_time": "1415381963"}, "other_key": {} }
{ "page": { "url": "url3", "update_time": "1415384938"}, "other_key": {} }

想要输出:

{ "page": { "url": "url1", "update_time": "1415387875"}, "other_key": {} }
{ "page": { "url": "url3", "update_time": "1415384938"}, "other_key": {} }
{ "page": { "url": "url2", "update_time": "1415381963"}, "other_key": {} }

我的代码:

#!/bin/env python
#coding: utf8

import sys
import os
import json
import operator

#load json from file
lines = []
while True:
    line = sys.stdin.readline()
    if not line: break
    line = line.strip()
    json_obj = json.loads(line)
    lines.append(json_obj)

#sort json
lines = sorted(lines, key=lambda k: k['page']['update_time'], reverse=True)

#output result
for line in lines:
    print line

代码适用于示例JSON文件,但如果JSON没有' update_time',则会引发KeyError异常。是否有非常规方法可以做到这一点?

4 个答案:

答案 0 :(得分:21)

编写一个使用try...except处理KeyError的函数,然后将其用作key参数而不是lambda。

def extract_time(json):
    try:
        # Also convert to int since update_time will be string.  When comparing
        # strings, "10" is smaller than "2".
        return int(json['page']['update_time'])
    except KeyError:
        return 0

# lines.sort() is more efficient than lines = lines.sorted()
lines.sort(key=extract_time, reverse=True)

答案 1 :(得分:13)

您可以将dict.get()与默认值一起使用:

lines = sorted(lines, key=lambda k: k['page'].get('update_time', 0), reverse=True)

示例:

>>> lines = [
...     {"page": {"url": "url1", "update_time": "1415387875"}, "other_key": {}},
...     {"page": {"url": "url2", "update_time": "1415381963"}, "other_key": {}},
...     {"page": {"url": "url3", "update_time": "1415384938"}, "other_key": {}},
...     {"page": {"url": "url4"}, "other_key": {}},
...     {"page": {"url": "url5"}, "other_key": {}}
... ]
>>> lines = sorted(lines, key=lambda k: k['page'].get('update_time', 0), reverse=True)
>>> for line in lines:
...     print line
... 
{'other_key': {}, 'page': {'url': 'url1', 'update_time': '1415387875'}}
{'other_key': {}, 'page': {'url': 'url3', 'update_time': '1415384938'}}
{'other_key': {}, 'page': {'url': 'url2', 'update_time': '1415381963'}}
{'other_key': {}, 'page': {'url': 'url4'}}
{'other_key': {}, 'page': {'url': 'url5'}}

尽管如此,我仍然会遵循Ferdinand建议的EAFP principle - 这样你也可以处理page密钥丢失的情况。比检查各种角落情况更容易让它失败并处理它。

答案 2 :(得分:8)

# sort json
lines = sorted(lines, key=lambda k: k['page'].get('update_time', 0), reverse=True)

答案 3 :(得分:2)

def get_sortest_key(a: dict, o: dict):
    v = None
    k = None
    for key, value in a.items():
        if v is None:
            v = value
            k = key
            continue
        if v > value:
            v = value
            k = key
    o.update({k: v})
    a.pop(k)
    if a:
        get_sortest_key(a, o)
    else:
        return


def call(o):
    a = {'a': 9, 'b': 1, 'c': 3, 'k': 3, 'l': -1, 's': 100}
    z = get_sortest_key(a, o)
    print(o)


o={}    
call(o)