python sort list of json by value

Question

I have a file consists of JSON, each a line, and want to sort the file by update_time reversed.

sample JSON file:

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

want output:

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

my code:

#!/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 

The code works fine with sample JSON file, but if a JSON has no 'update_time', it will raise KeyError exception. Are there non-exception ways to do this?

Answer 1

Write a function that uses try...except to handle the KeyError, then use this as the key argument instead of your 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)