从元组列表格式化JSON字符串的更多pythonic方法 [英] more pythonic way to format a JSON string from a list of tuples

问题描述

目前我正在这样做:

def getJSONString(lst):
    join = ""
    rs = "{"
    for i in lst:
        rs += join + '"' + str(i[0]) + '":"' + str(i[1]) + '"'
        join = ","
    return rs + "}"

我这样称呼:

rs = getJSONString([("name", "value"), ("name2", "value2")])

它不需要嵌套(它只会是一个简单的名称/值对列表).但是我愿意以不同的方式调用该函数.似乎有点笨拙，有没有更优雅的方式?这需要在2.x下运行.

It doesn't need to be nested (it's only ever going to be a simple list of name value pairs). But I am open to calling the function differently. It all seems a bit cludgy, is there a more elegant way? This needs to run under 2.x.

请注意，这不是以下项的重复项: Python-转换元组列表(除非可以修改答案以创建JSON字符串作为输出).

Note that this is not a duplicate of: Python - convert list of tuples to string (unless that answer can be modified to create a JSON string as output).

将名称值对作为字典传递会更好吗?

edit: would it be better to pass the name value pairs as a dictionary?

推荐答案

有一种更好的方法来生成JSON字符串:

There is a much better way to generate JSON strings: the json module.

import json
rs = json.dumps(dict(lst))

这利用了以下事实:dict()可以采用一系列键-值对(二值元组)并将其转换为映射，json模块直接将其转换为JSON对象结构.

This takes advantage of the fact that dict() can take a sequence of key-value pairs (two-value tuples) and turn that into a mapping, which the json module directly translates to a JSON object structure.

演示:

>>> import json
>>> lst = [("name", "value"), ("name2", "value2")]
>>> rs = json.dumps(dict(lst))
>>> print rs
{"name2": "value2", "name": "value"}

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