[英]shortening key names in a json object using python/bash
这是我编写的函数,它将更改/缩短json对象上的所有键
function replaceKeyWithNewKey(jsonObj, new_keys, old_keys){
console.log("test")
for(i=0;i<jsonObj.length;i++){
for(el in jsonObj){
//console.log(new_keys[el])
jsonObj[i][new_keys[el]]=jsonObj[i][old_keys[el]] //add new key
delete jsonObj[i][old_keys[el]] // delete old key
}
}
return jsonObj
}
我正在寻找的是一种在终端中使用bash或python或其他方式执行此操作的方法,但我想在终端中执行此操作。 因此,我将在file.json上运行脚本,结果将是file2.json,其中file2.json具有较短的键名。 我怎样才能做到这一点?
对于bash,我正在考虑使用sed
但是我认为这不会像替换我不想替换的值那样好。 我知道的python一点点可能是要走的路。
这是控制台打印输出:
>data[0]
Object {Rec_Open_Date: "2016-07-07", MSISDN: 123, IMEI: 223, Data_Volume_Bytes: "673", Device_Manufacturer: "Samsung Korea"…}
>old_keys=Object.keys(data[0])
["Rec_Open_Date", "MSISDN", "IMEI", "Data_Volume_Bytes", "Device_Manufacturer", "Device_Model", "Product_Description", "Data_Volume_MB"]
>new_keys=["r", "m", "i", "d", "f", "l", "s", "d2"]
["r", "m", "i", "d", "f", "l", "s", "d2"]
>function replaceKeyWithNewKey(jsonObj, new_keys, old_keys){
console.log("test")
for(i=0;i<jsonObj.length;i++){
for(el in jsonObj){
//console.log(new_keys[el])
jsonObj[i][new_keys[el]]=jsonObj[i][old_keys[el]] //add new key
delete jsonObj[i][old_keys[el]] // delete old key
}
}
return jsonObj
}
undefined
> replaceKeyWithNewKey(data, new_keys, old_keys)
VM129:2 test
[Objectd: "673"d2: "0.000641823"f: "Samsung Korea"i: 223l: "Samsung GT-I9505"m: 123r: "2016-07-07"s: "PREPAY PLUS - $1 - #33"__proto__: Object, Object, Object, Object, Object, Object, Object, Object, Object]
这是用于测试我的功能的示例数据:
var json = '[{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}]';
假设您知道如何将json加载到python数据结构中。 您提供的示例数据将是dict
的python list
。 我认为最直接的方法是建立一个将旧键映射到新键的字典,然后使用字典理解遍历列表,以从构建的键名映射中使用新的键名重建dict
。 使用样本:
In [8]: jobj = [{"_id":"5078c3a803ff4197dc81fbfb","email":"user1@gmail.com","image":"some_image_url","name":"Name 1"},{"_id":"5078c3a803ff4197dc81fbfc","email":"user2@gmail.com","image":"some_image_url","name":"Name 2"}]
In [9]: keymap = {'email':'e', 'image':'img', 'name':'n', '_id':'id'}
In [10]: for i in range(len(jobj)):
...: jobj[i] = {keymap[k]:jobj[i][k] for k in jobj[i]}
...:
In [11]: jobj
Out[11]:
[{'e': 'user1@gmail.com',
'id': '5078c3a803ff4197dc81fbfb',
'img': 'some_image_url',
'n': 'Name 1'},
{'e': 'user2@gmail.com',
'id': '5078c3a803ff4197dc81fbfc',
'img': 'some_image_url',
'n': 'Name 2'}]
编辑添加
如果要确保所有键都保持顺序,则需要使用collections
OrderedDict
import json
from collections import OrderedDict
keymap = {'email':'e', 'image':'img', 'name':'n', '_id':'id'}
with open('ordered_example.json') as f:
jobj = json.load(f, object_pairs_hook=OrderedDict)
for i in range(len(jobj)):
jobj[i] = OrderedDict((keymap[k],jobj[i][k]) for k in jobj[i])
print(jobj)
输出:
[OrderedDict([('id', '5078c3a803ff4197dc81fbfb'),
('e', 'user1@gmail.com'),
('img', 'some_image_url'),
('n', 'Name 1')]),
OrderedDict([('id', '5078c3a803ff4197dc81fbfc'),
('e', 'user2@gmail.com'),
('img', 'some_image_url'),
('n', 'Name 2')])]
如果要从命令行处理JSON
,建议安装jq
。
您可以将键映射放置在bash 4+中的关联数组中,如下所示:
declare -A map=([foobar]=foo [poohbah]=pooh [zoowicky]=zoo)
并循环遍历以构建一个jq
脚本来替换密钥。
jq_script=
for old in "${!map[@]}"; do
new="${map[$old]}"
jq_script+="${jq_script:+|}if has(\"$old\") then { \"$new\": .[\"$old\"] } + del(.[\"$old\"]) else . end"
done
您可以这样运行:
jq "$jq_script" <old.json >new.json
在我的示例JSON上:
{
"foobar": 1,
"zoowicky": 3,
"different": 4
}
它在地图中没有所有键,并且在地图中没有键,它会产生以下结果:
{
"zoo": 3,
"foo": 1,
"different": 4
}
仅使用sed替代键。
#!/bin/bash
function replaceKeyWithNewKey() {
json_file="$1"
old_key="$2"
new_key="$3"
# you'd better to backup first.
sed -i "" "s/${old_key}/${new_key}/g" ${json_file}
}
old_keys=("Rec_Open_Date" "MSISDN" "IMEI" "Data_Volume_Bytes" "Device_Manufacturer" "Device_Model" "Product_Description" "Data_Volume_MB")
new_keys=("r" "m" "i" "d" "f" "l" "s" "d2")
# input json file
json_file="input"
for ((i = 0; i < ${#old_keys[@]}; ++i)) do
replaceKeyWithNewKey ${json_file} "${old_keys[$i]}" "${new_keys[$i]}"
done
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