化学结构式识别OCR API

简介

用户需要先注册登录，然后才能正常使用。注意使用的时候需要带 token。

示例图片

接口/路由/入口

• 接口名称：化学结构式识别OCR
• 接口地址：https://www.docpartner-serve.com/python/bpInterfaceSmilesModel
• 请求方式：POST
• 响应类型：JSON（同步）
• content-type：application/json

输入说明

{
    "requestData": {
        "queryDataDict": "asdfasf", // base64 图片base64 数据，不需要 data 开头 
    }
}

输出说明

{
    "isSuccessful": True,
    "errorMsg" : "",
    "data" : {
        "respData":{
            "smiles":str
            "deepsmiles":str
        }
    }
}

程序示例

js 示例

// 基于 axios

import axios from "axios"

let token = "TODO" // TODO 

let bpInterfaceSmilesModel = "/bpInterfaceSmilesModel"

let requestJsonData = {
    requestData: {
        queryDataDict: "TODO", // base64 字符串
    }
}

axios.post(
    bpInterfaceSmilesModel, 
    requestJsonData,
    {
        baseURL: 'https://www.docpartner-serve.com/python/', // TODO 
        headers: {
            "Authorization":`Bearer ${token}`,
            "Content-Type":"application/json;charset=utf-8",
            
        },
        responseType: 'json'
    }
)
.then(resp => {
    if (resp.data.isSuccessful) {
        console.log(resp.data.data)
    } else {
        console.log(`请求api失败！${resp.data.errorMsg}`);
    }
})
.catch(err => {
    console.log(err)
})

python 示例

# 基于 requests - 已测试

import os 
import base64
import requests

token = "" # TODO


bp = "bpInterfaceSmilesModel"

url = "https://www.docpartner-serve.com/python/{bp}".format(
    bp=bp
) # 接口地址


imgPath = "test.png" # TODO 
with open(imgPath, "rb") as f:
    imgDataStr = base64.b64encode(f.read()) # --> b''
imgDataBase64 = imgDataStr.decode() # --> str  图片 base64 编码成字符串

jsondata = {
    "requestData": {
        "queryDataDict": imgDataBase64,
    }
}
headers = {
    'Authorization': "Bearer " + token
}
r = requests.post(url,headers=headers,data=None,json=jsondata)

assert type(r.json()) == dict
print(r.json())
# {'data': {'respData': {'deep_smiles': 'C=C[C@@H]C[C@@H]5COS=O)=O)O))))))NC=NC=C5N=CN)N=C6NCCC3', 'scores': 0.6, 'smiles': 'C1=C[C@@H](C[C@@H]1COS(=O)(=O)O)N2C=NC3=C2N=C(N)N=C3NC4CC4'}}, 'errorMsg': '', 'isSuccessful': True}