Question

我写了像这样的代码：

from urllib.request import urlopen as uReq
from bs4 import BeautifulSoup as soup

my_url = 'https://www.pro-football-reference.com/teams/nwe/2013_injuries.htm'

# opening up connection, grabbing the page
uClient = uReq(my_url)
page_html = uClient.read()
uClient.close()

# html parsing
page_soup = soup(page_html, "html.parser")

containers = page_soup.find("tbody")
head = page_soup.find("thead")


player = containers.find_all("tr")

weeks = head.find_all('th')
week_list = [i['data-stat'] for i in weeks][1:]

for week in week_list:
    for tr in player:
        th = tr.find_all("th")
        name = [i.text for i in th]

        td = tr.find('td', {'data-stat':week})
        week = td["data-stat"]

        try:
            injury = td["data-tip"]
            print([name, injury, week])
        except KeyError:
            injury = "NA"
            print([name, injury, week])

我得到的输出是：

enter image description here

，即所有数据均按列显示。但是我想按行显示数据。即每个条目分别

like;

import pandas as pd
import numpy as np
import json
from flask import Flask,request,jsonify
app = Flask(__name__)

@app.route('/df',methods=['POST','GET'])
def ff():
	df = pd.read_csv(r'dataframe_post.csv')
	row = [5, 'Sanjeev', 'AE']
	df.loc[len(df)] = row
	# print(dfs)
	ls=list(df.to_dict().values())
	return jsonify(ls)
if __name__ == '__main__':
    app.run(debug=True)

，依此类推。

Answer 1

在您的df.to_dict调用中，使用to_dict(orient='records')将按行构建json

Answer 2

要以所需的格式返回json，可以使用内置的dataframe方法，而不是列出并进行jsonifying：

df.to_json(orient="records")

这将为您提供一个json编码的字符串，如下例所示：

df = pd.DataFrame([[5, 'Sanjeev', 'AE'], [6, 'Sven', 'AA']], columns = ["id", "name", "2"])

哪个返回：

   id     name   2
0   5  Sanjeev  AE
1   6     Sven  AA

然后作为JSON：

df.to_json(orient="records")

'[{"id":5,"name":"Sanjeev","2":"AE"},{"id":6,"name":"Sven","2":"AA"}]'

如何使用flask按行从csv文件中获取json

2 个答案: