Question

我需要一些帮助才能将基本网络刮刀的输出保存为CSV文件。

以下是代码：

from urllib.request import urlopen
from bs4 import BeautifulSoup
import csv

html_ = urlopen("some_url")
bsObj_ = BeautifulSoup(html_, "html.parser")
nameList_ = bsObj_2.findAll("div", {"class":"row proyecto_name_venta"})

for name in nameList_:

    print(name.get_text())

具体来说，我想将name.get_text()结果保存在CSV文件中。

Answer 1

如果nameList_中的元素是由','分隔的列的行，请尝试以下操作：

import csv

with open('out.csv', 'w') as outf:
    writer = csv.writer(outf)
    writer.writerows(name.get_text().split(',') for name nameList_)

如果nameList_.get_text()只是一个字符串，并且您想要编写单个列CSV，则可以尝试这样做：

import csv

with open('out.csv', 'w') as outf:
    writer = csv.writer(outf)
    writer.writerows([name.get_text()] for name in nameList_)

Answer 2

这是你要求的一个非常全面的例子。。。

import urllib2

listOfStocks = ["AAPL", "MSFT", "GOOG", "FB", "AMZN"]

urls = []

for company in listOfStocks:
    urls.append('http://real-chart.finance.yahoo.com/table.csv?s=' + company + '&d=6&e=28&f=2015&g=m&a=11&b=12&c=1980&ignore=.csv')

Output_File = open('C:/Users/rshuell001/Historical_Prices.csv','w')

New_Format_Data = ''

for counter in range(0, len(urls)):

    Original_Data = urllib2.urlopen(urls[counter]).read()

    if counter == 0:
        New_Format_Data = "Company," + urllib2.urlopen(urls[counter]).readline()

    rows = Original_Data.splitlines(1)

    for row in range(1, len(rows)):

        New_Format_Data = New_Format_Data + listOfStocks[counter] + ',' + rows[row]

Output_File.write(New_Format_Data)
Output_File.close()

如何将抓取的内容写入CSV文件？

2 个答案: