使用urllib.request

Question

我尝试向Yahoo Finance发出多个链接请求，然后分别返回Income Statement，Balance Sheet和Cash Flow的数组。我发现自己陷入了很多for loops。我想知道下面这段代码是否有更好的版本：

def scrapper(symbol):

    htmls = []
    soup = []
    gen_table = []

    IS = "http://finance.yahoo.com/q/is?s={}+Income+Statement&annual".format(symbol)
    BS = "http://finance.yahoo.com/q/is?s={}+Balance+Sheet&annual".format(symbol)
    CF = "http://finance.yahoo.com/q/is?s={}+Cash+Flow&annual".format(symbol)
    urls = [IS, BS, CF]

    # read each link in urls
    for url in urls:
        with urllib.request.urlopen(url) as response:
            htmls.append(response.read())
    # parse data with BeautifulSoup
    for html in htmls:
        soup.append(BeautifulSoup(html))
    # store income statement, balance sheet and cash flow into soup
    for s in soup:
        gen_table.append(s.find_all("table", class_="yfnc_tabledata1"))

    return gen_table

Answer 1

我可能会这样做：

from bs4 import BeautifulSoup
import urllib

def fetch_table(symbol, table):
    url = "http://finance.yahoo.com/q/is?s={}+{}&annual".format(symbol, table)
    with urllib.request.urlopen(url) as response:
        result = response.read()
    result = BeautifulSoup(result)
    result = result.find_all("table", class_="yfnc_tabledata1")
    return result

def scrapper(symbol):
    return [fetch_table(symbol, table)
            for table in (
                "Income+Statement",
                "Balance+Sheet",
                "Cash+Flow")]

print (scrapper("X"))