python - 如果在列表中找不到 key ，如何获得默认值零？

Question

for url in urls:
            uClient = ureq(url)
            page_html = uClient.read()
            uClient.close()
            soup = BeautifulSoup(page_html, "html.parser")
            text = (''.join(s.findAll(text=True))for s in soup.findAll('p'))
            c = Counter((re.sub(r"[^a-zA-Z0-9 ]","",x)).strip(punctuation).lower() for y in text for x in y.split())
            for key in sorted(c.keys()):
                l.append([key, c[key]])

        d = collections.defaultdict(list)
        for k, v in l:
            d[k].append(v)

        print(d.items())

我得到的输出是:

([('', [3, 9, 4, 1]), ('1', [1, 2, 2]), ('1960', [1]), ('1974', [1]), ('1996', [1]), ('1997', [1]), ('1998', [1]), ('2001', [2]), ('2002', [1]), ...

如果在列表中找不到该键，我想要默认值 0。例如，如果 Key: g 在第一个列表中为 1 次，在第二个列表中为 0，在第三个列表中为 3，在第四个列表中为 6。它应该返回: 'g':[1,0,3,6]

编辑:

这对我的完整代码进行了注释，以显示未成功的试验:

        #m = list(map(dict, map(zip, list_1, list_2)))    
        #matrix = pd.DataFrame.from_dict(d, orient='index')
        matrix = pd.DataFrame({ key:pd.Series(value) for key, value in d.items() })

我有一个名为“urls.txt”的文本文件，其中包含 URL:

https://en.wikipedia.org/wiki/Data_science
https://datajobs.com/what-is-data-science

我需要一个包含所有独特字母数字的文档术语矩阵。让我们说一下数据和科学:
一行应为[文档编号，术语“数据”，术语“科学”]
它应该看起来像:

   data   science
1  96      65
2  105     22
3  0       16

我已经很接近了，但无法以正确的方式做到这一点。尝试了列表到数据框，字典到数据框，纯粹通过数据框，但没有任何效果。到处找了，没有找到类似的东西。

Answer 1

我正在回答我自己的问题，因为我可以找到一种方法，并将其发布在这里，以防有人需要帮助:

import requests
from bs4 import BeautifulSoup
import collections
from string import punctuation
from urllib.request import urlopen as ureq
import re
import pandas as pd
import numpy as np
import operator
Q1= open ("Q1.txt", "w") 
def web_parsing(filename):
    with open (filename, "r") as df:
        urls = df.readlines()
        url_number = 0 
        url_count = []
        l = {} 
        d = []
        a =[]
        b = []
        e=[]
        for url in urls:
            uClient = ureq(url)
            page_html = uClient.read()
            uClient.close()
            soup = BeautifulSoup(page_html, "html.parser")
            text = (''.join(s.findAll(text=True))for s in soup.findAll('p'))
            c = Counter((re.sub(r"[^a-zA-Z0-9 ]","",x)).strip(punctuation).lower() for y in text for x in y.split())
            for key in c.keys():
                if key in a:
                    continue
                else:
                    a.append(key)
            #print(sorted(a))
            a = list(filter(None, a))
            #print(sorted(a))
            stopfile = open('stop_words.txt', 'r')
            stopwords = [line.split(',') for line in stopfile.readlines()]
            #print(stopwords)
            a = [item for item in a if item not in stopwords]
            #print(len(a))
            l = [list(([word, c[word]])) for word in a]
            l =sorted(l)
            flat_list = [item for sublist in l for item in sublist]
            d.extend(flat_list)
            b = {d[i]: d[i+1] for i in range(0, len(d), 2)}
            e.append(b)
        j=0
        for url in urls:
            j = j+1
        #print(j)
        result = {}
        for key in a:
            for i in range(0,j):
                if key in e[i]: result.setdefault(key, []).append(e[i][key])
                if key not in e[i]: result.setdefault(key, []).append(0)
            #print (result)
            #print (result)
        od = collections.OrderedDict(sorted(result.items()))
        #print(od)
        df1 = pd.DataFrame(od)
        df2 =df1.loc[:, ['data', 'companies', 'business', 'action', 'mining', 'science']]
        #return(df2)
        df1.to_csv(Q1, header=True)
        df2.to_csv(Q1,  header=True)        
        print(len(a))
        return(df1)