如果在列表中找不到键，如何获取默认值零？

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]), ...

如果列表中找不到键，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：

这注释了我的完整代码中的行，以显示没有成功的试验：

#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"的文本文件包含 URL：

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

我需要所有唯一字母数字的文档术语矩阵。假设单词数据和科学：
一行应该是[Document number, term 'data', term 'science']
它应该看起来像：

data   science
1  96      65
2  105     22
3  0       16

我非常接近，但无法以正确的方式做到这一点。尝试列表到数据帧，字典到数据帧，纯粹通过数据帧，但没有任何效果。到处搜索，找不到类似的东西。

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在此处发布，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)