Python 检查文本/字符串是否出现预定义的列表元素
我有几个文本文件,我想将它们与由表达式和单个单词组成的词汇表进行比较。所需的输出应该是一个字典,其中包含该列表中的所有元素作为键,它们在文本文件中的各自频率作为值。要构建词汇表,我需要将两个列表匹配在一起Python 检查文本/字符串是否出现预定义的列表元素,python,list,frequency,custom-lists,vocabulary,Python,List,Frequency,Custom Lists,Vocabulary,我有几个文本文件,我想将它们与由表达式和单个单词组成的词汇表进行比较。所需的输出应该是一个字典,其中包含该列表中的所有元素作为键,它们在文本文件中的各自频率作为值。要构建词汇表,我需要将两个列表匹配在一起 list1 = ['accounting',..., 'yields', 'zero-bond'] list2 = ['accounting', 'actual cost', ..., 'zero-bond'] vocabulary_list = ['accounting', 'actual
list1 = ['accounting',..., 'yields', 'zero-bond']
list2 = ['accounting', 'actual cost', ..., 'zero-bond']
vocabulary_list = ['accounting', 'actual cost', ..., 'yields', 'zero-bond']
sample_text = "Accounting experts predict an increase in yields for zero-bond and yields for junk-bonds."
desired_output = ['accounting':1, 'actual cost':0, ..., 'yields':2, 'zero-bond':1]
我尝试的是:
def word_frequency(fileobj, words):
"""Build a Counter of specified words in fileobj"""
# initialise the counter to 0 for each word
ct = Counter(dict((w, 0) for w in words))
file_words = (word for line in fileobj for word in line)
filtered_words = (word for word in file_words if word in words)
return Counter(filtered_words)
def print_summary(filepath, ct):
words = sorted(ct.keys())
counts = [str(ct[k]) for k in words] with open(filepath[:-4] + '_dict' + '.txt', mode = 'w') as outfile:
outfile.write('{0}\n{1}\n{2}\n\n'.format(filepath,', '.join(words),', '.join(counts)))
return outfile
在Python中有什么方法可以做到这一点吗?我想出了如何用单个单词1的词汇表来管理这个问题,但却找不到解决多个单词大小写的方法如果你想考虑用标点符号结尾的单词,你需要清理文本,也就是“收益”和“收益”。p> 要将所有列表链接到一个单字目录中,请执行以下操作:
from collections import Counter
from itertools import chain
import re
c = Counter()
l1,l2 = ['accounting', 'actual cost'], ['yields', 'zero-bond']
vocabulary_dict = {k:0 for k in chain(l1,l2)}
print(vocabulary_dict)
sample_text = "Accounting experts predict actual costs an increase in yields for zero-bond and yields for junk-bonds.".lower()
splitted = sample_text.split()
c.update(splitted)
for k in vocabulary_dict:
spl = k.split()
ln = len(spl)
if ln > 1:
check = re.findall(r'\b{0}\b'.format(k),sample_text)
if check:
vocabulary_dict[k] += len(check)
elif k in sample_text.split():
vocabulary_dict[k] += c[k]
print(vocabulary_dict)
您可以创建两个dict,一个用于短语,另一个用于单词,并对每个dict进行遍历。您的单字解决方案是什么?它在哪些方面不适用于表达式?定义word_frequencyfileobj,words:在fileobj中构建指定单词的计数器将每个单词的计数器初始化为0 ct=Counterdictw,0表示w in words file\u words=word for line in fileobj表示word in line filtered\u words=word for word in file\u words如果words in words返回反过滤的单词def print\u summaryfilepath,ct:words=sortedct dct.keys counts=[strct[k]表示k in words],带openfilepath[:-4]+'.\u dict'+'+'.txt',mode='w'作为outfile:outfile.write'{0}\n{1}\n{2}\n\n'.formatfilepath','.joinwords',','.joincounts返回outfilewords=词汇表\u列表不幸的是,第一个函数只捕获单个标记,因此它只能将这些sinlge标记字与词汇表Nice解决方案Padraic进行比较,但这不适用于这样的示例:示例文本=会计专家…实际成本。。。预测零债券和收益率的收益率增加->实际成本:0,“会计”:1。。。非常感谢Padraic:缺少一点,你的脚本的输出是…'yields':1应该是…'yields':2吗?@DominikScheld,是的,需要将逻辑颠倒一秒现在它工作得很好,你是否也知道如何组合这两个列表以构建一个独特的词汇表?@DominikScheld,添加了一个如何链接列表和创建dict的示例。您还可以有两个dict,一个用于短语,另一个用于单个单词,只需在末尾合并,无需拆分和检查len
from collections import Counter
from itertools import chain
import re
c = Counter()
l1,l2 = ['accounting', 'actual cost'], ['yields', 'zero-bond']
vocabulary_dict = {k:0 for k in chain(l1,l2)}
print(vocabulary_dict)
sample_text = "Accounting experts predict actual costs an increase in yields for zero-bond and yields for junk-bonds.".lower()
splitted = sample_text.split()
c.update(splitted)
for k in vocabulary_dict:
spl = k.split()
ln = len(spl)
if ln > 1:
check = re.findall(r'\b{0}\b'.format(k),sample_text)
if check:
vocabulary_dict[k] += len(check)
elif k in sample_text.split():
vocabulary_dict[k] += c[k]
print(vocabulary_dict)