[英]Using directory as input for tf-idf with python `textblob`
我正在嘗試修改此代碼(在此處找到源代碼)以遍歷文件目錄,而不是對輸入進行硬編碼。
#!/usr/bin/python
# -*- coding: utf-8 -*-
from __future__ import division, unicode_literals
import math
from textblob import TextBlob as tb
def tf(word, blob):
return blob.words.count(word) / len(blob.words)
def n_containing(word, bloblist):
return sum(1 for blob in bloblist if word in blob)
def idf(word, bloblist):
return math.log(len(bloblist) / (1 + n_containing(word, bloblist)))
def tfidf(word, blob, bloblist):
return tf(word, blob) * idf(word, bloblist)
document1 = tb("""Today, the weather is 30 degrees in Celcius. It is really hot""")
document2 = tb("""I can't believe the traffic headed to the beach. It is really a circus out there.'""")
document3 = tb("""There are so many tolls on this road. I recommend taking the interstate.""")
bloblist = [document1, document2, document3]
for i, blob in enumerate(bloblist):
print("Document {}".format(i + 1))
scores = {word: tfidf(word, blob, bloblist) for word in blob.words}
sorted_words = sorted(scores.items(), key=lambda x: x[1], reverse=True)
for word, score in sorted_words:
score_weight = score * 100
print("\t{}, {}".format(word, round(score_weight, 5)))
我想在目錄中使用輸入txt文件,而不是每個硬編碼的document
。
例如,假設我有一個目錄foo
,其中包含三個文件file1
, file2
和file3
。
文件1包含document1
包含的內容,即
文件1:
Today, the weather is 30 degrees in Celcius. It is really hot
文件2包含document2
包含的內容,即
I can't believe the traffic headed to the beach. It is really a circus out there.
文件3包含document3
包含的內容,即
There are so many tolls on this road. I recommend taking the interstate.
我必須使用glob
來達到我想要的結果,並且我想出了以下代碼修改,它可以正確識別文件,但不會像原始代碼那樣單獨處理它們:
file_names = glob.glob("/path/to/foo/*")
files = map(open,file_names)
documents = [file.read() for file in files]
[file.close() for file in files]
bloblist = [documents]
for i, blob in enumerate(bloblist):
print("Document {}".format(i + 1))
scores = {word: tfidf(word, blob, bloblist) for word in blob.words}
sorted_words = sorted(scores.items(), key=lambda x: x[1], reverse=True)
for word, score in sorted_words:
score_weight = score * 100
print("\t{}, {}".format(word, round(score_weight, 5)))
如何使用glob
維護每個文件的分數?
將目錄中的文件用作輸入后,期望的結果將與原始代碼相同(結果被截斷為前3個空格):
Document 1
Celcius, 3.37888
30, 3.37888
hot, 3.37888
Document 2
there, 2.38509
out, 2.38509
headed, 2.38509
Document 3
on, 3.11896
this, 3.11896
many, 3.11896
這里的類似問題並未完全解決問題。 我想知道如何調用文件來計算idf
但分別維護它們以計算完整的tf-idf
?
@AnnaBonazzi在此處提供了代碼段, https: //gist.github.com/sloria/6407257,
import os, glob
folder = "/path/to/folder/"
os.chdir(folder)
files = glob.glob("*.txt") # Makes a list of all files in folder
bloblist = []
for file1 in files:
with open (file1, 'r') as f:
data = f.read() # Reads document content into a string
document = tb(data.decode("utf-8")) # Makes TextBlob object
bloblist.append(document)
我對其進行了修改(Python 3):
import os, glob
bloblist = []
def make_corpus(input_dir):
""" Based on code snippet from https://gist.github.com/sloria/6407257 """
global doc ## used outside this method
input_folder = "input"
os.chdir(input_folder)
files = glob.glob("*.*") ## or "*.txt", etc.
for doc in files:
# print('doc:', doc) ## prints filename (doc)
with open (doc, 'r') as f:
data = f.read() ## read document content into a string
document = tb(data) ## make TextBlob object
bloblist.append(document)
# print('bloblist:\n', bloblist) ## copious output ...
print('len(bloblist):', len(bloblist))
make_corpus('input') ## input directory 'input'
更新1:
我個人除了使用Python glob模塊外沒有其他困難,因為我經常(i)文件名不帶擴展名(例如01),並且(ii)想遞歸嵌套目錄。
乍一看,“全局”方法似乎是一個簡單的解決方案。 但是,當嘗試遍歷glob返回的文件時,我經常遇到錯誤(例如)
IsADirectoryError: [Errno 21] Is a directory: ...
當循環遇到glob返回的目錄(而不是文件)名稱時。
我認為,只需付出一點點努力,以下方法就會更加健壯:
import os
bloblist = []
def make_corpus(input_dir):
for root, subdirs, files in os.walk(input_dir):
for filename in files:
f = os.path.join(root, filename)
print('file:', f)
with open(os.path.join(root, filename)) as f:
for line in f:
# print(line, end='')
bloblist.append(line)
# print('bloblist:\n', bloblist)
print('len(bloblist):', len(bloblist), '\n')
make_corpus('input') ## 'input' = input dir
更新2:
最后一種方法(Linux shell find
命令,適用於Python 3):
import sh ## pip install sh
def make_corpus(input_dir):
'''find (here) matches filenames, excludes directory names'''
corpus = []
file_list = []
#FILES = sh.find(input_dir, '-type', 'f', '-iname', '*.txt') ## find all .txt files
FILES = sh.find(input_dir, '-type', 'f', '-iname', '*') ## find any file
print('FILES:', FILES) ## caveat: files in FILES are '\n'-terminated ...
for filename in FILES:
#print(filename, end='')
# file_list.append(filename) ## when printed, each filename ends with '\n'
filename = filename.rstrip('\n') ## ... this addresses that issue
file_list.append(filename)
with open(filename) as f:
#print('file:', filename)
# ----------------------------------------
# for general use:
#for line in f:
#print(line)
#corpus.append(line)
# ----------------------------------------
# for this particular example (Question, above):
data = f.read()
document = tb(data)
corpus.append(document)
print('file_list:', file_list)
print('corpus length (lines):', len(corpus))
with open('output/corpus', 'w') as f: ## write to file
for line in corpus:
f.write(line)
在第一個代碼示例中,用tb()
結果填充bloblist
,在第二個示例中,用tb()
輸入(僅字符串)填充。
嘗試將bloblist = [documents]
替換為bloblist = map(tb, documents)
。
您還可以像這樣對文件名列表file_names = sorted(glob.glob("/path/to/foo/*"))
以使兩個版本的輸出匹配。
我不確定您要實現的目標到底是什么。 您可以有一個數組並將結果附加到該數組:
scores = []
bloblist = [documents]
for i, blob in enumerate(bloblist):
... do your evaluation ..
scores.append(score_weight)
print scores
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