dan*_*man 7 python dictionary enumerate multidimensional-array readlines
我是Python新手,我目前正在使用Python 2.我有一些源文件,每个源文件都包含大量数据(大约1900万行).它看起来如下:
apple \t N \t apple
n&apos
garden \t N \t garden
b\ta\md
great \t Adj \t great
nice \t Adj \t (unknown)
etc
我的任务是在每个文件的第3列搜索一些目标词,并且每次在语料库中找到目标词时,必须将该词前后的10个词添加到多维词典中.
编辑:应排除包含'&','\'或字符串'(未知)'的行.
我尝试使用readlines()和enumerate()来解决这个问题,如下面的代码所示.代码执行它应该做的事情但显然对源文件中提供的数据量不够高效.
我知道readlines()或read()不应该用于大型数据集,因为它将整个文件加载到内存中.然而,逐行读取文件,我没有设法使用枚举方法来获取目标词之前和之后的10个单词.我也不能使用mmap,因为我没有权限在该文件上使用它.
所以,我认为具有一定大小限制的readlines方法将是最有效的解决方案.然而,为此,我不会做出一些错误,因为每次达到大小限制结束时,目标字不会被捕获,因为代码刚刚破坏了10个字？
def get_target_to_dict(file):
targets_dict = {}
with open(file) as f:
for line in f:
targets_dict[line.strip()] = {}
return targets_dict
targets_dict = get_target_to_dict('targets_uniq.txt')
# browse directory and process each file
# find the target words to include the 10 words before and after to the dictionary
# exclude lines starting with <,-,; to just have raw text
def get_co_occurence(path_file_dir, targets, results):
lines = []
for file in os.listdir(path_file_dir):
if file.startswith('corpus'):
path_file = os.path.join(path_file_dir, file)
with gzip.open(path_file) as corpusfile:
# PROBLEMATIC CODE HERE
# lines = corpusfile.readlines()
for line in corpusfile:
if re.match('[A-Z]|[a-z]', line):
if '(unknown)' in line:
continue
elif '\\' in line:
continue
elif '&' in line:
continue
lines.append(line)
for i, line in enumerate(lines):
line = line.strip()
if re.match('[A-Z][a-z]', line):
parts = line.split('\t')
lemma = parts[2]
if lemma in targets:
pos = parts[1]
if pos not in targets[lemma]:
targets[lemma][pos] = {}
counts = targets[lemma][pos]
context = []
# look at 10 previous lines
for j in range(max(0, i-10), i):
context.append(lines[j])
# look at the next 10 lines
for j in range(i+1, min(i+11, len(lines))):
context.append(lines[j])
# END OF PROBLEMATIC CODE
for context_line in context:
context_line = context_line.strip()
parts_context = context_line.split('\t')
context_lemma = parts_context[2]
if context_lemma not in counts:
counts[context_lemma] = {}
context_pos = parts_context[1]
if context_pos not in counts[context_lemma]:
counts[context_lemma][context_pos] = 0
counts[context_lemma][context_pos] += 1
csvwriter = csv.writer(results, delimiter='\t')
for k,v in targets.iteritems():
for k2,v2 in v.iteritems():
for k3,v3 in v2.iteritems():
for k4,v4 in v3.iteritems():
csvwriter.writerow([str(k), str(k2), str(k3), str(k4), str(v4)])
#print(str(k) + "\t" + str(k2) + "\t" + str(k3) + "\t" + str(k4) + "\t" + str(v4))
results = open('results_corpus.csv', 'wb')
word_occurrence = get_co_occurence(path_file_dir, targets_dict, results)
我复制整个代码的部分是出于完整性的原因,因为它是一个函数的一部分,它从所有提取的信息中创建一个多维字典,然后将其写入csv文件.
我真的很感激任何提示或建议使这个代码更有效.
编辑我更正了代码,因此它考虑了目标词之前和之后的确切10个单词
我的想法是创建一个缓冲区来存储 10 行之前的内容，另一个缓冲区来存储 10 行之后的内容，当文件被读取时，它将被推入缓冲区之前，如果大小超过 10，缓冲区将被弹出
对于后缓冲区，我从第一个文件迭代器克隆另一个迭代器。然后在循环内并行运行两个迭代器，克隆迭代器提前运行 10 次迭代以获得后 10 行。
这可以避免使用 readlines() 并将整个文件加载到内存中。希望它对您的实际情况有用
编辑： 如果第 3 列不包含任何“&”、“\”、“(未知)”，则仅填充前后缓冲区。还将 split('\t') 更改为 split()，这样它将照顾所有空格或制表符
import itertools
def get_co_occurence(path_file_dir, targets, results):
excluded_words = ['&', '\\', '(unknown)'] # modify excluded words here
for file in os.listdir(path_file_dir):
if file.startswith('testset'):
path_file = os.path.join(path_file_dir, file)
with open(path_file) as corpusfile:
# CHANGED CODE HERE
before_buf = [] # buffer to store before 10 lines
after_buf = [] # buffer to store after 10 lines
corpusfile, corpusfile_clone = itertools.tee(corpusfile) # clone file iterator to access next 10 lines
for line in corpusfile:
line = line.strip()
if re.match('[A-Z]|[a-z]', line):
parts = line.split()
lemma = parts[2]
# before buffer handling, fill buffer excluded line contains any of excluded words
if not any(w in line for w in excluded_words):
before_buf.append(line) # append to before buffer
if len(before_buf)>11:
before_buf.pop(0) # keep the buffer at size 10
# next buffer handling
while len(after_buf)<=10:
try:
after = next(corpusfile_clone) # advance 1 iterator
after_lemma = ''
after_tmp = after.split()
if re.match('[A-Z]|[a-z]', after) and len(after_tmp)>2:
after_lemma = after_tmp[2]
except StopIteration:
break # copy iterator will exhaust 1st coz its 10 iteration ahead
if after_lemma and not any(w in after for w in excluded_words):
after_buf.append(after) # append to buffer
# print 'after',z,after, ' - ',after_lemma
if (after_buf and line in after_buf[0]):
after_buf.pop(0) # pop off one ready for next
if lemma in targets:
pos = parts[1]
if pos not in targets[lemma]:
targets[lemma][pos] = {}
counts = targets[lemma][pos]
# context = []
# look at 10 previous lines
context= before_buf[:-1] # minus out current line
# look at the next 10 lines
context.extend(after_buf)
# END OF CHANGED CODE
# CONTINUE YOUR STUFF HERE WITH CONTEXT