python - 通过 readlines(size) 提高大文件搜索的效率

Question

我是 Python 新手，目前正在使用 Python 2。我有一些源文件，每个文件都包含大量数据（大约 1900 万行）。如下所示：

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，每次在语料库中找到目标词时，都必须将这个词之前和之后的 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。代码做了它应该做的事情，但对于源文件中提供的数据量来说，它显然不够高效。

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，逐行阅读文件，我没有设法使用 enumerate 方法来获取目标单词前后的 10 个单词。 我也无法使用 mmap，因为我没有在该文件上使用它的权限。

所以，我想具有一定大小限制的 readlines 方法将是最有效的解决方案。但是，为此，我会不会犯一些错误，因为每次到达大小限制的末尾时，由于代码刚刚中断，目标单词之后的 10 个单词不会被捕获？

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)

出于完整性的原因，我复制了整个代码部分，因为它是一个函数的一部分，该函数从提取的所有信息中创建一个多维字典，然后将其写入 csv 文件。

如果有任何提示或建议可以使这段代码更高效，我将不胜感激。

编辑我更正了代码，以便它考虑到目标单词前后的确切 10 个单词

Answer 1

我的想法是创建一个缓冲区在 10 行之前存储，另一个缓冲区在 10 行之后存储，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，我从文件迭代器 1st 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 次迭代以获得后面的 10 行。

这避免了使用 readlines() 并将整个文件加载到内存中。 希望它在实际情况下对你有用

已编辑： 如果第 3 列不包含“&”、“\”、“（未知）”中的任何一个，则仅填充前后缓冲区。还将 split('\t') 更改为 split() 以便处理所有空格或标签

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

Answer 2

用 Python 3.5 编写的功能替代方案。我简化了你的例子，两边只取 5 个字。关于垃圾值过滤还有其他简化，但只需要稍作修改。我将使用 PyPI 中的包 fn 使这个功能代码更易于阅读。

from typing import List, Tuple
from itertools import groupby, filterfalse
from fn import F

首先我们需要提取列：

def getcol3(line: str) -> str:
    return line.split("\t")[2]

然后我们需要将行拆分为由谓词分隔的块：

TARGET_WORDS = {"target1", "target2"}

# this is out predicate
def istarget(word: str) -> bool:
    return word in TARGET_WORDS        

让我们过滤垃圾并编写一个函数来获取最后和前 5 个单词：

def isjunk(word: str) -> bool:
    return word == "(unknown)"

def first_and_last(words: List[str]) -> (List[str], List[str]):
    first = words[:5]
    last = words[-5:]
    return first, last

现在，让我们获取组：

words = (F() >> (map, str.strip) >> (filter, bool) >> (map, getcol3) >> (filterfalse, isjunk))(lines)
groups = groupby(words, istarget)

现在，处理组

def is_target_group(group: Tuple[str, List[str]]) -> bool:
    return istarget(group[0])

def unpack_word_group(group: Tuple[str, List[str]]) -> List[str]:
    return [*group[1]]

def unpack_target_group(group: Tuple[str, List[str]]) -> List[str]:
    return [group[0]]

def process_group(group: Tuple[str, List[str]]):
    return (unpack_target_group(group) if is_target_group(group) 
            else first_and_last(unpack_word_group(group)))

最后的步骤是：

words = list(map(process_group, groups))

附言

这是我的测试用例：

from io import StringIO

buffer = """
_\t_\tword
_\t_\tword
_\t_\tword
_\t_\t(unknown)
_\t_\tword
_\t_\tword
_\t_\ttarget1
_\t_\tword
_\t_\t(unknown)
_\t_\tword
_\t_\tword
_\t_\tword
_\t_\ttarget2
_\t_\tword
_\t_\t(unknown)
_\t_\tword
_\t_\tword
_\t_\tword
_\t_\t(unknown)
_\t_\tword
_\t_\tword
_\t_\ttarget1
_\t_\tword
_\t_\t(unknown)
_\t_\tword
_\t_\tword
_\t_\tword
"""

# this simulates an opened file
lines = StringIO(buffer)

给定这个文件，你会得到这个输出：

[(['word', 'word', 'word', 'word', 'word'],
  ['word', 'word', 'word', 'word', 'word']),
 (['target1'], ['target1']),
 (['word', 'word', 'word', 'word'], ['word', 'word', 'word', 'word']),
 (['target2'], ['target2']),
 (['word', 'word', 'word', 'word', 'word'],
  ['word', 'word', 'word', 'word', 'word']),
 (['target1'], ['target1']),
 (['word', 'word', 'word', 'word'], ['word', 'word', 'word', 'word'])]

您可以从这里删除前 5 个单词和最后 5 个单词。