Question

我正在尝试确定是否可以访问当前所在元素周围的列表元素。我有一个很大的列表（超过20k行），我想查找字符串“名称”的每个实例。另外，我还希望每个“名称”元素周围有+/- 5个元素。因此，前5行和后5行。我正在使用的代码如下。

search_string = 'Name'

with open('test.txt', 'r') as infile, open ('textOut.txt','w') as outfile:
    for line in infile:
        if search_string in line:
            outfile.writelines([line, next(infile), next(infile), 
                               next(infile), next(infile), next(infile)])

在'Name'出现之后获取行很简单，但是在弄乱我之前弄清楚如何访问元素。有人有想法吗？

Answer 1

2万行不是太多，如果可以读取列表中的所有行，我们可以在找到匹配项的索引周围进行切片，如下所示：

with open('test.txt', 'r') as infile, open('textOut.txt','w') as outfile:
    lines = [line.strip() for line in infile.readlines()]
    n = len(lines)
    for i in range(n):
        if search_string in lines[i]:
            start = max(0, i - 5)
            end = min(n, i + 6)
            outfile.writelines(lines[start:end])

Answer 2

您可以使用函数enumerate来迭代元素和索引。

访问当前元素之前和之后的5个索引元素的示例：

GenericForeignKey

Answer 3

您需要跟踪列表中当前位置的索引

类似这样：

# Read the file into list_of_lines
index = 0
while index < len(list_of_lines):
    if list_of_lines[index] == 'Name':
        print(list_of_lines[index - 1]) # This is the previous line
        print(list_of_lines[index + 1]) # This is the next line
        # And so on...
    index += 1

Answer 4

假设您将行存储在列表中

lines  = ['line1', 'line2', 'line3', 'line4', 'line5', 'line6', 'line7', 'line8', 'line9']

您可以定义一个方法来返回按n个连续的元素分组的元素作为生成器：

def each_cons(iterable, n = 2):
  if n < 2: n = 1
  i, size = 0, len(iterable)
  while i < size-n+1:
    yield iterable[i:i+n]
    i += 1

青少年，只需调用该方法即可。要显示我正在呼叫的内容列表，但您可以对其进行迭代：

lines_by_3_cons = each_cons(lines, 3) # or any number of lines, 5 in your case
print(list(lines_by_3_cons))

#=> [['line1', 'line2', 'line3'], ['line2', 'line3', 'line4'], ['line3', 'line4', 'line5'], ['line4', 'line5', 'line6'], ['line5', 'line6', 'line7'], ['line6', 'line7', 'line8'], ['line7', 'line8', 'line9']]

Answer 5

我个人很喜欢这个问题。这里的所有人都是通过将整个文件存储到内存中来完成此操作的。我想我写了一个内存有效的代码。在这里，检查一下！

myfile = open('infile.txt')

stack_print_moments = []
expression = 'MYEXPRESSION'
neighbourhood_size = 5

def print_stack(stack):
    for line in stack:
        print(line.strip())
    print('-----')


current_stack = []
for index, line in enumerate(myfile):
    current_stack.append(line)
    if len(current_stack) > 2 * neighbourhood_size + 1:
        current_stack.pop(0)

    if expression in line:
        stack_print_moments.append(index + neighbourhood_size)

    if index in stack_print_moments:
        print_stack(current_stack)

last_index = index

for index in range(last_index, last_index + neighbourhood_size + 1):
    if index in stack_print_moments:
        print_stack(current_stack)
    current_stack.pop(0)

更多高级代码在这里：Github link

访问当前元素周围的列表元素？

5 个答案: