Question

我有一个输入文件，而我必须根据2个空白的新行提取几行。

例如：文本文件如下所示。

1. Sometext
Sometext 
Sometext

2. Sometext
Sometext
Sometext

3. Sometext
Sometext
Sometext

Sometext which is not needed
Sometext which is not needed
Sometext which is not needed

我必须从“1”中提取一个子串。在“2.”之前的所有人和来自“2”的第二个子串在“3.”之前的所有人等等基于数字。我有下面的脚本获取输出，但它也获得了我不想要的所有“不需要的文本”。请参阅以下代码：

file_path = open("filename", "r")
content = file_path.read()
size1 = len(content)
start =0
a=1
b=2
end =0
ext =0   

while (start<size):
   if (end !=-1):
   subString = content[content.find(str(a)+".")+0:content.find("\n"+str(b)+".")] 
   print (subString)
   end = content.find(str(b)+".",start)
                print ("\n")
                a = int(a)+1 # increment to find the next start number
                b = int(b)+1 # increment to find the next end number
                start = end+1 # continuing to search the next
            else:
                break

所以，我决定为最终位置找到2个连续的空白行，并使用下面的空白行，但是没有用。

subString = content[content.find (str(a)+".")+3:content.find("\n\n")]

如果您有任何疑问，请帮忙告诉我。提前谢谢。

Answer 1

我不确定我是否正确理解了您的问题，但以下是输出的代码：

['Sometext', 'Sometext', 'Sometext']
['Sometext', 'Sometext', 'Sometext']
['Sometext', 'Sometext', 'Sometext']

根据您问题中的文字。相反，你想要1到2是这样的整个子串：

['1. Sometext\nSometext\nSometext']
['2. Sometext\nSometext\nSometext']
['3. Sometext\nSometext\nSometext']

您应该将if语句更改为：

if is_number(i[0]):
            substring = []
            substring.append(i)
            print(substring)

否则你可以使用下面的代码

def is_number(string):
    try:
        float(string)
        return True
    except ValueError:
        return False

with open('testing.txt', 'r') as f:
content = f.read().split('\n\n')
for i in content:
    if is_number(i[0]):
        c = i.split('\n')
        substring = [line[3:] if is_number(line[0]) else line for line in c]
        print(substring)

Answer 2

您必须在结尾处过滤掉不需要的行，但这会让您想要：

from itertools import groupby
with open("in.txt") as f:
    grps = groupby(f, key=lambda x: bool(x.strip()))
    print([list(v) for k,v in grps if k])

输出：

[['1. Sometext\n', 'Sometext\n', 'Sometext\n'], ['2. Sometext\n', 'Sometext\n', 'Sometext\n'], ['3. Sometext\n', 'Sometext\n', 'Sometext\n'], ['Sometext which is not needed\n', 'Sometext which is not needed\n', 'Sometext which is not needed']]

由于您要保留的所有部分都以数字开头：

from itertools import groupby, takewhile

with open("in.txt") as f:
    grps = groupby(f, key=lambda x: bool(x.strip()))
    print (list(takewhile(lambda x: x[0][0].isdigit(),(list(v) for k,v in grps if k))))

输出：

[['1. Sometext\n', 'Sometext\n', 'Sometext\n'],
 ['2. Sometext\n', 'Sometext\n', 'Sometext\n'],
['3. Sometext\n', 'Sometext\n', 'Sometext\n']]

如果您知道有n个组可以切片：

from itertools import groupby, islice
with open("in.txt") as f:
    grps = groupby(f, key=lambda x: bool(x.strip()))
    print (list(islice((list(v) for k,v in grps if k),3)))

输出：

[['1. Sometext\n', 'Sometext\n', 'Sometext\n'],
 ['2. Sometext\n', 'Sometext\n', 'Sometext\n'], 
['3. Sometext\n', 'Sometext\n', 'Sometext\n']]

Python - 从文本文件中提取字符串，直到前2个新行空间

2 个答案: