[文本处理] 需要提取文本连续数字

ai8866

00001
00002
00003
00004
00008
00010
00013
00014
00015
00016
00021
00022
00023
00026
00030
00031
00035
00036
00049
00060
00061
00062
00065


提取出连续的4位数字
提取后的文本样子如下

00001
00002
00003
00004


00013
00014
00015
00016


00021
00022
00023
00024

went

数据保存为1.txt
for读取每一行,然后再一路判断下去

@echo off
setlocal enabledelayedexpansion
cd.>"out.txt"
(
for /f %%i in (1.txt) do (
	if "!v1!"=="" (
		set "v1=%%i"
	) else (
		if "!v2!"=="" (
			set /a "1/(1!v1!+1-1%%i)" >nul 2>nul && ( set "v1=%%i" ) || set "v2=%%i" 
		) else (
			if "!v3!"=="" (
				set /a "1/(1!v2!+1-1%%i)" >nul 2>nul && ( set "v1=%%i" & set "v2=" ) || set "v3=%%i" 
			) else (
				if "!v4!"=="" (
					set /a "1/(1!v3!+1-1%%i)" >nul 2>nul && ( set "v1=%%i" & set "v2=" & set "v3=" ) || set "v4=%%i" 
				) 
			)
		)
	)
	if not "!v4!"=="" (
		echo !v1!
		echo !v2!
		echo !v3!
		echo !v4!
		echo.
		set "v1=" & set "v2=" & set "v3=" & set "v4="
	)
)
)>"out.txt"
type "out.txt"
pause&exit
复制代码

ai8866

十分感谢大神

WHY

@echo off
setlocal enabledelayedexpansion
for /f %%i in (a.txt) do (
    set "s1=!s2!"
    set "s2=!s3!"
    set "s3=!s4!"
    set "s4=%%i"
    if defined s1 (
        set /a n1=1!s1! + 3, n2=1!s2! + 2, n3=1!s3! + 1, n4=1!s4!
        if "!n1!" == "!n2!" if "!n2!" == "!n3!" if "!n3!" == "!n4!" (
            echo;!s1!
            echo;!s2!
            echo;!s3!
            echo;!s4!
            Rem endlocal & setlocal enabledelayedexpansion
        )
    )   
)
pause
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flashercs

@echo off
REM 查找step=1的等差数列arithmetric progression,长度为arraylen
setlocal enabledelayedexpansion
set in=series.txt
set out=ArithmetricProgression.txt
set arraylen=4
set step=1
if %arraylen% lss 1 (
  echo 序列长度必须是大于0的整数
  goto end
)
REM init: clear array
for /f "delims==" %%A in ('2^>nul set arr[') do set %%A=
set /a i=begin=0,end=begin+arraylen-1
 (
  for /f "delims=" %%A in (%in%) do (
    if !i! gtr %begin% (
      set /a p=i-1
      for %%P in (!p!) do set /a sub=1%%A-1!arr[%%P]!
      if !sub! neq %step% (
        set /a i=begin
      )
    )
    set arr[!i!]=%%A
    if !i! equ %end% (
      for /l %%B in (%begin%,1,%end%) do echo !arr[%%B]!
      echo.
      set /a i=begin
    ) else (
      set /a i+=1
    )
  )
)

:end
endlocal
pause
exit /b
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went

各位可以试下这种情况,3种代码输出都不一样
插入00005
具体要看楼主怎么保留

00001
00002
00003
00004
00005
00008
00010
00013
00014
00015
00016
00021
00022
00023
00026
00030
00031
00035
00036
00049
00060
00061
00062
00065
复制代码

WHY

回复 6# went


    我倒是觉得，只要能够正确处理顶楼的样本就好。

flashercs

回复 6# went

原来忽略了一种情况,现在修改了.可以任意修改数列长度,步长可为负值

ivor

#! /usr/bin/env python3
# ! coding:utf-8

#auto detect step is negative or positive

arr = list(map(lambda x: str(x).replace('\n', ''), open('series.txt', 'r')))
result = list()
_next = None
nega_or_posi = (int(arr[1]) - int(arr[0])) /abs(int(arr[1]) - int(arr[0]))
for i in range(len(arr)):
    if i != len(arr)-1 and int(arr[i].lstrip('0')) + nega_or_posi == int(arr[i+1].lstrip('0')):
        result.append(arr[i])
        _next = arr[i + 1]
    elif isinstance(_next, str):
        result.append(_next)
        _next = None

print(result)
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ivor

第三方库

import more_itertools as mit

iterable = list(map(lambda x: int(str(x).strip('0\n')), open('num', 'r')))
result = map(lambda x: [str(i).zfill(5) for i in x], list(filter(lambda x:len(x) > 1, [list(group) for group in mit.consecutive_groups(iterable)])))
print(list(result))
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