Question

如何创建正则表达式以忽略以“空字符”，＃或字母开头的行。下面是我拥有的数据样本，我需要匹配仅以数字开头的行（负数或正数）：

0.000000    1.2712052472 0.8899021956 22.2458  265.2511402076  322.1539247218 -13.6281 -130.986 0.155342 0.889755 phaet_000227
0.000000    1.2712052462 0.8899021922 22.2458  265.2511430964  322.1539209801 -13.6281 -130.986 0.155342 0.889755 phaet_000090
0.000000    1.2712052476 0.8899022047 22.2458  265.2511396341  322.1539260295 -13.6281 -130.986 0.155342 0.889755 phaet_000111
0.000000    1.2712052465 0.8899022229 22.2458  265.2511497521  322.1539197205 -13.6281 -130.986 0.155342 0.889755 phaet_000059
Nplanets 9  Nparticles 500: alive 509/509  ejected 0  rmin 0  rmax 0
Full close app checks 0/0 (0.000000%)   BS fails 0
Close apps 1   bounces 0   accretions 0  Max n/step 0
Simulation time 0 going to -100000.
Real time 1 s   Force  0 s (0.00 %)   Coll 0 s  (0.00 %)
               E&L    0 s (0.00 %)    Kep 0 s (0.00 %)
CPU time 0.037627 s   Force  0 s (0.00 %)   Coll 0 s  (0.00 %)
               E&L     0 s (0.00 %)  Kep 0 s (0.00 %)
# Nplanets 9  Nparticles 500: alive 509/509  ejected 0  rmin 0  rmax 0
# Full close app checks 0/0 (0.000000%)   BS fails 0
# Close apps 1   bounces 0   accretions 0  Max n/step 0
# Simulation time 0 going to -100000.
# Real time 1 s   Force  0 s (0.00 %)   Coll 0 s  (0.00 %)
#                E&L    0 s (0.00 %)    Kep 0 s (0.00 %)
# CPU time 0.037627 s   Force  0 s (0.00 %)   Coll 0 s  (0.00 %)
#                E&L     0 s (0.00 %)  Kep 0 s (0.00 %)
Output step 1 at t=-10 going to -100000
-10.000000    1.2713031501 0.8900442847 22.1802  265.4033924020  322.0041354013 -5.32091 -102.357 0.155286 0.88482 phaet_000065
-10.000000    1.2713031508 0.8900443093 22.1802  265.4033954804  322.0041360861 -5.32091 -102.357 0.155286 0.88482 phaet_000299
-10.000000    1.2713031483 0.8900442977 22.1802  265.4033839221  322.0041469420 -5.32092 -102.357 0.155286 0.88482 phaet_000102
-10.000000    1.2713031486 0.8900442931 22.1802  265.4033724632  322.0041581369 -5.32092 -102.357 0.155286 0.884821 phaet_000371
-10.000000    1.2713031463 0.8900442910 22.1802  265.4033772870  322.0041532421 -5.32093 -102.357 0.155286 0.884821 phaet_000019

我想最终拥有：

0.000000    1.2712052472 0.8899021956 22.2458  265.2511402076  322.1539247218 -13.6281 -130.986 0.155342 0.889755 phaet_000227
0.000000    1.2712052462 0.8899021922 22.2458  265.2511430964  322.1539209801 -13.6281 -130.986 0.155342 0.889755 phaet_000090
0.000000    1.2712052476 0.8899022047 22.2458  265.2511396341  322.1539260295 -13.6281 -130.986 0.155342 0.889755 phaet_000111
0.000000    1.2712052465 0.8899022229 22.2458  265.2511497521  322.1539197205 -13.6281 -130.986 0.155342 0.889755 phaet_000059
-10.000000    1.2713031501 0.8900442847 22.1802  265.4033924020  322.0041354013 -5.32091 -102.357 0.155286 0.88482 phaet_000065
-10.000000    1.2713031508 0.8900443093 22.1802  265.4033954804  322.0041360861 -5.32091 -102.357 0.155286 0.88482 phaet_000299
-10.000000    1.2713031483 0.8900442977 22.1802  265.4033839221  322.0041469420 -5.32092 -102.357 0.155286 0.88482 phaet_000102
-10.000000    1.2713031486 0.8900442931 22.1802  265.4033724632  322.0041581369 -5.32092 -102.357 0.155286 0.884821 phaet_000371
-10.000000    1.2713031463 0.8900442910 22.1802  265.4033772870  322.0041532421 -5.32093 -102.357 0.155286 0.884821 phaet_000019

所以，我尝试了“grep”如下：

grep -v '^[a-z,A-Z,\s,\#]' file1.dat > file2.dat

它摆脱了以字母和'＃'开头的行，但是以空格开头的行仍然存在，即我无法删除：

       E&L    0 s (0.00 %)    Kep 0 s (0.00 %)
       E&L     0 s (0.00 %)  Kep 0 s (0.00 %)

请注意，“E＆amp; L”之前有空格。

知道如何摆脱这些吗？

Answer 1

由于前面的空白区域，这两条线不会消除。你可以先消除它们。

sed "s/^[ \t]*//" file1.dat > file3.dat

然后使用'grep'命令过滤文件。

grep -v '^[a-z,A-Z,\s,\#]' file3.dat > file2.dat

Answer 2

在grep中，[\s,\#]匹配反斜杠，逗号或井号。（反斜杠在括号表达式中没有特殊含义，逗号永远不会特殊。）匹配空格的最简单方法是使用[:space:]字符类。所以你的正则表达式将是：

^[a-zA-Z#[:space:]]

你也可以积极搜索做以数字开头的行：

^-\?[[:digit:]]\+\.[[:digit:]]\+

正则表达式：忽略以这些字符开头的行

2 个答案: