请问如何删除文件的某一列？

Ethanolfree

RT
本人做数据处理，用for循环grep出一系列数据，但是有的数据时没什么用的，请问该怎么样删除文件中某些特定的列？请指教。

比如删除如下文件的第一列和最后一列。

谢谢。


  3.0000000e+000  6.5936846e+001 -1.4979541e+001  7.7524469e+001  4.6088705e+000 -6.2799303e-002  3.7613030e-002
  3.0000000e+000  4.8110855e+001 -1.4875115e+001  1.3685352e+002  4.9868712e+000 -4.2744056e-002  8.3132162e-002
  3.0000000e+000  1.2031220e+001 -8.1583356e+000  1.6622129e+002  4.7542681e+000 -3.1356255e-002  4.1376053e-002
  3.0000000e+000  9.2129320e+001 -6.1178748e+000  3.5476271e+001  4.0275424e+000 -1.1041718e-001  4.1990421e-002
  3.0000000e+000  4.2758362e+001 -3.5779865e+001  2.8111568e+002  4.3803787e+000 -1.5701563e-002  9.9691881e-003
  3.0000000e+000  4.3766129e+001 -2.2280382e+001  1.9362229e+002  4.8885946e+000 -2.8752287e-002  7.3178508e-002
  3.0000000e+000  5.7864843e+001 -1.9549955e+001  8.3078974e+001  4.2548473e+000 -5.0658826e-002  4.4097765e-002
  3.0000000e+000  6.1063789e+001 -4.9747630e+000  6.8463298e+001  5.0891832e+000 -9.0039368e-002  1.1975120e-001
  3.0000000e+000  4.3714999e+001 -2.8359051e+001  2.0722720e+002  4.8330829e+000 -2.6135560e-002  4.6466144e-002
  3.0000000e+000  1.2983904e+002 -2.5761524e+001  3.9463712e+001  3.2922362e+000 -9.3302289e-002  9.1351866e-003
  3.0000000e+000  2.1810545e+002 -6.7293722e+000  1.1465653e+001  4.2503761e+000 -3.6648181e-001  1.9340095e-001

yangkyo821

1. awk '{a=$2;for(i=3;i<NF;i++)a=a FS $i;print a}' file
   

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Shell_HAT

1. awk '{for(i=2;i<NF;i++)printf $i FS;print ""}' urfile

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Ethanolfree

谢谢，见识了
第一次见到这样写awk的参数。

Shell_HAT

1. sed 's/^ \+[0-9.e+]\+[^ ] \+//; s/ [^ ]\+$//' urfile

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Ethanolfree

Shell_HAT 发表于 2012-02-27 10:51

谢谢您提供的帮助。

qinggzha

cat file|awk '{print $2,$3,$4,$5,$6}'

HH106

回复 1# Ethanolfree


     awk '{$1=$NF=""}1' urfile