如何提取字符串组，并在文件

我有显示这样的文件：

Bin 1 = 5778 669 753 731 743 752 727 736 667  
Bin 5 = 2  2 0 0 0 0 0 0 0  Bin 7 = 63  36 5 1 1 5 1 5 9  
Bin 9 = 0  0 0 0 0 0 0 0 0  Bin 24 = 2  0 0 0 0 0 0 0 2  
Bin 31 = 0  0 0 0 0 0 0 0 0  Bin 47 = 28  2 4 6 1 2 5 3 5  
Bin 70 = 1  0 0 0 0 0 1 0 0  Bin 95 = 1  0 1 0 0 0 0 0 0  
Bin 120 = 1  0 1 0 0 0 0 0 0  Bin 121 = 162  49 13 11 10 3 10 12 54  
Bin 131 = 7  1 1 2 0 0 1 0 2  Bin 133 = 3  0 0 0 0 0 0 0 3  
Bin 141 = 6  0 0 0 1 0 1 2 2  Bin 560 = 1  0 0 0 1 0 0 0 0  
Bin 620 = 1  0 0 0 1 0 0 0 0  Bin 630 = 2  0 0 1 0 0 0 0 1  
Bin 1250 = 1  0 0 0 0 0 1 0 0  Bin 1410 = 1  0 1 0 0 0 0 0 0  
Bin 2004 = 1  0 0 0 0 0 0 1 0  Bin 2240 = 1  0 0 0 0 1 0 0 0  
Bin 2247 = 1  0 0 1 0 0 0 0 0  Bin 2300 = 1  0 0 0 0 1 0 0 0  
Bin 2309 = 1  0 0 0 1 0 0 0 0  Bin 2310 = 3  1 1 0 0 0 0 0 1  
Bin 2602 = 1  0 0 0 1 0 0 0 0

我想削减第二届“宾“模式及其右侧的值并粘贴到文件底部，然后再进行排序。感谢您对我这个问题的帮助。我只是unix命令中的新成员，仍然在学习它。提前致谢。

我想有这样的输出。

Bin 1 = 5778 669 753 731 743 752 727 736 667  
Bin 5 = 2  2 0 0 0 0 0 0 0  
Bin 7 = 63  36 5 1 1 5 1 5 9  
Bin 9 = 0  0 0 0 0 0 0 0 0  
Bin 24 = 2  0 0 0 0 0 0 0 2  
Bin 31 = 0  0 0 0 0 0 0 0 0  
Bin 47 = 28  2 4 6 1 2 5 3 5  
Bin 70 = 1  0 0 0 0 0 1 0 0  
Bin 95 = 1  0 1 0 0 0 0 0 0  
Bin 120 = 1  0 1 0 0 0 0 0 0  
Bin 121 = 162  49 13 11 10 3 10 12 54  
Bin 131 = 7  1 1 2 0 0 1 0 2  
Bin 133 = 3  0 0 0 0 0 0 0 3  
Bin 141 = 6  0 0 0 1 0 1 2 2  
Bin 560 = 1  0 0 0 1 0 0 0 0  
Bin 620 = 1  0 0 0 1 0 0 0 0  
Bin 630 = 2  0 0 1 0 0 0 0 1  
Bin 1250 = 1  0 0 0 0 0 1 0 0  
Bin 1410 = 1  0 1 0 0 0 0 0 0  
Bin 2004 = 1  0 0 0 0 0 0 1 0  
Bin 2240 = 1  0 0 0 0 1 0 0 0  
Bin 2247 = 1  0 0 1 0 0 0 0 0  
Bin 2300 = 1  0 0 0 0 1 0 0 0  
Bin 2309 = 1  0 0 0 1 0 0 0 0  
Bin 2310 = 3  1 1 0 0 0 0 0 1  
Bin 2602 = 1  0 0 0 1 0 0 0 0

方面，麦克

来源

2017-06-13 Mike

显示您当前的工作和后期所期望的结果 – RomanPerekhrest

喜@RomanPerekhrest，抱歉地说，我没有拿出任何解决办法呢，我一直念叨的sed和awk，但我不知道如何开始，只是不知道如何开始，对不起 – Mike

然后做第二个动作：*发布所需的结果* – RomanPerekhrest

短awk的方法：

awk 'NF>12{ $13= ORS $13 }1' file | column -t

NF - 字段
的总数
NF>12 - 根据预期的结果字段的最大数目为12。非空白序列被视为一个字段
$13= ORS $13 - 打破记录在第13场的记录分隔符ORS（换行符charac之三）

输出：

Bin 1  = 5778 669 753 731 743 752 727 736 667 
Bin 5  = 2  2 0 0 0 0 0 0 0 
Bin 7  = 63 36 5 1 1 5 1 5 9 
Bin 9  = 0  0 0 0 0 0 0 0 0 
Bin 24 = 2  0 0 0 0 0 0 0 2 
Bin 31 = 0  0 0 0 0 0 0 0 0 
Bin 47 = 28 2 4 6 1 2 5 3 5 
Bin 70 = 1  0 0 0 0 0 1 0 0 
Bin 95 = 1  0 1 0 0 0 0 0 0 
Bin 120 = 1  0 1 0 0 0 0 0 0 
Bin 121 = 162 49 13 11 10 3 10 12 54 
Bin 131 = 7  1 1 2 0 0 1 0 2 
Bin 133 = 3  0 0 0 0 0 0 0 3 
Bin 141 = 6  0 0 0 1 0 1 2 2 
Bin 560 = 1  0 0 0 1 0 0 0 0 
Bin 620 = 1  0 0 0 1 0 0 0 0 
Bin 630 = 2  0 0 1 0 0 0 0 1 
Bin 1250 = 1  0 0 0 0 0 1 0 0 
Bin 1410 = 1  0 1 0 0 0 0 0 0 
Bin 2004 = 1  0 0 0 0 0 0 1 0 
Bin 2240 = 1  0 0 0 0 1 0 0 0 
Bin 2247 = 1  0 0 1 0 0 0 0 0 
Bin 2300 = 1  0 0 0 0 1 0 0 0 
Bin 2309 = 1  0 0 0 1 0 0 0 0 
Bin 2310 = 3  1 1 0 0 0 0 0 1 
Bin 2602 = 1  0 0 0 1 0 0 0 0

来源

2017-06-13 08:07:29 RomanPerekhrest

嗨@RomanPerekhrest，你的代码工作，非常感谢，介意如果你能解释它是如何工作的，再次感谢 – Mike

@Mike，看我的详细信息 – RomanPerekhrest

非常感谢帮助，我现在明白了。再一次，真的很感谢帮助 – Mike

sed 's/ \(Bin.*\)$/\n\1/g' file | sort -k 2 -n > outputfile

此搜索具有空间，其次是“宾”和任何字符，直到行的其余部分开头的字符串。
放置在逃逸的部分，可以在命令的替换部分引用\1。你在那里插入一个换行符。

排序部分取参数

-k 2
-n（数值排序）（在第二列排序）

输出：

# sed 's/ \(Bin.*\)$/\n\1/g' your_file | sort -k 2 -n 
Bin 1 = 5778 669 753 731 743 752 727 736 667 
Bin 5 = 2  2 0 0 0 0 0 0 0 
Bin 7 = 63  36 5 1 1 5 1 5 9 
Bin 9 = 0  0 0 0 0 0 0 0 0 
Bin 24 = 2  0 0 0 0 0 0 0 2 
Bin 31 = 0  0 0 0 0 0 0 0 0 
Bin 47 = 28  2 4 6 1 2 5 3 5 
Bin 70 = 1  0 0 0 0 0 1 0 0 
Bin 95 = 1  0 1 0 0 0 0 0 0 
Bin 120 = 1  0 1 0 0 0 0 0 0 
Bin 121 = 162  49 13 11 10 3 10 12 54 
Bin 131 = 7  1 1 2 0 0 1 0 2 
Bin 133 = 3  0 0 0 0 0 0 0 3 
Bin 141 = 6  0 0 0 1 0 1 2 2 
Bin 560 = 1  0 0 0 1 0 0 0 0 
Bin 620 = 1  0 0 0 1 0 0 0 0 
Bin 630 = 2  0 0 1 0 0 0 0 1 
Bin 1250 = 1  0 0 0 0 0 1 0 0 
Bin 1410 = 1  0 1 0 0 0 0 0 0 
Bin 2004 = 1  0 0 0 0 0 0 1 0 
Bin 2240 = 1  0 0 0 0 1 0 0 0 
Bin 2247 = 1  0 0 1 0 0 0 0 0 
Bin 2300 = 1  0 0 0 0 1 0 0 0 
Bin 2309 = 1  0 0 0 1 0 0 0 0 
Bin 2310 = 3  1 1 0 0 0 0 0 1 
Bin 2602 = 1  0 0 0 1 0 0 0 0

来源

2017-06-13 08:02:29 fancyPants

嗨@fancyPants，我试过你的代码，但它只是添加了“n”在该行的开始。 Bin字符串前面没有空格，您可以修改它，谢谢输入 – Mike

您是否复制并粘贴了我的代码？听起来你在n之前忘了反斜杠。 – fancyPants

是的，我复制并粘贴它们。我再次做了，但结果是一样的 – Mike

如果文本的结构，它可能是足以与cut，例如：

(cut -c1-56 infile; cut -c62- infile) | sort -k2n

输出：

Bin 1 = 5778 669 753 731 743 752 727 736 667 
Bin 5 = 2  2 0 0 0 0 0 0 0 
Bin 7 = 63  36 5 1 1 5 1 5 9  
Bin 9 = 0  0 0 0 0 0 0 0 0 
Bin 24 = 2  0 0 0 0 0 0 0 2  
Bin 31 = 0  0 0 0 0 0 0 0 0 
Bin 47 = 28  2 4 6 1 2 5 3 5  
Bin 70 = 1  0 0 0 0 0 1 0 0 
Bin 95 = 1  0 1 0 0 0 0 0 0  
Bin 120 = 1  0 1 0 0 0 0 0 0 
Bin 121 = 162  49 13 11 10 3 10 12 54  
Bin 131 = 7  1 1 2 0 0 1 0 2 
Bin 133 = 3  0 0 0 0 0 0 0 3  
Bin 141 = 6  0 0 0 1 0 1 2 2 
Bin 560 = 1  0 0 0 1 0 0 0 0  
Bin 620 = 1  0 0 0 1 0 0 0 0 
Bin 630 = 2  0 0 1 0 0 0 0 1  
Bin 1250 = 1  0 0 0 0 0 1 0 0 
Bin 1410 = 1  0 1 0 0 0 0 0 0  
Bin 2004 = 1  0 0 0 0 0 0 1 0 
Bin 2240 = 1  0 0 0 0 1 0 0 0  
Bin 2247 = 1  0 0 1 0 0 0 0 0 
Bin 2300 = 1  0 0 0 0 1 0 0 0  
Bin 2309 = 1  0 0 0 1 0 0 0 0 
Bin 2310 = 3  1 1 0 0 0 0 0 1  
Bin 2602 = 1  0 0 0 1 0 0 0 0

来源

2017-06-13 08:14:04 Thor

这也适用，非常感谢您的意见，非常感谢 – Mike

如何提取字符串组，并在文件

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