Question

如果有一个文件，行格式如下：

SOME_ATTRIBUTE_1 XYZ; IMPORTANT_ATTRIBUTE_1 1234; SOME_ATTRIBUTE_2 XYZ; IMPORTANT_ATTRIBUTE_2 AB;

现在我想将其转换为以下形式，即两个重要的属性值产生一个新属性：

JOIN_IMPORTANT_ATTRIBUTE AB1234; SOME_ATTRIBUTE_1 XYZ; IMPORTANT_ATTRIBUTE_1 1234; SOME_ATTRIBUTE_2 XYZ; IMPORTANT_ATTRIBUTE_2 AB;

这可以用一些带有awk或类似的单线程来完成吗？我不知道如何解决这个问题，而没有抓住java技巧框。

Answer 1

使用awk，您可以在分号+任意数量的空格中拆分输入，并进一步拆分重要字段，如下所示：

awk -F'; *' '{ split($2, a1, / +/); split($4, a2, / +/); print "JOIN_IMPORTANT_ATTRIBUTE", a2[2] a1[2] ";", $0 }' infile

输出：

JOIN_IMPORTANT_ATTRIBUTE AB1234; SOME_ATTRIBUTE_1 XYZ; IMPORTANT_ATTRIBUTE_1 1234; SOME_ATTRIBUTE_2 XYZ; IMPORTANT_ATTRIBUTE_2 AB;

这假设您知道重要属性所在的列。

Answer 2

Perl解决方案：

perl -lane 'print join " ", "JOIN_IMPORTANT_ATTRIBUTE", substr($F[7], 0, -1) . $F[3], @F'

Answer 3

awk -F'[; ]+' '{print "JOIN_IMPORTANT_ATTRIBUTE", $8 $4 "; " $0}' file

Answer 4

这是我的bash + awk替代方案。

cat attrs.awk
# Awk script to get joined attributes for one line of attributes

BEGIN {
RS=";";     
PROCINFO["sorted_in"]="@ind_num_asc"; #gawk only: sort attributes on their attr id (so that IMPORTANT_ATTRIBUTE_n comes before IMPORTANT_ATTRIBUTE_n+1
}

$1 ~ /^IMPORTANT_ATTRIBUTE_/ {
            attrId=substr($1, 1 + length("IMPORTANT_ATTRIBUTE_"));
    if ($2 ~ /^[0-9]/) 
            impAttrsNum[attrId]=$2;
    else
            impAttrsAlpha[attrId]=$2;
}

END {
    #alpha attribs come before num attribs
    for(i in impAttrsAlpha)
            alphaVals = alphaVals impAttrsAlpha[i];
    for(i in impAttrsNum)
            numVals = numVals impAttrsNum[i];

    printf("JOIN_IMPORTANT_ATTRIBUTE %s%s%s", alphaVals, numVals, RS);
}

cat joinattrs
#!/bin/bash
#
# Applies joined attributes for each input line

while read l
do
    if [[ -n "$l" ]]
    then   
            joinAttrs=$(echo "$l" | awk -f attrs.awk)
            echo "$joinAttrs $l"
    fi
done

如何使用它： ./joinattrs＆lt;数据文件

不是单行：）

Linux：从行中提取和转换属性

4 个答案: