bash连接多个文件，并且空替换(-e选项) [英] bash join multiple files with empty replacement (-e option)

问题描述

我有以下代码将多个文件连接在一起.它工作正常，但我想将空值替换为0，所以我使用了-e"0".但这是行不通的. 有什么想法吗?

I have following code to join multiple files together. It works fine but I want to replace the empty values to 0, so I used -e "0". But it doesn't work. Any ideas?

for k in `ls file?`
do
    if [ -a final.results ]
    then
            join -a1 -a2 -e "0" final.results $k  > tmp.res
            mv tmp.res final.results
    else
            cp $k final.results
    fi

done

示例:

file1: 
a 1 
b 2
file2:
a 1 
c 2
file3:
b 1 
d 2

Results:
a 1 0 1 0
b 2 1 0
c 2
d 2

expected:
a 1 1 0
b 2 0 1
c 0 2 0
d 0 0 2

推荐答案

它的文献资料很少，但是当使用join时，-e选项只能与-o选项一起使用.每次循环时都需要修改订单字符串.以下代码将生成所需的输出.

It's poorly documented, but when using join the -e option only works in conjunction with the -o option. The order string needs to be amended each time around the loop. The following code should generate your desired output.

i=3
orderl='0,1.2'
orderr=',2.2'
for k in $(ls file?)
do
    if [ -a final.results ]
    then
            join -a1 -a2 -e "0" -o "$orderl$orderr" final.results $k  > tmp.res
            orderl="$orderl,1.$i"
            i=$((i+1))
            mv tmp.res final.results
    else
            cp $k final.results
    fi
done

如您所见，它开始变得凌乱.如果您需要进一步扩展此功能，则可能值得使用功能更强大的工具，例如awk或python.

As you can see, it starts to become messy. If you need to extend this much further it might be worth deferring to a beefier tool such as awk or python.

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