bash - 替换文件中的列但保持空格格式

Question

下面的代码根据文件 B 中的数据替换了文件 A 中的第 4 列，但输出没有保留原始文件的空间。有办法吗？

 tr , " " <fileB | awk 'NR==FNR{a[$2]=$1;next} {$4=a[$4];print}' - fileA

文件A

 xxx    xxx   xxx Z0002

文件B

 3100,3000
 W0002,Z0002

使用上面的代码输出:

 xxx xxx xxx W0002

预期输出:

xxx    xxx   xxx W0002

Answer 1

应该这样做:

awk 'FNR==NR {split($0,a,",");b[a[2]]=a[1];next} {n=split($0,d,/[^[:space:]]*/);if(b[$4])$4=b[$4];for(i=1;i<=n;i++) printf("%s%s",d[i],$i);print ""}' fileB fileA

它将空格存储在一个数组中，以便以后可以重用

例子:

cat fileA
xxx    xxx   xxx Z0002   not change this
xxx   xxx  Z0002 zzz
xxx Z000223213 xxx Z0002 xxx xxx xxx Z0002

cat fileB
3100,3000
W0002,Z0002

awk 'FNR==NR {split($0,a,",");b[a[2]]=a[1];next} {n=split($0,d,/[^[:space:]]*/);if(b[$4])$4=b[$4];for(i=1;i<=n;i++) printf("%s%s",d[i],$i);print ""}' fileB fileA
xxx    xxx   xxx  W0002   not change this
xxx   xxx  Z0002 zzz
xxx Z000223213 xxx  W0002 xxx xxx xxx Z0002

---

一些更具可读性的以及它是如何工作的:

awk '
FNR==NR {                           # For the first file "fileB"
    split($0,a,",")                 # Split it to an array "a" using "," as separator 
    b[a[2]]=a[1]                    # Store the data in array "b" using second column as index
    next                            # Skip to next record
    }
    {                               # Then for the file "fileA"
    n=split($0,d,/[^[:space:]]*/)   # Split the spaces inn group and store them in array "d"
    if(b[$4])                       # If array "b" as data for field "4"
        $4=b[$4]                    # Change filed "4" to data found in array "b"
    for(i=1;i<=n;i++)               # Loop trough all field in the line
        printf("%s%s",d[i],$i)      # print correct separator and data
    print ""                        # Add new line at the end
    }
' fileB fileA                       # Read the files.