linux - 使用 shell 脚本和 awk 格式化然后将 txt 转换为 csv

Question

我有一个文本文件:

ifile.txt
x       y       z       t              value
1       1       5       01hr01Jan2018   3
1       1       5       02hr01Jan2018   3.1
1       1       5       03hr01Jan2018   3.2
1       3.4     3       01hr01Jan2018   4.1
1       3.4     3       02hr01Jan2018   6.1
1       3.4     3       03hr01Jan2018   1.1
1       4.2     6       01hr01Jan2018   6.33
1       4.2     6       02hr01Jan2018   8.33
1       4.2     6       03hr01Jan2018   5.33
3.4     1       2       01hr01Jan2018   3.5
3.4     1       2       02hr01Jan2018   5.65
3.4     1       2       03hr01Jan2018   3.66
3.4     3.4     4       01hr01Jan2018   6.32
3.4     3.4     4       02hr01Jan2018   9.32
3.4     3.4     4       03hr01Jan2018   12.32
3.4     4.2     8.1     01hr01Jan2018   7.43
3.4     4.2     8.1     02hr01Jan2018   7.93
3.4     4.2     8.1     03hr01Jan2018   5.43
4.2     1       3.4     01hr01Jan2018   6.12
4.2     1       3.4     02hr01Jan2018   7.15
4.2     1       3.4     03hr01Jan2018   9.12
4.2     3.4     5.5     01hr01Jan2018   2.2
4.2     3.4     5.5     02hr01Jan2018   3.42
4.2     3.4     5.5     03hr01Jan2018   3.21
4.2     4.2     6.2     01hr01Jan2018   1.3
4.2     4.2     6.2     02hr01Jan2018   3.4
4.2     4.2     6.2     03hr01Jan2018   1

解释:每个坐标 (x,y) 都有一个 z 值和三个时间值。空格不是制表符。它们是空格序列。

我想将 t 列格式化为行，然后转换为 csv 文件。我的预期输出是:

ofile.txt
x,y,z,01hr01Jan2018,02hr01Jan2018,03hr01Jan2018
1,1,5,3,3.1,3.2
1,3.4,3,4.1,6.1,1.1
1,4.2,6,6.33,8.33,5.33
3.4,1,2,3.5,5.65,3.66
3.4,3.4,4,6.32,9.32,12.32
3.4,4.2,8.1,7.43,7.93,5.43
4.2,1,3.4,6.12,7.15,9.12
4.2,3.4,5.5,2.2,3.42,3.21
4.2,4.2,6.2,1.3,3.4,1

我正在按照以下方式尝试，但仍然没有得到想要的输出。我的脚本在末尾打印了一些额外的逗号 (,)。

我的算法和脚本是:

    #Step1:- Split into two files: one with x,y,z (0001.txt) and
    #        another with t,value (0002.txt).

    awk '{n=3; for (i=1;i<=n;i++) printf "%s ", $i; print "";}' ifile.txt > 0001.txt
    awk '{n=5; for (i=4;i<=n;i++) printf "%s ", $i; print "";}' ifile.txt > 0002.txt

    #Setp2:- In 0001.txt: Delete the repetition rows. 

    awk '!seen[$1,$2,$3]++' 0001.txt > 00011.txt

    #Step3:- In 0002.txt: Delete the first row. For each 3 rows in t-column,
    #        write the value-column as rows. Add the t-row at top
    #        this is very manual. I am wondering for some command

    grep -E "^[0-9].*" 0002.txt > 0003.txt
   awk -v n=3 '{ row = row $2 " "; if (NR % n == 0) { print row; row = "" } }' 0003.txt > 0004.txt
    (echo "01hr01Jan2018,02hr01Jan2018,03hr01Jan2018";cat 0004.txt) > 00022.txt  

    #Step4:- Paste output of two and convert to csv.
    paste 00011.txt 00022.txt > 0005.txt
    cat 0005.txt | tr -s '[:blank:]' ',' > ofile.txt

Answer 1

你可以使用这个awk:

awk -v OFS=, '{k=$1 OFS $2 OFS $3}
!($4 in hdr){hn[++h]=$4; hdr[$4]}
k in row{row[k]=row[k] OFS $5; next}
{rn[++n]=k; row[k]=$5}
END {
   printf "%s", rn[1]
   for(i=1; i<=h; i++)
      printf "%s", OFS hn[i]
   print ""
   for (i=2; i<=n; i++)
      print rn[i], row[rn[i]]
}' file

x,y,z,t,01hr01Jan2018,02hr01Jan2018,03hr01Jan2018
1,1,5,3,3.1,3.2
1,3.4,3,4.1,6.1,1.1
1,4.2,6,6.33,8.33,5.33
3.4,1,2,3.5,5.65,3.66
3.4,3.4,4,6.32,9.32,12.32
3.4,4.2,8.1,7.43,7.93,5.43
4.2,1,3.4,6.12,7.15,9.12
4.2,3.4,5.5,2.2,3.42,3.21
4.2,4.2,6.2,1.3,3.4,1