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
说明: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进行尝试,但仍未获得期望的输出。我的脚本在末尾打印了一些多余的逗号(,)。
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
答案 0 :(得分:1)
您可以使用此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
答案 1 :(得分:1)
单个awk程序可以生成所需的输出:使用GNU awk
gawk '
BEGIN {SUBSEP = OFS = ","}
NR==1 {next}
{ groups[$4]; value[$1,$2,$3][$4] = $5 }
END {
PROCINFO["sorted_in"] = "@ind_str_asc"
printf "x,y,z"
for (g in groups) printf ",%s", g
printf "\n"
for (a in value) {
printf "%s", a
for (g in groups) printf "%s%s", OFS, 0+value[a][g]
printf "\n"
}
}
' ifile.txt
答案 2 :(得分:1)
另一个类似的awk
,但标题不正确
$ awk -v OFS=, '{k=$1 OFS $2 OFS $3}
p!=k {if(p) print line; p=k; line=k}
{line=line OFS $NF}
END {print line}' file
x,y,z,value
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