[英]Aggregating csv file in bash script
我有多行的csv文件。 每行具有相同的列數。 我需要做的是將這些行按一些指定的列進行分組,並匯總來自其他列的數據。 輸入文件示例:
proces1,pathA,5-May-2011,10-Sep-2017,5
proces2,pathB,6-Jun-2014,7-Jun-2015,2
proces1,pathB,6-Jun-2017,7-Jun-2017,1
proces1,pathA,11-Sep-2017,15-Oct-2017,2
對於上面的示例,我需要按前兩列對行進行分組。 從第3列中,我需要選擇最小值,第4列中的最大值,第5列應具有總和。 因此,對於這樣的輸入文件,我需要輸出:
proces1,pathA,5-May-2011,15-Oct-2017,7
proces1,pathB,6-Jun-2017,7-Jun-2017,1
proces2,pathB,6-Jun-2014,7-Jun-2015,2
我需要在bash中處理它(我也可以使用awk或sed)。
使用bash和sort:
#!/bin/bash
# create associative arrays
declare -A month2num=([Jan]=1 [Feb]=2 [Mar]=3 [Apr]=4 [May]=5 [Jun]=6 [Jul]=7 [Aug]=8 [Sep]=9 [Oct]=10 [Nov]=11 [Dec]=12])
declare -A p ds de # date start and date end
declare -A -i sum # set integer attribute
# function to convert 5-Jun-2011 to 20110605
date2num() { local d m y; IFS="-" read -r d m y <<< "$1"; printf "%d%.2d%.2d\n" $y ${month2num[$m]} $d; }
# read all columns to variables p1 p2 d1 d2 s
while IFS="," read -r p1 p2 d1 d2 s; do
# if associative array is still empty for this entry
# fill with current strings/value
if [[ -z ${p[$p1,$p2]} ]]; then
p[$p1,$p2]="$p1,$p2"
ds[$p1,$p2]="$d1"
de[$p1,$p2]="$d2"
sum[$p1,$p2]="$s"
continue
fi
# compare strings, set new strings and sum value
if [[ ${p[$p1,$p2]} == "$p1,$p2" ]]; then
[[ $(date2num "$d1") < $(date2num ${ds[$p1,$p2]}) ]] && ds[$p1,$p2]="$d1"
[[ $(date2num "$d2") > $(date2num ${de[$p1,$p2]}) ]] && de[$p1,$p2]="$d2"
sum[$p1,$p2]=sum[$p1,$p2]+s
fi
done < file
# print content of all associative arrays with key vom associative array p
for i in "${!p[@]}"; do echo "${p[$i]},${ds[$i]},${de[$i]},${sum[$i]}"; done
用法: ./script.sh | sort
./script.sh | sort
輸出到標准輸出:
proces1,pathA,5-May-2011,15-Oct-2017,7 proces1,pathB,6-Jun-2017,7-Jun-2017,1 proces2,pathB,6-Jun-2014,7-Jun-2015,2
請參閱: help declare
, help read
,當然還有man bash
用awk +排序
awk -F',|-' '
BEGIN{
A["Jan"]="01"
A["Feb"]="02"
A["Mar"]="03"
A["Apr"]="04"
A["May"]="05"
A["Jun"]="06"
A["July"]="07"
A["Aug"]="08"
A["Sep"]="09"
A["Oct"]="10"
A["Nov"]="11"
A["Dec"]="12"
}
{
B[$1","$2]=B[$1","$2]+$9
z=sprintf( "%.2d",$3)
y=sprintf("%s",$5 A[$4] z)
if(!start[$1$2])
{
end[$1$2]=0
start[$1$2]=99999999
}
if (y < start[$1$2])
{
start[$1$2]=y
C[$1","$2]=$3"-"$4"-"$5
}
x=sprintf( "%.2d",$6)
w=sprintf("%s",$8 A[$7] x)
if(w > end[$1$2] )
{
end[$1$2]=w
D[$1","$2]=$6"-"$7"-"$8
}
}
END{
for (i in B)print i "," C[i] "," D[i] "," B[i]
}
' infile | sort
擴展的GNU awk
解決方案:
awk -F, 'function parse_date(d_str){
split(d_str, d, "-");
t = mktime(sprintf("%d %d %d 00 00 00", d[3], m[d[2]], d[1]));
return t
}
BEGIN{ m["Jan"]=1; m["Feb"]=2; m["Mar"]=3; m["Apr"]=4; m["May"]=5; m["Jun"]=6;
m["Jul"]=7; m["Aug"]=8; m["Sep"]=9; m["Oct"]=10; m["Nov"]=11; m["Dec"]=12;
}
{
k=$1 SUBSEP $2;
if (k in a){
if (parse_date(a[k]["min"]) > parse_date($3)) { a[k]["min"]=$3 }
if (parse_date(a[k]["max"]) < parse_date($4)) { a[k]["max"]=$4 }
} else {
a[k]["min"]=$3; a[k]["max"]=$4
}
a[k]["sum"]+= $5
}
END{
for (i in a) {
split(i, j, SUBSEP);
print j[1], j[2], a[i]["min"], a[i]["max"], a[i]["sum"]
}
}' OFS=',' file
輸出:
proces1,pathA,5-May-2011,15-Oct-2017,7
proces1,pathB,6-Jun-2017,7-Jun-2017,1
proces2,pathB,6-Jun-2014,7-Jun-2015,2
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