我有一个包含在第一列任务名称以及完成任务的第二列如下时间文件：unix文件中任务的最小值和最大值？
Task2, 3421
Task3, 3300
Task1, 1000
Task2, 1100
Task3, 1200
Task3, 1209
Task4, 1299
Task3, 1289
Task1, 1389
Task2, 1211
Task5, 1216
Task2, 1416
Task1, 2100
Task6, 2416
Task5, 2216
Task7, 1116
现在，我必须找到采取每个任务的最小和最大时间并以下面的格式输出
task , maxtime , min time
eg
Task1, 1000, 2100 (from the data given above)
您可以awk
awk '
BEGIN{FS=","; OFS=", "}
!($1 in max) || $2>max[$1]{max[$1]=$2}
!($1 in min) || $2<min[$1]{min[$1]=$2}
END{
for(k in max){print k, min[k], max[k]}
}' input.txt
你试试，
Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116
另一种方式来做到这一点是通过列1，然后通过列2排序，并采取了第一个和最后一个值像这样的每个任务
awk -F, '{arr[$1]=arr[$1] $2} END {for(key in arr) print key, arr[key]}' <(sort -t 1 -k 1,2 file) | awk '{OFS=", "; print $1, $2, $NF}'
样品运行：
$ cat file
Task2, 3421
Task3, 3300
Task1, 1000
Task2, 1100
Task3, 1200
Task3, 1209
Task4, 1299
Task3, 1289
Task1, 1389
Task2, 1211
Task5, 1216
Task2, 1416
Task1, 2100
Task6, 2416
Task5, 2216
Task7, 1116
$ sort -t 1 -k 1,2 file
Task1, 1000
Task1, 1389
Task1, 2100
Task2, 1100
Task2, 1211
Task2, 1416
Task2, 3421
Task3, 1200
Task3, 1209
Task3, 1289
Task3, 3300
Task4, 1299
Task5, 1216
Task5, 2216
Task6, 2416
Task7, 1116
$ awk -F, '{arr[$1]=arr[$1] $2} END {for(key in arr) print key, arr[key]}' <(sort -t 1 -k 1,2 file) | awk '{OFS=", "; print $1, $2, $NF}'
Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116
使用gawk的array of arrays：
gawk 'BEGIN{OFS=FS=","}
$2>a[$1]["max"]{a[$1]["max"]=$2}
$2<a[$1]["min"] || !a[$1]["min"] {a[$1]["min"]=$2}
END {for (i in a){
print i, a[i]["min"],a[i]["max"]
}
}' file
例here。
这里是另一替代
$ join -t, <(sort file){,} | sort -k1,1 -k2n -k3nr | rev | uniq -2 | rev
sort它上的第一和第二列，然后在awk它。这个解决方案（awk部分）的好处在于它不会将数据存储在内存中并最终将其转储出去，而是一旦找到新数据就会输出以前的$1的数据。在这里：
$ sort -t, -k1 foo -k2n | \ # sort
awk '!($1 in min) {min[$1]=$2} # first of each is always min (and max)
($1 in min) {max[$1]=$2} # every current one is always max
$1!=p && NR>1 {print p, min[p], max[p]} # if $1 differs from previous, print previous
{p=$1} # p is current for next round
END {print p, min[p], max[p]}' # dump buffer
Task1, 1000 2100
Task2, 1100 3421
Task3, 1200 3300
Task4, 1299 1299
Task5, 1216 2216
Task6, 2416 2416
Task7, 1116 1116
使用sort，sed和awk
sort -k1,1 -k2n input.txt | sed -r ':a;N;$!ba;:b;s/(Task[0-9]+,)([0-9 ,]+)\n?\1([0-9]+)/\1\2, \3/g;tb;' | awk 'BEGIN{FS=OFS=", ";}{print $1, $2, $NF}'
使用sort和sed替代解决方案的另一个答案只有
sort -k1,1 -k2n input.txt | sed -r ':a;N;$!ba;:b;s/(Task[0-9]+,)([0-9 ,]+)\n?\1([0-9]+)/\1\2, \3/g;tb;' | sed -r -e 's/^([^ ]+)\s([^ ]+)\s.*\s([^ ]+)/\1 \2 \3/' -e 's/^([^ ]+)\s([^ ]+)$/\1 \2, \2/'
你，
Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116
这主要是bash，如果你有这方面的问题，我可以用别的东西替代awk命令......（例如，如果时间始终在同一列中，则为colrm）。
# Keep a list of already processed task names
already_processed=""
# Use read to read only the first column from the data file
while IFS=',' read -ra task; do
# If the task has already been processed, skip it and go to the next line
if echo "$already_processed" | grep $task > /dev/null; then
continue
else
# Select all the task with the same name from the data file, take the
#+second column and sort it to find the max and the minimum.
MIN=`grep $task $1 | awk -F',' '{print $2}' | sort -n | head -1`
MAX=`grep $task $1 | awk -F',' '{print $2}' | sort -n | tail -1`
# Add the task to the "already_processed" tasks (to be sure each task will
#+appear only once in the output
already_processed="$already_processed:$task"
# Print the output in the wanted format.
echo "${task}, ${MIN}, ${MAX}"
fi
done < $1
只要确保您的数据文件以空行结束。
实施例：
bash <name_of_script_file> <name_of_data_file> | sort
Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116
见[这里]（http://stackoverflow.com/a/40780716/6769931）用于准 “自由AWK-” 的答案。;） –